Class 10 Maths - Triangles

In the given figure you find two triangles. Indicate whether the triangles are similar. Give reasons in support of your answer.

Let the bigger triangle be ABC and smaller be ADE with A being the common vertex and

$\angle ADE = 60^{\circ}$ ,

$\angle ABC = 60^{\circ}$
In

$\triangle ABC$ and

$\triangle ADE$ ,

$\angle A = \angle A$ (Common)

$\angle ABC = \angle ADE$ (Both

$60^{\circ}$ )

$\angle ACB = \angle AED$ (Third angle)
Thus

$\triangle ABC \sim \triangle ADE$ (AAA rule)

In fig $PQ = 5, XY = 7.5, QR = 4,YZ =PR = 6, XZ = 9.$ Are $\Delta PQR$ and $\Delta XYZ$ similar? If yes, state the test and the correspondence of similarity.

Given:

$PQ = 5, XY = 7.5, QR = 4,YZ = 6. PR = 6, XZ = 9$

Now,

$\dfrac{PQ}{XY} = \dfrac{5}{7.5} = \dfrac{2}{3}$

$\dfrac{QR}{YZ} = \dfrac{4}{6} = \dfrac{2}{3}$

$\dfrac{PR}{XZ} = \dfrac{6}{9} = \dfrac{2}{3}$

Since,

$\dfrac{PQ}{XY} = \dfrac{QR}{YZ} = \dfrac{PR}{XZ}$

Hence,

$\triangle PQR \sim \triangle XYZ$ (SSS rule)

Without drawing exact triangles, state, giving reasons, whether the given pairs of triangle are congruent or not :
In $\Delta ABC \,\,\, and \,\,\, \Delta PBC; AB=BP, AC=PC$

$\triangle$ ABC and PBC,

$AB = BP$ (Given)

$AC = PC$ (Given)

$BC = BC$ (Common)

Thus,

$\triangle ABC \cong \triangle PBC$ (SSS rule)

BP = 2AC
If the above statement is true then mention answer as 1, else mention 0 if false

$\triangle APB$ and

$\triangle ECB$ ,

$\angle ABP = \angle EBC$ (Common angle)

$\angle PAB = \angle CEB$ (Corresponding angle of parallel lines)

$\angle BPA = \angle ECB$ (third angle)
Thus,

$\triangle APB \sim \triangle ECB$

$\frac{AB}{EB} = \frac{BP}{BC}$

$\frac{2 EB}{EB} = \frac{PB}{BC}$ (E is the mid point of AB)

$BP = 2 BC$

$BP = 2 AD$ (Since, BC = AD)

In the figure above, straight lines $AB$ and $CD$ intersect at $P$ , and $AC \ || \ BD$ . Then $\Delta\,APC$ and $\Delta\,BPD$ are similar.If true enter 1 else 0.

$\triangle APC$ and

$\triangle BPD$

$\angle APC = \angle BPD$ (Vertically opposite angles)

$\angle PAC = \angle PBD$ (Alternate angles of parallel lines AC and BD)

$\angle PCA = \angle PDB$ (Third angle)
Thus,

$\triangle APC \sim \triangle BPD$ (AAA rule)

A vertical stick $12$ m long casts a shadow $8$ m long on the ground. At the same time a tower casts the shadow of length $40$ m on the ground. Determine the height of the tower.

length of vertical stick corresponds to length of tower and shadow of vertical stick corresponds to shadow of tower

Let height of tower = h

$\cfrac{Length Tower}{Length Stick} = \cfrac{Shadow Tower}{Shadow Stick}$

$\cfrac{h}{12} = \cfrac{40}{8}$

$h = \cfrac{40 \times 12}{8}$

$h = 60$ m

Height of tower = 60 m

$DE=4\,cm$ , $BC=8\,cm$ , $A\displaystyle(\triangle ADE)=25\,{ cm }^{ 2 }$ , find $A\displaystyle(\triangle ABC)$

$\triangle$ s, ABC and ADE,

$\angle BAC = \angle DAE$ (Common)

$\angle ADE = \angle ABC$ (Corresponding angles of parallel lines)

$\angle AED = \angle ACB$ (corresponding angles of parallel lines)

Therefore,

$\triangle ABC \sim \triangle ADE$

For similar triangles,

ratio of area of triangles = ratio of square of corresponding sides

Hence,

$\cfrac{Ar. \triangle ABC}{Ar. \triangle ADE} = \cfrac{BC^2}{DE^2}$

$\cfrac{Ar. \triangle ABC}{25} = \cfrac{8 \times 8}{4\times 4}$

$Ar. \triangle ABC = 25 \times 4$

$Ar. \triangle ABC = 100 cm^2$

f $A\,\triangle ABC=\,9\,{ cm }^{ 2 }$ , $A\,\triangle DEF=\,64\,{ cm }^{ 2 }$ , $DE=5.6\,cm$ , then find $AB$ .

For similar triangles,

ratio of area of triangles = ratio of square of corresponding sides

Hence,

$\cfrac{Ar. \triangle ABC}{Ar. \triangle DEF} = \cfrac{AB^2}{DE^2}$

$\cfrac{9}{64} = \cfrac{AB^2}{(5.6)^2}$

$AB^2 = \cfrac{9}{64} (5.6\times 5.6)$

$AB^2 = 4.41$

$AB = 2.1 cm$

In $\Delta ABC$ , the bisector of $\angle A$ intersects the base $BC$ at the point $D$ . Prove that $AB\times AC=BD\times DC+AD^2$ .

$Given:In\quad \triangle ABC,,the\quad bisector\quad of\angle A\quad intersects\quad the\quad base\quad BC\quad at\quad the\quad pointD\\ To\quad Prove:AB\times AC={ AD }^{ 2 }+BD\times BC\\ Proof:\\ Draw\quad a\quad circumcircle\quad of\quad \triangle ABC.Produce\quad AD\quad to\quad meet\quad the\quad circle\quad at\quad E.Join\quad EC.\\ In\triangle ABD\quad and\quad \triangle AEC,\\ \angle BAD=\angle EAC\quad [\because AD\quad is\quad the\quad bisector\quad of\angle BAC]\\ \angle ABD=\angle AEC[Angles\quad in\quad the\quad same\quad segment\quad are\quad equal]\\ \therefore \triangle ABD\sim \triangle AEC.(By\quad AA\quad similarity)\\ \therefore \dfrac { AD }{ AC } =\dfrac { AB }{ AE } \\ \Rightarrow AB\times AC=AD\times AE\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =AD\times (AD+DE)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad ={ AD }^{ 2 }+AD\times DE\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad ={ AD }^{ 2 }+BD\times DC[\because AD\times DE=BD\times DC]\\ \Rightarrow AB\times AC={ AD }^{ 2 }+BD\times DC\\ Hence\quad Proved$

In the given figure if, line $PY$ $\parallel$ side $BC$ , $AP=3$ , $PB=6$ , $AY=6$ and $YC=x$ , then find $x$ .

$\Delta ABC$ , line

$l$

$\parallel$ side

$BC$ ,
By B.P.T., we have

$\dfrac { AY }{ YC } =\dfrac { AP }{ PB }$

$\Rightarrow \dfrac { 6 }{ x } =\dfrac { 3 }{ 6 }$

$\Rightarrow x\times 3=6\times 6$

$\Rightarrow x=\dfrac { 36 }{ 3 }$

$\Rightarrow x=12$ units.

Write the conditions that are required for "Two polygons to be similar to each other".

There are two conditions that are required for two polygons to be similar to each other. They are:

1. Their corresponding angles must be equal.

2. The length of their corresponding sides must be proportional.

In $\triangle PQR \ 2PM = 3PN$ and $2PQ = 3PR$ . Prove that MQRN is a trapezium

Given :

$2PM=3PN$ and

$2PQ=3PR$

Proof:

(i)

$\dfrac{2 PM}{2 PQ} = \dfrac{3PN}{3 PR}$ .......... [Given]

$\implies \dfrac{PM}{PQ} = \dfrac{PN}{PR}$

$\therefore MN \parallel QR$ .......... [Using Converse of Basic Proportionality theorem]
(ii) Since

$QP$ and

$RP$ intersects at

$P$ ,

$QP$ cannot be parallel to

$RP$

$\implies$

$MQRN$ is a trapezium
[

$\because$ If only one pair of opposite sides of a quadrilateral is parallel then it is a trapezium]

In the following figure drawn, line $PY$ $\parallel$ side $BC$ , $AP=5, PB=10, AY=8, YC=x$ , then find $x$ .

Since

$PQ\parallel BC$ ,

Thus using Basic Proportionality Theorem

$\dfrac{AY}{YC}=\dfrac{AP}{PB}$

$\Rightarrow \dfrac{8}{x}=\dfrac{5}{10}$

$\Rightarrow 5x=80$ , cross multiply

$\therefore x=16$

In $\triangle ABC, \angle ABC = \angle DAC, AB = 8\ cm, AC = 4\ cm, AD = 5\ cm$ .
Find $\dfrac{\text{area of}\, \triangle ABC}{\text{area of}\, \triangle ACD}$

$\triangle ACD$ and

$\triangle BCA$

$\angle C = \angle C$ (common)

$\angle ABC = \angle CAD$ (given)

$\therefore \triangle ACD \sim \triangle BCA$ (

$A-A$ postulates)

$\therefore \dfrac {\text{area} (\triangle ACD)}{\text{area} (\triangle ABC)} = \dfrac {AC^{2}}{AB^{2}}$

$= \dfrac {(4)^{2}}{(8)^{2}} = \dfrac {16}{64} = \dfrac {1}{4}$

$\text{area of}\, \triangle ABC : \text{area of}\, \triangle ACD = 4 : 1$

What is similar triangles?

Two triangles are similar if and only if corresponding angles have the same measure: this implies they are similar if and only if the length of corresponding sides are proportional.

Note: Triangles are similar if they have the same shape, but not necessarily the same size.

Example of similar triangles : From above,

$\triangle PQR \sim \triangle{P}{'}{Q}{'}{R}{'}$

Prove that "In a trapezium, the line joining the mid points of non-parallel sides is (i) parallel to the parallel sides and (ii) Half of the sum of the parallel sides"

Data: In the trapezium ABCD,
AD || BC, AX = XB and DY = YC
To Prove: (i) XY || AD or XY || BC
(ii) XY =

$\dfrac{1}{2}$ (AD + BC)
Construction: Extend BA and CD to meet at Z.
Join A and C. Let it cut XY at P
Proof: (i) In

$\triangle ZBC, AD || BC$ [

$\because$ Data]

$\therefore \dfrac{ZA}{AB} = \dfrac{ZD}{DC}$ [

$\because$ BPT]

$\therefore \dfrac{ZA}{2AX} = \dfrac{ZD}{2DY}$ [

$\because$ X & Y are mid points of AB and DC]

$\therefore \dfrac{ZA}{AX} = \dfrac{ZD}{DY}$

$\Rightarrow XY || AD$ [

$\because$ Converse of B.P.T.]
(ii) In

$\triangle ABC, AX = XB$ [

$\because$ Data]

$XP || BC$ [

$\because$ Proved]

$\therefore AP = PC$ [

$\because$ Converse of mid point theorem]

$\therefore XP = \dfrac{1}{2} BC$ [

$\because$ Midpoint Theorem]
In

$\triangle ADC, PY = \dfrac{1}{2} AD$
By adding, we get

$XP + PY = \dfrac{1}{2} BC + \dfrac{1}{2} AD$

$\therefore XY = \dfrac{1}{2} (BC + AD)$

In the figure, line $PQ$ is parallel to side $BC$ . Find the value of ' $x$ '. If $AP=3$ , $PB=6$ , $AQ=5$ , $QC=x$ .

Since

$PQ\parallel BC$ ,

Thus using Basic Proportionality Theorem

$\dfrac{AQ}{QC}=\dfrac{AP}{PB}$

$\Rightarrow \dfrac{5}{x}=\dfrac{3}{6}$

$\Rightarrow 3x=30$ , cross multiply

$\therefore x=10$

In $\triangle ABC$ , line $PQ \parallel$ side $BC$ . Find the value of $x$ from the information given in the following figure.

Using Basic Proportionality Theorem,

$\dfrac{AP}{PB}=\dfrac{AQ}{QC}$

$\Rightarrow \dfrac{x}{4}=\dfrac{15}{5}=3$

$\Rightarrow x=4\times 3=12$

In the figure, it is given that $BR = PC$ and $\angle ACB = \angle QPR$ and $AB \parallel PQ$ . Prove that $AC= QR$ .

Given

$BR=PC$ and

$\angle ACB = \angle QRP$ and

$AB || PQ$

Join the points

$P$ and

$B$

$\Rightarrow BC=PR$

Then

$\angle CBA=\angle RPQ$

$\angle ACB=\angle QRP$ (given )

$\therefore$ In

$\bigtriangleup ABC$ and

$\bigtriangleup PQR$

$\angle BAC=\angle QPR$

$\therefore \bigtriangleup ABC\cong PQR$ ...... [By AAA condition of similarity]

$\Rightarrow AC=QR$

State the converse of Pythagorean Theorem.

Converse of Pythagorean Theorem:

Statement: In a triangle, if the square of one side of triangle is equal to sum of the squares of the other two sides, then the triangle is a right triangle.

Prove that
"If two triangles are equiangular then their corresponding sides are in proportion"

Data : In

$\triangle ABC$ and

$\triangle DEF$

$\angle BAC = \angle EDF$

$\angle ABC = \angle DEF$
To prove :

$\dfrac {AB}{DE} = \dfrac {BC}{EF} = \dfrac {CA}{FD}$
Construction : Mark points

$G$ and

$H$ on

$AB$ and

$AC$ such that

$AG = DE$ and

$AH = DF$ . Join

$G$ and

$H$ .
Proof : In

$\triangle AGH$ and

$\triangle DEF$

$AG = DE\because Construction$

$\angle GAH = \angle EDF \because Data$

$AH = DF \therefore Construction$

$\therefore \triangle AGH \cong \triangle DEF \because SAS$

$\angle AGH = \angle DEF$
But,

$\angle ABC = \angle DEF$

$\Rightarrow \angle AGH = \angle ABC$

$\therefore GH\parallel BC$ .
In

$\triangle ABC = \dfrac {AB}{AG} = \dfrac {BC}{GH} = \dfrac {CA}{HA}$
Hence

$\dfrac {AB}{DE} = \dfrac {BC}{EF} = \dfrac {CA}{FD}$ .

In the following figure, $DE \parallel AB$ . If $AD = 7\ cm, CD = 5\ cm$ and $BC = 18\ cm$ , find $CE$ .

$\triangle ABC, DE\parallel AB$

$\dfrac {CD}{CA} = \dfrac {CE}{CB}$ (Corollary to B.P.T.)

$\dfrac {5}{12} = \dfrac {CE}{18}$

$12\times CE = 5\times 18$

$CE = \dfrac {5\times 18}{12} = \dfrac {15}{2}$

$CE = 7.5\ cm$

$\triangle {ABC}\sim \triangle{DEF}$ . Their areas are $64\ {cm}^{2}$ and $121\ {cm}^{2}$ .
Then ratio of corresponding sides $=$

Given that:

$\triangle ABC\sim \triangle DEF$

$A(\triangle ABC)=64\ cm^2, A(\triangle DEF)=121\ cm^2$

Solution:
We know that,

$\cfrac{A(\triangle ABC)}{A(\triangle DEF)}=\cfrac{AB^2}{DE^2}$

or,

$\cfrac{AB^2}{DE^2}=\dfrac{64\ cm^2}{121\ cm^2}$

or,

$\cfrac{AB}{DE}=\sqrt{\dfrac{64\ cm^2}{121\ cm^2}}$

or,

$\cfrac{AB}{DE}=\dfrac8{11}$

Therefore, ratio of corresponding sides of the two triangles is

$8:11.$

In $\triangle ABC, DE\parallel BC$ and $\dfrac {AD}{DB} = \dfrac {2}{3}$ . If $AE = 3.7\ cm$ , find $EC$

$\triangle ABC, DE\parallel BC$

$\therefore \dfrac {AD}{DB} = \dfrac {AE}{EC}$

$\Rightarrow EC = \dfrac {AE\times DB}{AD}$
Thus,

$EC = \dfrac {3.7\times 3}{2} = 5.55\ cm$
1070038_622300_ans_e755aae0921043b4be4d518cc2cb4977.jpg

State Thale's theorem.

Thales theorem states that:

If a line is drawn parallel to one side of a triangle, then it divides the other two sides in the same ratio.

This theorem is also called as basic proportionality theorem.

If the ratio of corresponding medians of two similar triangles are $9:16$ , then find the ratio of their areas.

Ratio of medians

$=9:16=\dfrac{9}{16}$

Ratio of their areas

$={ \left( \dfrac { 9 }{ 16 } \right) }^{ 2 }=\dfrac { 81 }{ 256 }$

Fill in the blanks using the correct word given in brackets:
All .......... triangles are similar (isosceles, equilateral).

All equilateral triangles are similar.

The geometric shapes are said to be similar if its shape is same and there size is different.

The geometric shapes are said to be congruent if its shape and size are same .

All equilateral triangles shape are same but sizes are different.

Isosceles triangle can be a right angled isosceles triangle or a normal Isosceles triangle which differ by shape.

$\therefore$ all equilateral triangles are similar .

The corresponding sides of two similar triangles are in the ratio 1 :If the area of the smaller triangle in 40 $\text{ cm }^{ 2 }$ , find the area of the larger triangle.

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\therefore \left(\dfrac{\text{Side of the smaller triangle}}{\text{Side of the larger triangle}}\right)^2=\left(\dfrac{\text {Area of the smaller triangle}}{\text{Area of the larger triangle}}\right)$

Let, the area of the larger triangle be

$x$ , then by substituting the given values in above expression, we get:

$\left(\dfrac{1}{3}\right)^{2}=\dfrac{40}{x}$

$\dfrac{1}{9}=\dfrac{40}{x}$

$x=40\times 9$

$x=360\ \text{cm}^{2}$

Area of the larger triangle is

$360\ \text{cm}^{2}$ .

Prove that, in a right-angled triangle, the square of hypotenuse is equal to the sum of the square of remaining two sides.

Consider a right angled triangle with length of hypotenuse as

$c$ and the remaining two sides as

$a$ and

$b$ respectively.

We arrange 4 such triangles in such a fashion that they form a square of side

$a+b$ (as shown in the figure).

Area of the figure

$A=$

$(a+b)(a+b)$

$.......(1)$

Also area of the figure can be written as,

$A = 4 \times ar(Blue\, right\, triangle) + ar(central\, yellow\, square)$

$A = 4 \times \frac{1}{2}ab + c^2$

$.......(2)$

Equating (1) and (2) we get,

$c^2 + 2ab = a^2 + b^2 + 2ab$

Consequently,

$c^2 = a^2 + b^2$ . Hence, Pythagoras theorem is proved.

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively find the ratio of their areas.

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding altitudes.

Hence the ratio of the areas

$=\dfrac{(6)^{2}}{(9)^{2}}=\dfrac{36}{81}=\dfrac{4}{9}$

Hence the ratio of areas

$=\dfrac{4}{9}$

What values of x will make DE $\parallel$ AB in the given figure?

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Hence

$\dfrac{AD}{DC}=\dfrac{BE}{EC}$

$\dfrac{3x+19}{x+3}=\dfrac{3x+4}{x}$

$3x^{2}+19x=3x^{2}+4x+9x+12$

$19x=13x+12$

$6x=12$

$x=\dfrac{12}{6}$

Hence

$x=2$

The area of two similar triangles $ABC$ and $PQR$ are in the ratio $9 : 16$ . If $BC = 4.5 cm$ , find the length of $QR$ .

Given

$\triangle ABC \sim \triangle PQR$

$\therefore \dfrac {area(\triangle ABC)} {area(\triangle PQR)} =\dfrac {BC^{2}}{QR^{2}}$

$\implies \dfrac {9}{16}=\dfrac {4.5^{2}}{QR^{2}}$

$QR=\dfrac {4\times 4.5}{3}$

$\therefore QR =6$ cm

In $\Delta$ PQR, MN is parallel to QR and $\displaystyle\frac{PM}{MQ}=\frac{2}{3}$ . Find, Area of $\Delta$ OMN:Area of $\Delta$ ORQ.

$\dfrac{PM}{PQ} = \dfrac{PM}{PM+MQ} = \dfrac{2}{2+3} = \dfrac{2}{5}$

Since

$MN||QR$ , we have,

$\triangle PMN$ is similar to

$\triangle PQR$

$\Rightarrow \dfrac{PM}{PQ} = \dfrac{MN}{QR} = \dfrac{2}{5}$

$MN||QR$

Alternate interior angles are equal, so

$\Rightarrow \angle OMN = \angle ORQ$

$\Rightarrow \angle ONM = \angle OQR$

Vertically opposite angles are equal, so

$\angle MON = \angle QOR$

Thus, we have that

$\triangle OMN \sim \triangle ORQ$

So,

$\dfrac{Area \ of \ \triangle OMN}{Area \ of \ \triangle ORQ} = \dfrac{MN^2}{QR^2} = \dfrac{4}{25}$

Given $\Delta$ ABC $\sim$ $\Delta$ PQR, if $\displaystyle\frac{AB}{PQ}=\frac{1}{3}$ , then find $\displaystyle\frac{ar \Delta ABC}{ar \Delta PQR}$ .

Let

$AD$ be the perpendicular dropped from

$A$ to the side

$BC$ and

$PS$ be the perpendicular dropped from

$P$ to the side

$QR$ respectively.

Since,

$\Delta ABC \sim \Delta PQR$ , we have

$\Delta ABD \sim \Delta PQS$ .

Therefore,

$\dfrac{AB}{PQ} = \dfrac{AD}{PS} = \dfrac{BC}{QR}$

$\Rightarrow \dfrac{ar \Delta ABC}{ar \Delta PQR} = \dfrac{AD\times BC}{PS \times QR} = \dfrac{AB}{PQ}\times \dfrac{AB}{PQ} = \dfrac{AB^2}{PQ^2} = \dfrac{1}{9}$

$ABC$ is a triangle in which $\angle A= 90^{o}$ , $AN \bot BC$ , $BC=12\ cm$ and $AC=5\ cm$ . Find the ratio of the areas of $\triangle ANC$ and $\triangle ABC$ .

$\triangle ANC$ and

$\triangle BAC$

$\angle ANC=\angle BAC=90^{\circ}$

$\angle ACN=\angle BCA$

So by AA similarity,

$\triangle ANC \sim \triangle BAC$

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\dfrac{area(\triangle ANC)}{area(\triangle BAC)}=\dfrac{AC^{2}}{BC^{2}}$

$\dfrac{area(\triangle ANC)}{area(\triangle BAC)}=\dfrac{5^{2}}{12^{2}}$

$\dfrac{area(\triangle ANC)}{area(\triangle BAC)}=\dfrac{25}{144}$

Hence the ratio of areas of

$\triangle ANC$ and

$\triangle BAC$ is

$25:144$

In the given Figure, $\triangle ABC \sim\triangle$ $PQR$ and quad $ABCD$ $\sim$ quad $PQRS$ . Determine the values of $x, y, z$ in each case.

$\triangle ABC\sim\triangle PQR$

$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AC}{PR}$

$\therefore$

$\dfrac{12}{9}=\dfrac{7}{x}=\dfrac{10}{y}$

$\Rightarrow$

$\dfrac{12}{9}=\dfrac{7}{x}$

$\therefore$

$x=5.25$

$\Rightarrow$

$\dfrac{12}{9}=\dfrac{10}{y}$

$\therefore$

$y=7.5$

Now,

$Quad\,ABCD\sim Quad\,PQRS$

Then,

$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{CD}{RS}=\dfrac{AD}{PS}$

$\dfrac{16}{x}=\dfrac{50}{y}=\dfrac{\dfrac{50}{3}}{z}=\dfrac{20}{7}$

$\Rightarrow$

$\dfrac{16}{x}=\dfrac{20}{7}$

$\therefore$

$x=5.6$

$\Rightarrow$

$\dfrac{50}{y}=\dfrac{20}{7}$

$\therefore$

$y=17.5$

$\Rightarrow$

$\dfrac{50}{3z}=\dfrac{20}{7}$

$\Rightarrow$

$3z=17.5$

$\Rightarrow$

$z=5.83$

Give two different examples of pair of
(ii) non-similar figures.

(1) Circle and rectangle

(2) Triangle and Hexagon

1161814_1084058_ans_021e92523a9f45a19e42b8b300d1ea9f.png

Give two different examples of pair of (i) similar figures

(1) Any two rectangles.

(2) Any two squares.

Hence, this is the answer.

1174507_1084056_ans_0241e62db10649968dc69d2b080333ea.jpg

In the figure, $DE \parallel BC$
(i) Prove that $\triangle ADE$ and $\triangle ABC$ are similar

$\Delta ADE$ and

$\Delta ABC$ we have

$1) \angle ADE=\angle ABC\quad [DE||BC\quad and AB$ is transverse]

$2) \angle A=\angle A$

$\therefore \Delta ADE$ and

$ABC$ are similar

In the figure $PS=3,SQ=6.QR=5,PT=x$ and $TR=y$ .Give any two pairs of value of $x$ and $y$ such that line $ST||QR$

$ST || SR$ , then

$\Delta PST \sim \Delta PQR$ .

$\implies \dfrac{PT}{PR}=\dfrac{PS}{PQ}$

$\implies \dfrac{x}{x+y}=\dfrac{3}{9}=\dfrac{1}{3}$

$\implies 3x=x+y$

$\implies 2x=y$

Two pairs satisfying the above are:

$x=3, y=6$

$x=4, y=8$

Area of similar triangles are in the ratio $25:36$ then ratio of their similar sides is _________?

The areas and sides of similar triangles are related as

$\dfrac{Ar(\Delta ABC)}{Ar(\Delta PQR)}=\left(\dfrac {AB}{PQ}\right)^2\\\dfrac {25}{36}=\left(\dfrac {AB}{PQ}\right)^2\\\dfrac{AB}{PQ}=\sqrt {\dfrac {25}{36}}=\dfrac 56$

Write the properties of similar triangles.

Two triangles are said to be similar if -

(i) corresponding angles are equal

(ii) corresponding sides are proportional

In figure if $AD=6\ cm, DB=9\ cm, AE=8\ cm$ and $EC=12\ cm$ and $\angle ADE=48^{o}$ . Find $\angle ABC$ .

$\triangle ADE, \ and \ \triangle ABC$

$AD=6,AB=AD+DB=6+9=15\implies \dfrac{AD}{AB}=\dfrac{6}{15}=\dfrac{2}{5}$

$AE=8,AC=AE+EC=8+12=20\implies \dfrac{AE}{AC}=\dfrac{8}{20}=\dfrac{2}{5}$

$\Rightarrow \dfrac{AD}{AB}=\dfrac{AE}{AC}, \angle DAE=\angle BAC$ (common angle)

As sides are proportional and an angle is equal

$\Rightarrow \triangle ABC \sim \triangle ADE$

$\therefore \angle ABC=\angle ADE=48^{\circ}$

Given that $\Delta ABC \sim \Delta DEF,BC = 3\,cm,EF = 4\,cm$ ans the area of $\Delta ABC$ is $54\,sq.cm.$ Then find the area of $\Delta DEF.$

Solution -

REF. Image

$\dfrac{Area\, \triangle ABC}{Area\, \triangle DEF}= \left ( \dfrac{BC}{EF} \right )^{2}$

Area

$\triangle DEF = 54\times \dfrac{4^{2}}{3^{2}}$

Area

$(\triangle DEF)= 6\times 4^{2}$

$= 96\, cm^{2}$

1130800_1101662_ans_5a58b7570dfd4d7d968060a197781cf7.jpg

In figure $\angle 1=\angle 2$ and $\Delta NSQ\cong \Delta MTR$ , then prove that $\Delta PTS\sim \Delta PRQ$ .

In figure triangle NSQ

$\cong$ triangle MTR

Therefore,

$\angle SQN=\angle TRM$

i.e., PQ=PR (sides opposite to equal angles are equal ) ...............1

And already it is given that

$\angle 1=\angle 2$

I.e.,PS=PT (sides opposite to equal angles are equal ) .............2

Subtracting 2 from 1 we get QS=TR

Therefore ST

$\parallel$ QR

Now as ST

$\parallel$ QR Therefore

$\angle 1=\angle SQR$ and

$\angle 2=\angle TRQ$ (corresponding angles)

And therefore

$\angle SPT=\angle QPR$

Then by AAA similarity criteria Triangle PTS

$\sim$ triangle PRQ.

$DA,CB,OM$ are each perpendicular to line segment $AB$ where $O$ is the point of intersection of $AC$ and $DB$ .If $OA=2.4$ cm, $OC=3.6$ cm then find the ratio of $AM:BM$

Given:

$AD\parallel\,OM\parallel\,BC$

By intercept theorem,

$\dfrac{OA}{OC}=\dfrac{AM}{MB}$

$\Rightarrow\,\dfrac{2.4}{3.6}=\dfrac{AM}{MB}$

$\Rightarrow\,\dfrac{AM}{MB}=\dfrac{2}{3}$

$\therefore\,AM:MB=2:3$

In $\Delta ABC, \bar{PQ} || \bar{BC}$ and $AP = 3x - 19, PB = x - 5, AQ = x - 3, QC = 3 \,cm$ . Find $x$ .

Given:-

$ABC$ is a

$\triangle$ in which

$PQ \parallel BC$ and

$AP = 3x - 19, PB = x - 5, AQ = x - 3, QC = 3 \; cm$

To find:-

$x = ?$

Solution:-

From basic proportionality theorem,

$\cfrac{AP}{PB} = \cfrac{AQ}{QC}$

$\cfrac{3x - 19}{x - 5} = \cfrac{x-3}{3}$

$3 \left( 3x - 19 \right) = \left( x - 3 \right) \left( x - 5 \right)$

$9x - 57 = {x}^{2} - 3x - 5x + 15$

${x}^{2} - 8x + 15 - 9x + 57 = 0$

${x}^{2} - 17x + 72 = 0$

${x}^{2} - 9x - 8x + 72 = 0$

$\left( x - 9 \right) \left( x - 8 \right) = 0$

$x = 8$ or

$x = 9$

Hence the valuen of

$x$ is

$8$ or

$9$ .

Ler $\Delta A B C \sim \Delta D E F$ and their areas be respectively, 64 $\mathrm { cm } ^ { 2 }$ and 121 $\mathrm { cm } ^ { 2 } .$ If EF $=15.4$ find BC

$\Delta ABC\sim \Delta DEF$

$\dfrac{area of \Delta ABC}{area of \Delta DEF}=\dfrac{64}{121}$

$\dfrac{BC^2}{EF^2}=\dfrac{64}{121}=\left(\dfrac{8}{11}\right)^2$

$\dfrac{BC}{EF}=\dfrac{8}{11}$

$\dfrac{BC}{15.4}=\dfrac{8}{11}$

$BC=\dfrac{8\times 15.4}{11}$

$=8\times 1.4$

$=11.2$

$\therefore BC=11.2cm$ .

In figure, $AC\parallel BD$ and $CE\parallel DF$ Find CD,where OD=7,OA=10.5,AB=4.5

Solution:Since

$AC\parallel BD\Rightarrow \dfrac{OA}{AB}=\dfrac{OC}{CD}$

Since

$CE\parallel DF\Rightarrow \dfrac{OC}{CD}=\dfrac{OE}{EF}$

Let

$CD=x,\,OD=7$ cm then

$OC=7-x$

$\Rightarrow \dfrac{OA}{AB}=\dfrac{OC}{CD}$

$\Rightarrow \dfrac{10.5}{4.5}=\dfrac{7-x}{x}$

$\Rightarrow 10.5x=31.5-4.5x$

$\Rightarrow 15x=31.5$

$\therefore x=\dfrac{31.5}{15}=2.1$ cm

Prove that the ratio of the area of two similar triangle is equal to the square of the ratio of their corresponding sides.

Given:- Let

$\triangle{ABC} \sim \triangle{PQR}$

$AD$ and

$PS$ are corresponding medians.

To prove:-

$\cfrac{ar{\left( \triangle{ABC} \right)}}{ar{\left( \triangle{PQR} \right)}} = {\left( \cfrac{AD}{PS} \right)}^{2}$

Proof:- In

$\triangle{ABC}$

$\because AD$ is median.

$\therefore BD = CD = \cfrac{1}{2} BC$

Similarly, in

$\triangle{PQR}$

$PS$ is median.

$\therefore QS = RS = \cfrac{1}{2} QR$

Now,

$\triangle{ABC} \sim \triangle{PQR} \left( \text{Given} \right)$

$\angle{B} = \angle{Q} ..... \left( 1 \right) \quad \left( \text{Corresponding angles of similar triangles are equal} \right)$

$\cfrac{AB}{PQ} = \cfrac{BC}{QR} \quad \left( \text{Corresponding sides of similar triangles are proportional} \right)$

$\Rightarrow \cfrac{AB}{PQ} = \cfrac{2 BD}{2 QS} \quad \left( \because \text{AD and PS are medians} \right)$

$\cfrac{AB}{PQ} = \cfrac{BD}{QS} ..... \left( 2 \right)$

Now, in

$\triangle{ABD}$ and

$\triangle{PQS}$ ,

$\angle{B} = \angle{Q} \quad \left( \text{From } \left( 1 \right) \right)$

$\cfrac{AB}{PQ} = \cfrac{BD}{QS} \quad \left( \text{From } \left( 2 \right) \right)$

$\therefore \triangle{ABD} \sim \triangle{PQS} \quad \left( \text{By SAS Property} \right)$

Therefore,

$\cfrac{AB}{PQ} = \cfrac{AD}{PS} ..... \left( 3 \right) \quad \left( \because \text{Corresponding sides of similar triangles are proportonal} \right)$

Now,

$\because \triangle{ABC} \sim \triangle{PQR}$

As we know that ratio of area of similar triangles is always equal to the square of ratio of their corresponding side.

Therefore,

$\cfrac{ar{\left( \triangle{ABC} \right)}}{ar{\left( \triangle{PQR} \right)}} = {\left( \cfrac{AB}{PQ} \right)}^{2}$

$\cfrac{ar{\left( \triangle{ABC} \right)}}{ar{\left( \triangle{PQR} \right)}} = {\left( \cfrac{AD}{PS} \right)}^{2} \quad \left( \text{From } \left( 3 \right) \right)$

Hence proved.

1199250_1508446_ans_4357ef4186f34358bcd7e49fadc37431.jpeg

If D and E are points on sides AB and AC respectively of ABC and AB- 12 cm, AD= 8 cm,AE = 12 cm, AC =18 cm, then prove that $DE \parallel BC$ .

Given: AB = 12cm, AD =8cm, AB = 12cm and AC = 18cm
To Prove:

$DE \parallel BC$
in ∆ABC,

$\frac {AD}{DB} = \frac {AD}{AB-AD} = \frac {8}{12-8} =2$
And

${AE}{AC} = \frac {AE}{AC-EC} = \frac {12}{18-12}= 2$
Then,

$\frac{AD}{DB} = \frac {AE}{EC}$
Basic Proportionality theorem which states that
if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

Hence,

$DE \parallel BC$
[by converse of basic
proportionality theorem]

Hence, Proved.

State whether the following pair of triangle is similar or not.

$\Delta MNL$ and

$\Delta PQR$

$\angle NML = \angle PQR = 70^o$

$\dfrac{MN}{PQ} = \dfrac{2.5}{6} = \dfrac{5}{6×2} = \dfrac{5}{12}$

$\dfrac{ML}{QR} = \dfrac{5}{10} = \dfrac {1}{2}$

$\implies \dfrac {MN}{PQ} \neq \dfrac {ML}{QR}$

No, the two triangles are not similar.

In the adjoining figure, $DE \parallel BC$ ,find x.

Given:

$AD=4, DB=x-4, AE=x-3$ and

$EC=3x-19$ and

$DE \parallel BC$
Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two-sided are divided in the same ratio.So, by basic proportionality theorem.

$therefore \frac{AD}{DB} = \frac {AE}{EC}$

$\implies \frac{4}{x-1} = \frac{x-3}{3x-19}$

$\implies 12x-76 = x^2 - 3x - 4x +12$

$\implies x^2 - 7x +12 - 12x + 76 = 0$

$\implies x^2 - 19x + 88 = 0$
Solving the Quadratic equation by splitting the
middle term, we get,

$\implies x^2 - 11x - 8x + 88 = 0$

$\implies (x-8)(x-11) = 0$

$\implies x = 8 , 11$

State whether the following pair of triangle is similar or not. Write the similarity criterion used and write it in symbolic form.

$\Delta ABC$ and

$\Delta PQR$

Hence,

$\dfrac{AB}{PQ} = \dfrac{2}{4} = \dfrac{1}{2}$

$\dfrac{BC}{QR} = \dfrac{1}{2}, \dfrac{AC}{PR} = \dfrac{1}{2}$

As,

$\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR}$

So,

$\Delta ABC \sim \Delta PQR$ [by SSS similarity criterion]

In the adjoining figure, $P$ and $Q$ are points on sides $AB$ and $AC$ respectively of mix such that $PQ \parallel BC$ and $$AP = 8\ cm, AB =12\ cm,AQ = 3x\ cm, QC = (x+2)\ cm.$$
Find x.

Given:

$AP = 8cm ,AB =12cm, AQ = (3x)cm$ and

$QC= (x+2)cm$ and

$PQ \parallel BC$
Basic Proportionality theorem which states that if aline is drawn parallel to one side of a triangle theother two sides in distinct points, then the other two sides are divided in the same ratio.

$\therefore d\frac {AP}{PB} = \dfrac {AQ}{QC}$
[by basic proportionality theorem]

$\implies \dfrac {AP}{AB-AP} = \dfrac{AQ}{QC}$

$\implies \dfrac {8}{4} = \dfrac{3x}{x+2}$

$\implies 2 = \dfrac {3x}{x+2}$

$\implies 2x +4 = 3x$

$\implies x = 4$

P and Q are points on sides AB and AC respectivelyof $\triangle$ ABC. For each of the following cases, statewhether PQ $\parallel$ BC.
AB = 5 cm, AC =10 cm, AP: 4 cm, AQ = 8 cm.

Given: AB = 5 cm, AC =10 cm, AP= 4 cm,AQ = 8 cm

To find: PQ

$\parallel$ BC

$\triangle$ ABC,

$\frac {AP} {PB} = \frac {4}{AB - AP} = \frac {4}{5-4} = \frac 41$

and

$\frac {AQ} {QC} = \frac {8}{AC - AQ} = \frac {8}{10 - 8} = \frac 41$

Thus,

$\frac {AP} {PB} = \frac {AQ} {QC}$

Basic proportionality theorem which states that if aline is drawn parallel to one side of a triangle theother two sides in distinct points, then the other twosides are divided in the same ratio.

Hence, PQ

$\parallel$ BC [by converse of basic proportionality theorem]

Hence, Proved.

State whether the following pair of triangle is similar or not. Write the similarity criterion used and write it in symbolic form.

In $\Delta PQR$ and $\Delta LMN$

$\angle PQR = \angle LNM = 50°$

$\dfrac {PQ}{LN} = \dfrac {4}{8} = \dfrac{1}{2}$

$\dfrac {QR}{MN} = \dfrac {10}{20} = \dfrac {1}{2}$

$\implies \dfrac {PQ}{LN} = \dfrac{QR}{MN}$

$\therefore \Delta PQR \sim \Delta LMN$ [by SAS similarity criterion]

If DE has been drawn parallel to side BC of ABC cutting AB and AC at points D and E respectively , such that $\frac{AD}{DB} = \frac {3}{4}$ then and the value of $\frac {AE}{EC}$

Given: DE || BC

and

$\frac {AD}{DB} = \frac{3}{4}$

To find

$\frac{AE}{EC}$

Given:

$DE \parallel BC$

Basic Proportionality theorem which states that if a

line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two
sides are divided in the same ratio.

$\frac {AD}{DB} = \frac {AE}{EC}$
[by basic proportionality theorem]

$\implies \frac {AD}{DB} = \frac {AE}{EC} = \frac {3}{4}$

P and Q are points on sides AB and AC respectivelyof $\triangle$ ABC. For each of the following cases, statewhether PQ $\parallel$ BC.
AP: 8 cm, PB = 3 cm, AC = 22 cm and AQ =16 cm.

HintGiven: AP: 8 cm, PB = 3 cm, AC = 22 cm
and AQ =16 cm
To lind: PQ

$\parallel$ BC

$\triangle$ ABC,

$\frac {AP} {PB} = \frac 83$ and

$\frac {AQ} {QC} = \frac {16}{AC - AQ} = \frac {16}{22-16} = \frac {16}{6} = \frac 83$

$\frac {AP} {PB} = \frac {AQ} {QC}$

Basic Proportionality theorem which states that if a
line is drawn parallel to one side of a triangle the
other two sides in distinct points, then the other two
sides are divided in the same ratio.
Hence, PQ

$\parallel$ BC [by converse of basic
proportionality theorem]
Hence, Proved.

State whether the following pair of triangle is similar or not. Write the similarity criterion used and write it in symbolic form.

$\Delta ABC$ ,

$\angle A = 70^o$ and

$\angle B = 50^o$
And we know that, sum of the angles = 180°

$\implies \angle A + \angle B + \angle C = 180^o$

$\implies 70^o + 50^o + \angle C = 180^o$

$\implies 120^o + \angle C = 180^o$

$\angle C = 60^o$
And in

$\Delta DEF$

$\angle F = 70^o$ and

$\angle E = 50^o$
And we know that, sum of the angles = 180°

$\angle D + \angle F + \angle E = 180^o$

$\implies \angle D = 60^o$
Yes,

$\Delta ABC \sim \Delta DEF$ [by AAA similarity criterion]

Give two different examples of pair of similar figures.

$Coin \space and \space Wheel \space of \space cart$

$Both \space are \space similar \space to \space circle$

In the following figure , $ABCD$ is a trapezium with AB || DC. If $AB = 9$ $cm$ , $DC = 18$ $cm$ , $CF = 13.5$ $cm$ , $AP = 6$ $cm$ and $BE = 15$ $cm$ . Find the value of $AF$

In $\Delta$ APB and $\Delta$ FPD,

$\angle{APB} = \angle{FPD}$ [Vertically opposite angles]

$\angle{BAP} = \angle{DFP}$ [Since, AB||DF]

Hence, $\Delta$ APB ~ $\Delta$ FPD by AA criterion for similarity.

So, we have

$AP/FP = AB/FD$

$6/FP = 9/31.5$

$FP = 21\,cm$

So, $AF = AP + PF = 6 + 21 = 27\,cm$

In triangle $ABC, D$ and $E$ are points on side $AB$ such that $AD=DE=EB$ . Through $D$ and $E$ lines are drawn parallel to $BC$ which meet side $AC$ at points $F$ and $G$ respectively.
Through $F$ and $G$ , lines are drawn parallel to $AB$ which meet side $BC$ at points $M$ and $N$ respectively. Prove that : $BM=MN=NC$ .

The figure is shown below
For triangle

$AEG$

$D$ is the midpoint of

$AE$ and

$DF\parallel EG\parallel BC$
Therefore

$F$ is the midpoint of

$AG$ .

$AF=GF$ ... (1)
Again

$DF\parallel EG\parallel BC\ \ \ DE=BE$ , therefore

$GF=GC$ .....(2)

$(1), (2)$ we get

$AF=GF=GC$ .
Similarly

$GN\parallel FM\parallel AB$ and

$AF=GF$ ,
therefore

$BM=MN=NC$
Hence proved

In the given figure, $XY \parallel BC$ . Find the length of XY, given BC = 6 cm

Given: XY || BC

To find: XY

Since, XY || BC, AB is transversal, then,

$\Delta AXY = \Delta ABC$ [by corresponding angles] Since, XY || BC, AC is transversal, then,

$\Delta AYX = \Delta ABC$ [by corresponding angles] In $\Delta AXY$ and $\Delta$ ABC

$\angle AXY = \angle ABC$

$\angle AYX = \angle ACB$

$\Delta$ AXY a $\Delta$ ABC [by AA similarity] Since, triangles are similar, hence corresponding sides will be proportional.

$\therefore \frac {AX}{AB} = \frac {XY}{BC} = \frac {AY}{AC}$

$XY = \frac{6}{4}$

XY = 1.5cm

In the figure given below, AB EF CD. If $AB = 22.5\,cm$ , $EP = 7.5\,cm$ , $PC = 15\,cm$ and $DC = 27\,cm$ . Calculate EF.

In $\Delta$ PCD and $\Delta$ PEF,

$\angle{CPD} = \angle{EPF}$ [Vertically opposite angles]

$\angle{DCE} = \angle{FEP}$ [As DC || EF, alternate angles.]

Hence, $\Delta$ PCD ~ $\Delta$ PEF by AA criterion for similarity.

So, we have

$27/EF = 15/7.5$

Thus,

$EF = 13.5$

In a $\Delta ABC , BD$ is the median to the side AC , BD is produce to E such that BD = DE
Prove that AE is parallel to BC.

Given : A

$\Delta ABC$ in which BD is the median to AC
BD is produce to E such that BD = DE
We need to prove that AE || BC
Construction: join AE

Proof:

$\Delta BDC$ and

$\Delta ADE$

$AD = DC$ [

$BD$ is median]

$BD = DE$ [Given]

$\angle BDC = \angle ADE = 90^{o}$ [Vertically opposite angles]

So,

$\Delta BDC \cong \Delta ADE$

The corresponding parts of the congruent triangles are congruent

$\angle EAD = \angle BCD$ [cpct]
But these are alternate angles and

$AC$ is the transversal
Thus,

$AE || BC$
1785686_1841195_ans_107ab5e274ea4d7f8473ab6595502034.png

The following figure shows a triangle ABC in which AB = AC . M is a point on AB and N is a point AC such that BM = CN
Prove that
BN = CM

$\Delta ABC ,\ AB = AC ,\ M$ and

$N$ are point on

$AB$ and

$AC$ such that

$BM = CN$

$BN$ and

$CM$ are joined.

In

$\Delta AMC$ and

$\Delta ANB$

$AB = AC$ [given] ...(1)

$BM = CN$ [given] ...(2)
Subtracting (2) from (1) we have

$AB - BM - AC - CN$

$AM = AN$ ....(3)

Consider the triangles

$\Delta AMC$ and

$\Delta ANB$

$AC = AB$ [given]

$\angle A = \angle A$ [common]

$AM = AN$ [From (3)]
By Side - Angle _ Side criterion of congruence, we have

$\Delta AMC \cong \Delta ANB$

The corresponding parts of the congruent triangles are congruent

$CM = BN$ [cpct] ...(4)

Statements in the List I have to be matched with statements in the List II.

(A)

$AB^2=AC^2+BC^2$

Since,

$\triangle ABC$ is an isosceles right angled triangle.

$\therefore AC=BC$

Now,

$AB^2=AC^2+AC^2=2AC^2$

(B)

$\dfrac{area(\triangle ABC)}{area(\triangle DEF)}=\dfrac{(AB)^2}{(DE)^2}=\dfrac{(1.2)^2}{(1.4)^2}=\dfrac{1.44}{1.96}$

$=\dfrac{36}{49}=\dfrac{(36\times 2)}{(49\times 2)}=\dfrac{72}{98}$

(C)

$\dfrac{area(\triangle ABC)}{area(\triangle APQ)}=\dfrac{(BC)^2}{(PQ)^2}=\dfrac{36}{49}$

$\dfrac{BC}{PQ}=\dfrac{6}{7}$

(D)

$\because DE\parallel BC$

$\therefore \dfrac{AD}{DB}=\dfrac{AE}{EC}=\dfrac{6}{7}$

In trapezium $ABCD$ , sides $AB$ and $DC$ are parallel to each other. $F$ is mid-point of $AD$ and $E$ is mid-point of $BC$ . Can it be concluded that $AB + DC= 2EF$ ?
If the above statement is true then mention answer as 1, else mention 0 if false

$\triangle FDP$ and

$\triangle ADB$ ,

$\angle ADB = \angle FDP$ (Common angle)

$\angle DAB = \angle DFP$ (Corresponding angles)

$\angle DBA = \angle DPF$ (Corresponding angles)
Thus,

$\triangle FDP \sim \triangle ADB$ (AAA rule)
hence,

$\frac{FP}{AB} = \frac{FD}{AD}$

$\frac{FP}{AB} = \frac{1}{2}$ (Since, AF = FD)
or

$FP = \frac{1}{2} AB$ ...(I)

Since,

$AB \parallel FB \parallel CD$
By equal Intercept theorem,

$BE = EC$
Hence,

$PE = \frac{1}{2} CD$ ...(II)
Thus, adding I and II

$PF + PE = \frac{1}{2} (AB + CD)$

$2 EF = AB + CD$

In figure, $\Delta$ UVW is right angled triangle and $\angle$ UVW $=$ 90 $^o$ , UV $=$ 6cm, and UW $=$ 8 cm, $\sin$ W = $p:m$ , then $p+m$ =

Given:

$\triangle UVW$ ,

$\angle UVW = 90^{\circ}$ ,

$UV = 6$ cm,

$UW = 8$ cm
By Pythagoras theorem,

$UV^2 + VW^2 = UW^2$

$6^2 + VW^2 = 8^2$

$VW^2 = 28$

$VW = 2\sqrt{7}$ cm

Trigonometric ratio w.r.t

$\angle W$

$SinW = \frac{P}{H} = \frac{UV}{UW} = \frac{6}{8} = \frac{3}{4}$

Show that $\displaystyle \Delta ADC$ and $\displaystyle \Delta BAC$ are similar.

$\triangle ADC$ and

$\triangle BAC$ ,

$\angle ADC = \angle BAC$ (Each

$90^{\circ}$ )

$\angle DCA = \angle BCA$ (Common)

$\angle DAC=\angle ABC$ (Third angle)

Thus,

$\triangle ADC \sim \triangle BAC$ (AAA rule)

In a triangle, $\displaystyle \angle BAC= 90^{\circ}$ and $\displaystyle AD\perp BC.$
If AC=13.5 cm and CD =4.5cm; calculate the length of 2BC

$\triangle ADC$ and

$\triangle BAC$ ,

$\angle ADC = \angle BAC$ (Each

$90^{\circ}$ )

$\angle DCA = \angle BCA$ (Common)

$\angle DAC = \angle ABC$ (Third angle)
Thus,

$\triangle ADC \sim \triangle BAC$ (AAA rule)

$\dfrac{AC}{BC} = \dfrac{CD}{AC}$ (Corresponding sides)

$AC^2 = CD \times BC$

$(13.5)^2 = 4.5 \times BC$

$BC = 40.5 cm$

$2BC=81 cm$

$9$ cm, $12$ cm and $15$ cm

In a right angled triangle, sum of squares of two sides should be equal to the square of the third side.

Here,

${9}^{2} = 81$

${12}^{2} = 144$

${15}^{2} = 225$

Since,

$81 + 144 = 225$ , the sides form a right angled triangle.

If a line parallel to $BC$ intersects side $AB$ and $AC$ of triangle $ABC$ at $P$ and $Q,$ then prove that $\dfrac {AP}{AB}=\dfrac {AQ}{AC}.$

$\triangle APQ$ and

$\triangle ABC$ ,

$\angle A$ is common.

$\\ \angle APQ = \angle ABC$ ...[corresponding angles]

$\\ \angle AQP = \angle ACB$ ...[corresponding angles]

$\\ \therefore \triangle APQ \sim \triangle ABC$ ...(By

$AAA$ )

So, once the triangles are similar, their corresponding sides would be proportional.

$\dfrac { AP }{ AB } =\dfrac { AQ }{ AC }$

Hence proved.

In the given figure you find two triangles. Indicate whether the triangles are similar. Give reasons in support of your answer.

Let the vertices of triangle be:

$ABO$ and

$CDO$ , where

$AO = 5 cm$ and

$BO = 5 cm$ and

$CO = 3 cm$ and

$DO = 3 cm$
In triangles AOB and COD,

$\angle AOB = \angle COD$ (Vertically opposite angles)

$\cfrac {AO}{OC} = \cfrac{OB} {OD}$ (Both the ratios are

$\cfrac{5}{3}$ )
Hence,

$\triangle AOB \sim \triangle COD$ (SAS postulate)

In the following figure, $\displaystyle \angle ABC=90^{\circ},AB =\left ( x+8 \right )\ cm, BC=\left ( x+1 \right )\ cm$ and $\displaystyle AC=(x + 15)\ cm$ .
Find the value of $x$ .

In a right angled triangle,

${\text{Hypotenuse}}^{2} = {\text{base}}^{2} + {\text{height}}^{2}$
So,

${AC}^{2} = {AB}^{2} + {BC}^{2}$

$\Rightarrow {( x + 15 )}^{2}= {(x +8)}^{2} + {(x + 1)}^{2}$

$\Rightarrow {x}^{2} + 225 + 30x = {x}^{2} + 64 + 16x + {x}^{2} + 1 + 2x$

$\Rightarrow 225 + 30x = {x}^{2} + 65 + 18x$

$\Rightarrow {x}^{2} - 160 - 12x = 0$

$\Rightarrow {x}^{2} - 20x + 8x - (20 \times 8) = 0$

$\Rightarrow x(x-20) + 8 (x - 20) = 0$

$\Rightarrow (x-20)(x+8) = 0$

$\Rightarrow x = 20$ or

$x=- 8$
Since, length cannot be negative,

$\Rightarrow x = 20$

$\displaystyle \Delta ABD\equiv \Delta ADC$
If the above statement is true then mention answer as 1, else mention 0 if false

$\triangle$ s, ABD and ACD,

$AD = AD$ (Common)

$AB = AC$ (Given, ABC is an isosceles triangle)

$DB = DC$ (Given, DBC is an isosceles triangle)
Thus,

$\triangle ABD \cong \triangle ACD$ (SSS postulate)

O is mid-point of AP
If the above statement is true then mention answer as 1, else mention 0 if false

$\triangle APB$ and

$\triangle ECB$ ,

$\angle ABP = \angle EBC$ (Common angle)

$\angle PAB = \angle CEB$ (Corresponding angle of parallel lines)

$\angle BPA = \angle ECB$ (third angle)
Thus,

$\triangle APB \sim \triangle ECB$ by AAA

$\cfrac{AB}{EB} = \cfrac{BP}{BC}$

$\cfrac{2 EB}{EB} = \cfrac{PB}{BC}$ (E is the mid point of AB)

$BP = 2 BC$

$BC + CP = 2 BC$

$CP = BC$ or

$PC = AD$ (Since BC = AD)

Now, In

$\triangle$ s AOD and COP,

$\angle AOD = \angle COP$ (Vertically opposite angles)

$AD = PC$ (Proved above)

$\angle ADO = \angle PCO$ (Alternate angles for parallel lines BC and AD)
Thus,

$\triangle AOD \cong \triangle POC$ (SAS rule)
Hence,

$OA = OP$
or O is the mid point of AP.

Find lengths of $ME(in\ cm)$ .

Given:

$DE \parallel BC$ ,
In

$\triangle ADM$ and

$\triangle ABN$ ,

$\angle DAM = \angle BAN$ (Common angle)

$\angle ADM = \angle ABN$ (Corresponding angles)

$\angle AMD = \angle ANB$ (Corresponding angles)
Thus,

$\triangle ADM \sim \triangle ABN$ (AAA rule)

Similarly,

$\triangle AME \sim \triangle ANC$ and

$\triangle ADE \sim \triangle ABC$

Since,

$\triangle AME \sim \triangle ANC$

$\dfrac{AE}{AC} = \dfrac{ME}{NC}$ (Corresponding sides)

$\dfrac{15}{24} = \dfrac{ME}{6}$

$ME = \dfrac{15 \times 6}{24}$

$ME = 3.75$ cm

Two angles of one triangle are $85^{\circ}$ and $65^{\circ}$ is equal to angles of the other. Are they similar? Prove it.

Two angles of one triangle are 85° and 65°.

$\angle A+\angle B=85°+65°=150°<180°$

The sum of the interior angles of a triangle are equal to 180°.

and sum of two angle should be less then 180°.

So, another triangle of same angles is possible.

Answer : yes

$\Delta\,APB$ is similar to $\Delta\,CPD$ .
Enter 1 if true , else 0

Given: ABCD is a trapezium with

$AB \parallel CD$
In

$\triangle APB$ and

$\triangle CPD$ ,

$\angle APB = \angle CPD$ (Vertically opposite angles)

$\angle PAB = \angle PCD$ (Alternate angles)

$\angle PBA = \angle PDC$ (Alternate angles)
Thus,

$\triangle APB \sim \triangle CPD$ (AAA rule)

Prove: 3DF = EF

Given: In

$\triangle ABC$ , D is mid point of BC and

$DP \parallel BA \parallel CR$
To Prove: 3 DF = EF

In

$\triangle ABC$ , D is the midpoint of BC and DP are drawn parallel to BA.
Therefore, P is the midpoint of AC.

$AP = PC$

Now,

$FA \parallel DP \parallel RC$ and APC is transversal such that AP = PC and FDR is the another transversal
Hence,

$FD = DR$ .........(I) (by intercept theorem)

$EC = \dfrac{1}{2} AC = PC$

In

$\triangle EPD$ ,
C is the midpoint of EP and

$CR \parallel DP$ .
R must be the midpoint of DE.
Thus,

$DR = RE$ .....(II)

Hence,

$FD = DR = RE$ (from (I) and (II))

$FE = 3 DF$

Write all possible pairs of similar triangles.

Given:

$DE \parallel BC$ ,
In

$\triangle ADM$ and

$\triangle ABN$ ,

$\angle DAM = \angle BAN$ (Common angle)

$\angle ADM = \angle ABN$ (Corresponding angles)

$\angle AMD = \angle ANB$ (Corresponding angles)
Thus,

$\triangle ADM \sim \triangle ABN$ (AAA rule)

Similarly,

$\triangle AME \sim \triangle ANC$ and

$\triangle ADE \sim \triangle ABC$

In the figure given below, straight lines $AB$ and $CD$ intersect at $P$ , and $AC || BD$ . If $BD = 2.4\ cm$ , $AC=4.8\ cm$ ; find the value of $\cfrac { AP }{ PB }$ .

Given:

$AC \parallel BD$

$BD=2.4\ cm$ and

$AC=4.8\ cm$

$\triangle ACP$ and

$\triangle BDP$

$\angle ACP=\angle BDP$ [alternate angles]

$\angle CAP=\angle DBP$ [alternate angles]

and

$\angle APC=\angle BPD$ [vertically opposite angles]

So, by

$AAA$ similarity

$\triangle ACP$ and

$\triangle BDP$ are similar which means their sides are proportional.

So,

$\cfrac { AC }{ BD } =\cfrac { AP }{ BP } =\cfrac { CP }{ PD } .....(i)$

Now,

$\Rightarrow \cfrac { AC }{ BD } =\cfrac { AP }{ BP }$

$\Rightarrow \cfrac { 4.8 }{ 2.4 } =\cfrac { AP }{ PB }$

$\Rightarrow \cfrac { AP }{ PB }=2$

Hence, the required value is

$2$ .

$\angle APB$ .

(i) Let AD = AB = 2AD = 2x

Also AP is the bisector ∠A∴∠1 = ∠2

Now, ∠2 = ∠5 (alternate angles)

∴∠1 = ∠5Now AD = DP = x [∵ Sides opposite to equal angles are also equal]

∵ AB = CD (opposite sides of parallelogram are equal)

∴ CD = 2x⇒ DP + PC = 2x⇒ x + PC = 2x⇒ PC = x

Also, BC = x In ΔBPC,∠6 = ∠4 (Angles opposite to equal sides are equal)

Also, ∠6 = ∠3 (alternate angles)

∵ ∠6 = ∠4 and ∠6 = ∠3⇒∠3 = ∠4

Hence, BP bisects ∠B.

(ii) To prove ∠APB = 90°∵ Opposite angles are supplementary..

Angle sum property,

DE = FB

From

$\triangle DEA$ and

$\triangle BFC$

$AD=BC; AE=FC$

$\therefore \angle DAE = \angle BCF$ ......... [Diagonal biesct]

So, congruent by SAS

$\Rightarrow DE=FB$ ..... [CPCT]

$\displaystyle\,\dfrac{CP}{PA}=\dfrac m{3}$ .Find $m$

$\triangle CAB$ and

$\triangle CPQ$ ,

$\angle ACB = \angle PCQ$ (Common angle)

$\angle CAB = \angle CPQ$ (Corresponding angles)

$\angle CBA = \angle CQP$ (Corresponding angles)
Thus,

$\triangle CAB \sim \triangle CPQ$ (AAA rule)

$\dfrac{CP}{CA} = \dfrac{CQ}{CB}$

$\dfrac{CP}{CA} = \dfrac{4.8}{4.8 + 3.6}$

$\dfrac{CP}{CP + AP} = \dfrac{4.8}{8.4}$

$84 CP = 48 CP + 48 AP$

$36 CP = 48 AP$

$\dfrac{CP}{AP} = \dfrac{4}{3}$

PQ

$\triangle CAB$ and

$\triangle CPQ$ ,

$\angle ACB = \angle PCQ$ (Common angle)

$\angle CAB = \angle CPQ$ (Corresponding angles)

$\angle CBA = \angle CQP$ (Corresponding angles)
Thus,

$\triangle CAB \sim \triangle CPQ$ (AAA rule)

$\dfrac{CQ}{CB} = \dfrac{PQ}{AB}$

$\dfrac{CQ}{CB + BQ} = \dfrac{PQ}{6.3}$

$\dfrac{4.8}{4.8 + 3.6} = \dfrac{PQ}{6.3}$

$PQ = \dfrac{4.8 \times 6.3}{8.4}$

$PQ = 3.6$ cm

$DE$ is parallel to $FB$

Consider

$\triangle DEA$ and

$\triangle BFC$ ,

$\Rightarrow AD=BC$ and

$AE=FC$

$\therefore \angle DAE=\angle BCF$ ...... (Diagonal bisect)

So it is congruent by SAS

$\therefore DE\parallel FB$ and

$DE=FB$ .... [By CPCT]

$DEBF$ is a parallelogram.

ABCD is a parallelogram,

To prove

$DE||FB$ ,

given

$AE=EF=FC$ (given)

$\Rightarrow DE||FB$

Given

$AE=EF=FC$ (given)

$\Rightarrow DE||FB$ [by vertically opposite side]

$\therefore DEBF$ is a parallelogram and

$DE=FB$

As the opposite sides are equal and parallel

$\displaystyle\,\frac{AE}{EC}$

Given

$\frac{AD}{BD} = 3 : 5$
In

$\triangle ADE$ and

$\triangle ABC$

$\angle DAE = \angle BAC$ (Common angle)

$\angle ADE = \angle ABC$ (Corresponding angles)

$\angle AED = \angle ACB$ (Corresponding angles)
Thus,

$\triangle ADE \sim \triangle ABC$ (AAA rule)

$\frac{AB}{AD} = \frac{AC}{AE}$ (Corresponding sides)

$\frac{AB}{AD} - 1 = \frac{AC}{AE} - 1$

$\frac{AB - AD}{AD} = \frac{AC - AE}{AE}$

$\frac{BD}{AD} = \frac{EC}{AE}$

$\frac{AD}{BD} = \frac{AE}{EC}$ (Reciprocating)

$\frac{AE}{EC} = \frac{3}{5}$

In the given figure, AD = AE and $AD^{2}\,=\,BD\,\times\,EC$ .
Hence , $\triangle ABD$ $\sim$ $\triangle CAE$ .
If the above statement is true then mention answer as 1, else mention 0 if false

GIven,

$AD = AE$

$\angle ADE = \angle AED$ (Isosceles triangle propery)

$180 - \angle ADE = 180 - \angle AED$

$\angle ADB = \angle AEC.....(1)$

Given:

$AD^2 = BD \times EC$

$\dfrac{AD}{EC} = \dfrac{BD}{AD}... (2)$

Now, in

$\triangle ABD$ and

$\triangle ACE$ ,

$\angle ADB = \angle AEC$ (From

$(1)$ )

$\dfrac{AD}{EC} = \dfrac{BD}{AD}$ (From

$(2)$ )

$\dfrac{AD}{BD} = \dfrac{EC}{AE}$

Corresponding sides of two triangles are proportional.

Thus,

$\triangle ABD \sim \triangle ACE$ (SAS criterion)

DE = 2.4 cm, find the length of BC.

Given

$\dfrac{AD}{BD} = 3 : 5$

$\dfrac{BD}{AD} = \dfrac{5}{3}$

$\dfrac{BD}{AD} + 1 = \dfrac{5}{3} + 1$

$\dfrac{BD + AD}{AD} = \dfrac{5 + 3}{3}$

$\dfrac{AB}{AD} = \dfrac{8}{3}$

$\dfrac{AD}{AB} = \dfrac{3}{8}$

In

$\triangle ADE$ and

$\triangle ABC$

$\angle DAE = \angle BAC$ (Common angle)

$\angle ADE = \angle ABC$ (Corresponding angles)

$\angle AED = \angle ACB$ (Corresponding angles)
Thus,

$\triangle ADE \sim \triangle ABC$ (AAA rule)

$\dfrac{AD}{AB} = \dfrac{DE}{BC}$ (Corresponding sides)

$\dfrac{3}{8} = \dfrac{2.4}{BC}$

$BC = \dfrac{2.4 \times 8}{3}$

$BC = 6.4$ cm

BC = 4.8 cm, the length of DE = 1.8cm (Enter 1 f true otherwise 0)

Given

$\dfrac{AD}{BD} = 3 : 5$

$\dfrac{BD}{AD} = \dfrac{5}{3}$

$\dfrac{BD}{AD} + 1 = \dfrac{5}{3} + 1$

$\dfrac{BD + AD}{AD} = \dfrac{5 + 3}{3}$

$\dfrac{AB}{AD} = \dfrac{8}{3}$

$\dfrac{AD}{AB} = \dfrac{3}{8}$

In

$\triangle ADE$ and

$\triangle ABC$

$\angle DAE = \angle BAC$ (Common angle)

$\angle ADE = \angle ABC$ (Corresponding angles)

$\angle AED = \angle ACB$ (Corresponding angles)
Thus,

$\triangle ADE \sim \triangle ABC$ (AAA rule)

$\dfrac{AD}{AB} = \dfrac{DE}{BC}$ (Corresponding sides)

$\dfrac{3}{8} = \dfrac{DE}{4.8}$

$DE = \dfrac{4.8 \times 3}{8}$

$DE = 1.8$ cm

PQR is a right-angle triangle right angled at Q. XY is parallel to QR. PQ = 6 cm and PX : XQ = 1:Calculate the lengths of PR and QR.

Given,

$\dfrac {PX}{XQ} = \dfrac{1}{2}$

$\dfrac{XQ}{PX} = 2$

$\dfrac{PX + XQ}{PX} = 2+ 1$

$\dfrac{PQ}{PX} = 3$

$\triangle PQR$ and

$\triangle PXY$ ,

$\angle XPY = \angle QPR$ (Common angle)

$\angle PXY = \angle PQR = 90^{\circ}$ (XY II QR)

$\angle XYP = \angle QRP$ (third angle)
hence,

$\triangle PQR \sim \triangle PXY$ (AAA rule)

Hence,

$\dfrac{PQ}{PX} = \dfrac{QR}{XY} = \dfrac{PR}{PY}$ (Corresponding sides)

$\dfrac{PR}{PY} = 3$

therefore,

$PR = 3 \times PR = 3\times 4 = 12 cm$

Now, In

$\triangle PQR$ ,

$PQ^2 + QR^2 = PR^2$ (Pythagoras theorem)

$6^2 + QR^2 = 12^2$

$QR^2 = 144 - 36$

$QR = \sqrt{108}$

$QR = 10.392 cm$

In the following figure, $XY$ is parallel to $BC$ , $AX = 9\ cm$ . $XB = 4.5 cm$ and $BC = 18\ cm$ .

What is the value of $\displaystyle\,\dfrac{AY}{YC}$ ?

$\triangle$ s AXY and ABC,

$\angle XAY = \angle BAC$ (Common angle)

$\angle AXY = \angle ABC$ (Corresponding angles for parallel lines, XY II BC)

$\angle AYX = \angle ACB$ (Corresponding angles for parallel lines, XY II BC)

Thus,

$\triangle AXY \sim \triangle ABC$

Hence,

$\dfrac{AX}{AB} = \dfrac{AY}{AC}$ (Using similar triangle property)

$\dfrac{AX}{AX + XB} = \dfrac{AY}{AY + YC}$

$\dfrac{9}{9 + 4.5} = \dfrac{AY}{AY + YC}$

$9 AY + 9 YC = 13.5 AY$

$9 YC = 4.5 AY$

$\dfrac{AY}{YC} = 2$

$\triangle ABC \sim \triangle PQR$ . If AD and PM are corresponding medians of the two triangles , then $\frac{AB}{PQ}\,=\,\frac{AD}{PM}$ .

Given:

$\triangle ABC \sim \triangle PQR$

$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ (Corresponding sides)
or

$\frac{AB}{PQ} = \frac{\frac{1}{2} BC}{\frac{1}{2} QR}$

$\angle ABC = \angle PQR$ (Corresponding angles)

Now, AD and PM are medians of

$\triangle ABC$ and

$\triangle PQR$ respectively,
thus, in

$\triangle ABD$ and

$\triangle PQM$

$\angle ABD = \angle PQM$ (Since,

$\triangle ABC \sim \triangle PQR$ )

$\frac{AB}{PQ} = \frac{BD}{QM}$ (AD and PM are medians)
Thus,

$\triangle ABD \sim \triangle PQM$ (SAS rule)
Hence,

$\frac{AB}{PQ} = \frac{AD}{PM}$

In triangle $ABC$ , the bisector of angle $BAC$ meets opposite side $BC$ at point $D$ . If $BD = CD$ , prove that $\Delta ABC$ is isosceles.

Given

$AD$ is angle bisector and

$BD=CD.$

Construction:

Draw a line from

$C$ parallel to

$AB$ with length

$AC.$

Now, in

$\triangle ABD$ and

$\triangle FCD,$

$\angle ABD=\angle FCD$ [Corresponding Angles]

$\angle BAD=\angle CFD$ [Corresponding Angles]

So, by AAA criteria of similarity,

$\triangle ABD\sim \triangle FCD$

$\Rightarrow \cfrac { AB }{ BD } =\cfrac { CF }{ CD }$

But given

$BD=CD$ and we took

$CF=AC$

$\Rightarrow AB=AC$

$\therefore \triangle ABC$ is isosceles.

1003659_194930_ans_2354630407bb42378146662e9563979a.png

What is the value of $EC$ in mm?

$\triangle AEB$ and

$\triangle ECF$ ,

$\angle AEB = \angle CEF$ (vertically opposite angles)

$\angle ABE = \angle ECF$ (Alternate angles of parallel lines AB and CF)

$\angle BAE =\angle EFC$ (Alternate angles of parallel lines AB and CF)
Thus,

$\triangle AEB \sim \triangle \angle FEC$ (AAA rule)
Hence,

$\dfrac{AB}{CF} = \dfrac{BE}{EC}$

$\dfrac{9}{13.5} = \dfrac{15}{EC}$

$EC = \dfrac{15 \times 13.5}{9}$

$EC = 22.5$ cm=

$225\ mm$

Find $XY$ in cm.

$\triangle$ s AXY and ABC,

$\angle XAY = \angle BAC$ (Common angle)

$\angle AXY = \angle ABC$ (Corresponding angles for parallel lines, XY II BC)

$\angle AYX = \angle ACB$ (Corresponding angles for parallel lines, XY II BC)

Thus,

$\triangle AXY \sim \triangle ABC$

Hence,

$\dfrac{AX}{AB} = \dfrac{XY}{BC}$ (Using similar triangle property)

$\dfrac{AX}{AX + XB} = \dfrac{XY}{18}$

$\dfrac{9}{9 + 4.5} = \dfrac{XY}{18}$

$XY = \dfrac{18 \times 9}{13.5}$

$XY = 12$

In the following figure , $ABCD$ is a trapezium with $AB \parallel DC.$ If $AB = 9$ $cm$ , $DC = 18$ $cm$ , $CF = 13.5$ $cm$ , $AP = 6$ $cm$ and $BE = 15$ $cm$ . Find the value of $PE.$

Given: Trapezium ABCD,

$AB \parallel CD$ ,

Now, In

$\triangle APB$ and

$\triangle DPF$ ,

$\angle APB = \angle DPF$ [Vertically Opposite angles]

$\angle ABP = \angle PDF$ [Alternate angles]

$\angle PAB = \angle PFD$ [Alternate angles]

So, by

$AAA$ criteria of similarity,

$\triangle APB \sim \triangle FPD$

Hence,

$\dfrac{AP}{PF} = \dfrac{AB}{DF}$

$\Rightarrow \dfrac{AP}{PF} = \dfrac{AB}{DC + CF}$

$\Rightarrow \dfrac{6}{PF} = \dfrac{9}{18 + 13.5}$

$\Rightarrow PF = \dfrac{6 \times 31.5}{9}$

$\Rightarrow PF = 21\ cm$

Now,

$AF = AP + PF$

$\Rightarrow AF = 6 + 21$

$\Rightarrow AF = 27\ cm$

In

$\triangle AEB$ and

$\triangle ECF$ ,

$\angle AEB = \angle CEF$ [Vertically opposite angles]

$\angle ABE = \angle ECF$ [Alternate angles of parallel lines

$AB$ and

$CF$ ]

$\angle BAE =\angle EFC$ [Alternate angles of parallel lines

$AB$ and

$CF$ ]

So, by

$AAA$ criteria of similarity,

$\triangle AEB \sim \triangle \angle FEC$

Hence,

$\dfrac{AB}{CF} = \dfrac{AE}{EF}$

$\Rightarrow \dfrac{9}{13.5} = \dfrac{15}{EC}$

$\Rightarrow \dfrac{AE}{EF} = \dfrac{2}{3}$

Then, let

$AE = 2x$ and

$EF = 3x$

$AF = AE + EF$

$\Rightarrow 21 = 2x + 3x$

$\Rightarrow x = 4.2$

Thus,

$AE = 8.4\ cm$ and

$EF = 12.6\ cm.$

And,

$AE = AP + PE$

$\Rightarrow 8.4 = 6 + PE$

$\Rightarrow PE = 2.4\ cm$

In the given figure, $\Delta \,ABC\,\sim\,\Delta\,ADE$ . If
$AE : EC = 4 : 7$ and $DE = 6.6\ cm,$ find $BC$ . If x be the length of the perpendicular from $A$ to $DE$ , find the length of perpendicular from $A$ and $BC$ in terms of $x.$

Given

$\dfrac{AE}{EC} = \dfrac{4}{7}$

$\dfrac{EC}{AE} = \dfrac{7}{4}$

$\dfrac{EC + AE} {AE} = \dfrac{7 + 4} {4}$

$\dfrac{AC}{AE} = \dfrac{11}{4}$

Given,

$\triangle ABC \sim \triangle ADE$

$\dfrac{AC}{AE} = \dfrac{BC}{DE}$

$\dfrac{11}{4} = \dfrac{BC}{6.6}$

$\therefore BC = 18.15$ cm

The perpendiculars of similar triangles are in proportion to their sides, Hence, perpendicular from A on BC =

$\dfrac{AC}{AE}$ (perpendicular from A on DE)
Perpendicular from A on BC =

$\dfrac{11}{4} x$

ABCD a parallelogram. $E$ is a point on $AD$ and $CE$ is produced to meet $BA$ at $F$ . If $AE = 4$ $cm$ , $AF = 8$ $cm$ and $AB = 12$ $cm$ . Find the perimeter (in $cm$ ) of parallelogram $ABCD$ .

$\triangle AFE$ and

$\triangle BFC$ ,

$\angle AFE = \angle BFC$ (Common angle)

$\angle AEF = \angle BCF$ (Corresponding angles of parallel lines)

$\angle FAE = \angle CBF$ (Corresponding angles of parallel lines)
Thus,

$\triangle AFE \sim \triangle CFB$ (AAA rule)
Hence,

$\frac{AF}{BF} = \frac{AE}{BC}$

$\frac{AF}{AF + AB} = \frac{AE}{BC}$

$\frac{8}{8 + 12} = \frac{4}{BC}$

$BC = \frac{4 \times 20}{8}$

$BC = 10$ cm
Perimeter of parallelogram =

$AB + BC + BC+ AD$
Perimeter of parallelogram =

$2 (AB + BC)$
Perimeter of parallelogram =

$2 (12 + 10)$ (Opposite sides are equal)
Perimeter of parallelogram =

$44$ cm

Calculate $AF$ .

Given: Trapezium ABCD,

$AB \parallel CD$ ,

Now, In

$\triangle APB$ and

$\triangle DPF$ ,

$\angle APB = \angle DPF$ (Vertically Opposite angles)

$\angle ABP = \angle PDF$ (Alternate angles)

$\angle PAB = \angle PFD$ (Alternate angles)

Thus,

$\triangle APB \sim \triangle FPD$ (AAA rule)

Hence,

$\dfrac{AP}{PF} = \dfrac{AB}{DF}$

$\dfrac{AP}{PF} = \dfrac{AB}{DC + CF}$

$\dfrac{6}{PF} = \dfrac{9}{18 + 13.5}$

$PF = \dfrac{6 \times 31.5}{9}$

$PF = 21$ cm

Now,

$AF = AP + PF$

$AF = 6 + 21$

$AF = 27$ cm

ABCD is a trapezium. Further if CD = 4.5 cm; find the length of 2AB in cm.

$\triangle$ s OAB and OCD,

$\angle AOB = \angle COD$ (Vertically opposite angles)

$\frac{OC}{OA} = \frac{OD}{OB}$ (Given)
Thus,

$\triangle OAB \sim \triangle OCD$ (SAS postulate)
Now,

$\angle OAB = \angle OCD$ (Corresponding angles of similar triangles)

$\angle OBA = \angle ODC$ (Corresponding angles of similar triangles)
Also, they are alternate angles for lines AB and CD. Hence, AB II CD.
Thus, ABCD is a trapezium with a parallel pair of lines AB and CD.

Also,

$\frac{OC}{OA} = \frac{CD}{AB} = \frac{1}{3}$ (ratio of corresponding sides is equal for similar triangles)

$\frac{4.5}{AB} = \frac{1}{3}$

$AB = 4.5 \times 3$

$AB = 13.5$ cm

$2AB=27$ cm

Area $(\Delta AGB) = \frac{2}{3} \times\, Area (\Delta ADB)$

Given

$AD$ and

$BC$ are the medians

$\triangle AGB,AG$ will be the median

$\triangle AGC ,AG$ will be the median

$\therefore Ar(\triangle AGB)=Ar(\triangle AGC)$

Similarly

$BG$ is the median

$Ar(\triangle AGB)=Ar(\triangle BGD)$

$\implies Ar(\triangle AGB)=Ar( \triangle AGC)=Ar(\triangle BGD)$

$Ar(\triangle ABC)=2\times Ar(\triangle ADB)$

$Ar(\triangle ABC)=Ar(\triangle AGB)+Ar(\triangle AGC)+Ar(\triangle BGD)$

$Ar(\triangle AGB)=\dfrac{1}{3}\times Ar(\triangle ABC)=\dfrac{2}{3}\times Ar(\triangle ABC)$

1166163_196523_ans_f2c448bffe6c44b1a6c8162d9eb8fb9b.png

Show that a diagonal divides a parallelogram into two triangles of equal area.

Let

$ABCD$ be a parallelogram and

$BD$ is one of its diagonal. As we know that the opposite sides of a parallelogram are equal.

$\therefore AB=CD$ and

$AD=BC$

$\triangle ABD$ and

$\triangle CDB$ ,

$AB=CD, AD=BC$ [given]

$BD=BD$ [common]

$\therefore \triangle ABD \cong \triangle ADC$

$\Rightarrow ar(\triangle ABD)=ar(\triangle CBD)$

$\Rightarrow ar(\triangle ABC)=ar(\triangle CDA)$

i.e congruent triangles have equal area

In the figure, given alongside, $\angle \,APQ\,=\,\angle\,C$ . Hence, If AQ = 9 cm, BP = 3 cm and AP = 15 cm; find AC in cm

Given:

$\angle APQ = \angle C$

$\triangle APQ$ and

$\triangle ABC$

$\angle PAQ = \angle BAC$ (Common angle)

$\angle APQ = \angle C$ (Given)

$\angle AQP = \angle B$ (third angle)

Thus,

$\triangle APQ \sim \triangle ACB$ (AAA rule)

Hence,

$\dfrac{AP}{AC} = \dfrac{AQ}{AB}$

$\dfrac{15}{AC} = \dfrac{9}{15 + 3}$

$AC = \dfrac{15 \times 18}{9}$

$AC = 30$ cm

$\Delta\,OAB\,\sim\,\Delta\,OAD$
If the above statement is true then mention answer as 1, else mention 0 if false

1906185_195497_ans_2266dd0df984402f80a6dc8e8d62c78b.jpeg

$D$ is a point on the side $BC$ of $\triangle ABC$ such that $\angle ADC=\angle BAC.$ Prove that $\displaystyle \frac{CA}{CD}=\frac{CB}{CA}$ or $CA^{2}=CB\times CD$

$\triangle ABC$ and

$\triangle DAC$ , we have

$\angle ADC=\angle BAC$ and

$\angle C=\angle C$

$\therefore$ By AA-axiom of similarity

$\triangle ABC\sim \triangle DAC$

$\Rightarrow$

$\displaystyle \frac{AB}{DA}=\frac{BC}{AC}=\frac{AC}{DC}$

$\Rightarrow$

$\displaystyle \frac{CB}{CA}=\frac{CA}{CD}$ ....Hence, proved

In $\Delta ABC,\;AN\;\bot\;BC\;and\;BM\;\bot\;AC$ if $AC=8\;cm\;and\;BC=7.5\;cm$ , If $AN=8.0\;cm$ , and length of $BM$ is given by $\dfrac pq$ , where p and q are co-primes, then find the value of $p - 7q$ .

In triangle

$ANC$ and

$BCM$

$\angle ANC=\angle BMC$

$\angle ACB=\angle ACN$

$\triangle ANC\sim \triangle BCM$
So,

$\dfrac{AC}{AN}=\dfrac{BC}{BM}$

$\dfrac88=\dfrac{7.5}{BM}$

$BM=7.5\ cm = \dfrac{15}2\ cm = \dfrac pq$

$p=15, q = 2$

$p-7q = 15-7\times2 = 1$

PQ = 1.28 cm; PR = 2.56 cm,
PM =0.16 cm; PN =0.32 cm.

PQ = 1.28, PM =0.16
$\Rightarrow$ MQ = PQ - PM = 1.28 - 0.16 = 1.12
NR = PR - PN = 2.56 - 0.32 = 2.24
Now

$\displaystyle \dfrac{PM}{MQ}= \dfrac{0.16}{1.12} = \dfrac{1}{7}$

$\displaystyle \dfrac{PN}{NR} = \dfrac{0.32}{2.24} = \dfrac{1}{7}$

$\displaystyle \therefore \dfrac{PM}{MQ} = \dfrac{PN}{NR}$ $\displaystyle \left[Each\, ratio = \dfrac{1}{7} \right]$
$\Rightarrow$ By the converse of Basic Proportionality Theorem,
$MN \parallel QR$

In the following figure, LM $\parallel$ AB. If AL = x - 3, AC = 2x, BM = x - 2, BC = 2x + 3, then find the value of x.

In $\Delta ABC, LM \parallel AB$

$\displaystyle \therefore \dfrac{AL}{LC} = \dfrac{BM}{MC}$ [By Basic Proportionality Theorem]

$\displaystyle \Rightarrow \dfrac{x - 3}{2x - (x -3)} = \dfrac{x - 2}{(2x + 3) - (x - 2)}$

$\displaystyle \Rightarrow \dfrac{x - 3}{x + 3} = \dfrac{x - 2}{x + 5}$

$\displaystyle \Rightarrow (x - 3)\times(x+5) = (x - 2)\times(x + 3)$

$\displaystyle \Rightarrow x^2 + 2x -15 = x^2 + x - 6$

$\displaystyle \Rightarrow x = 9$

If in fig. (i) and (ii), PQ $\parallel$ BC, find QC in (i) and AQ in (ii).

$\Delta ABC , PQ \parallel BC$

$\therefore$ By basic Proportionality Theorem, we get

$\displaystyle \frac{AP}{PB} = \frac{AQ}{QC}$

$\displaystyle \Rightarrow \frac{1.5}{3} = \frac{1.3}{QC}$

$\displaystyle \Rightarrow QC = \frac{1.3 \times 3}{15} = \frac{13}{5} = 2.6 cm$
Hence,

$QC = 2.6 cm$

In fig. (ii) given above
In

$\Delta ABC, PQ \parallel BC$

$\therefore$ By Basic Proportionality Theorem, we get

$\displaystyle \frac{AP}{PB} = \frac{AQ}{QC} \Rightarrow \frac{3}{6} = \frac{AQ}{5.3}$

$\displaystyle \Rightarrow AQ = \frac{3 \times 5.3}{6} = \frac{5.3}{2} = 2.65 cm$

1031922_345873_ans_e1cb364f2cce4ad2bb869e4395641e9c.jpg

PM =4 cm; QM =4.5 cm;
PN = 4 cm; NR = 4.5 cm

$\displaystyle \dfrac{PM}{QM} = \dfrac{4}{4.5}$ and $\displaystyle \dfrac{PN}{NR} = \dfrac{4}{4.5}$
$\displaystyle \therefore \dfrac{PM}{QM} = \dfrac{PN}{NR}$ $\displaystyle \left[ Each\, ratio = \dfrac{4}{4.5} \right]$
$\Rightarrow$ By the converse of Basic Proportinality theorem, we get, MN $\parallel$ QR.

1152216_345866_ans_eb8206b7afb74c2e9a1f42d5aa781bcc.jpg

In triangle $ABC,\;\angle BAC=90^{\circ},$ and $AD$ is its bisector. If $DE$ is drawn $\bot\;AC$ , prove that $DE\times (AB+AC)=AB\times AC$ .

It is given that

$AD$ is the bisector of

$\angle A$ of

$\Delta ABC$ .

$\therefore\;\displaystyle\frac{AB}{AC}=\displaystyle\frac{BD}{DC}$

$\Rightarrow\;\displaystyle\frac{AB}{AC}+1=\displaystyle\frac{BD}{DC}+1\;\;\;\;\;\;\;[Adding\;1\;on\;both\;sides]$

$\Rightarrow\;\displaystyle\frac{AB+AC}{AC}=\displaystyle\frac{BD+DC}{DC}$

$\Rightarrow\;\displaystyle\frac{AB+AC}{AC}=\displaystyle\frac{BC}{DC}\;\;\;\;\;\;\;\;\;\;\;\;\;......(i)$
In

$\Delta's CDE\;and\;CBA$ , we have

$\angle DCE=\angle BCA\;\;\;\;\;\;\;\;\;\;\;[Common]$

$\angle DEC=\angle BAC\;\;\;\;\;\;\;\;\;\;\;[Each\;equal\;to\;90^{\circ}$
So, by

$AA$ -criterion of similarity

$\Delta CDE\;\sim\;\Delta CBA$

$\Rightarrow\;\displaystyle\frac{CD}{CB}=\displaystyle\frac{DE}{BA}$

$\Rightarrow\;\displaystyle\frac{AB}{DE}=\displaystyle\frac{BC}{DC}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;.....(ii)$
From (i) and (ii) we have

$\Rightarrow\;\displaystyle\frac{AB+AC}{AC}=\displaystyle\frac{AB}{DE}$

$\Rightarrow\;DE\times (AB+AC)=AB\times AC$ .

In the given fig. DE $\parallel$ BC. If AD = x, DB = x - 2, AE = x + 2 and EC = x - 1, find the value of x.

$\Delta$ ABC, DE

$\parallel$ BC

$\displaystyle \therefore \dfrac{AD}{DB} = \dfrac{AE}{EC}$
(By basic proportionality theorem)

$\displaystyle \Rightarrow \dfrac{x}{x - 2} = \dfrac{x + 2}{x - 1}$

$\Rightarrow x(x - 1) = (x + 2)(x - 2)$

$\Rightarrow x^2 - x = x^2 - 4$

$\Rightarrow x = 4.$

The ratio of the corresponding sides of similar triangles $ABC$ and $A'B'C'$ is $2 : 1$ . Also, the altitudes $CD$ and $C'D',$ that we have drawn in these triangles are also in the same ratio $2: 1$ . Then, prove that the ratio of their areas is equal to the square of the ratio of their corresponding sides.

According to the given condition let the length of

$CD$ and

$C'D'$ be

$2x$ and

$x$ respectievly.

$\displaystyle \frac{CD}{C'D'} = \frac{2}{1}$
Area of a triangle

$= \dfrac{1}{2} \times base \times altitude$

$\displaystyle \therefore \frac{Area of \Delta ABC}{Area of \Delta A'B'C'} = \frac{\displaystyle \frac{1}{2} \times 42 \times 2x}{\displaystyle \frac{1}{2} \times 21 \times x}$

$=\displaystyle \frac{42 \times 2}{21 \times 1} \\= \dfrac{84}{21} \\= \dfrac{4}{1} \\= \left(\dfrac{2}{1} \right)^2$

$\therefore$ The ratio of the areas of these similar triangles is the square of the ratio of the measures of the corresponding sides.

1032085_345939_ans_77914d49385743d8af4f5eedaa0d9d1e.png

In a trapezium ABCD, O is the point of intersection of AC and BD, AB $\parallel$ CD and $AB = 2 CD$ . If the area of $\Delta AOB = 84 cm^2$ , find the area of $\Delta COD$ .

ABCD is a trapezium in which, AB

$\parallel$ CD and diagonals AC and BD intersect each other at O.
In

$\Delta AOB$ and

$\Delta COD$ we have

$\angle 1 = \angle 2$ [Alt

$\angle s]$

$\angle 3 = \angle 4$ [vert opp. \angle s

\therefor \Delta AOB \sim \Delta COD$$
[By AA criterion of similarity]

$\displaystyle \Rightarrow \frac{ar(\Delta AOB)}{ar(\Delta COD)} = \frac{(2CD)^2}{(CD)^2}$
[AB = 2CD (Given)]

$\displaystyle = \frac{4CD^2}{CD^2} = 4$

$\displaystyle = \frac{84}{ar(\Delta COD)} = 4$ ;

$\therefore ar(\Delta COD) = 84 + 4 = 21 cm^2$
1029249_348181_ans_f8b5aa834f7c419faa3de0ac75048162.jpg

$\Delta ABC \sim \Delta PQR$ . Also, $ar(\Delta ABC) = 4 ar(\Delta PQR)$ . If $BC = 12\ cm,$ then find $QR$ .

Given:

$ar(\Delta ABC) = 4 ar(\Delta PQR)$

$\displaystyle \Rightarrow \frac{ar(\Delta ABC)}{ar(\Delta PQR)} = \frac{4}{1}$

$\displaystyle \Rightarrow \frac{BC^2}{QR^2} = \frac{4}{1}$
[

$\therefore$ The ratio of the areas of two similar

$\Delta s$ is equal to the ratio of the squares of any two corresponding sides]

$\displaystyle \Rightarrow \frac{(12)^2}{QR^2} = \frac{4}{1}$ [

$\therefore$ QR= 12 cm]

$\displaystyle \Rightarrow QR^2 = \frac{12 \times 12}{4} = 36$

$\therefore QR = \sqrt{36} = 6cm$
1032329_348070_ans_71d24e229762460f833213815c94de1f.jpg

$\Delta ABC$ is similar to $\Delta PQR$ , ar( $\Delta ABC$ ) = 36 sq. cm. and ar( $\Delta PQR$ ) =49 sq.cm. If BC = 12 cm, find QR.

$\because \Delta ABC \sim \Delta PQR$
$\displaystyle \therefore \dfrac{ar(\Delta ABC)}{ar(\Delta PQR)} = \dfrac{BC^2}{QR^2}$
$\displaystyle \Rightarrow \dfrac{36}{49} = \dfrac{12^2}{QR^2}$
$\displaystyle \Rightarrow (QR)^2 = \dfrac{12 \times 12 \times 49}{36} = 4 \times 49$
$\Rightarrow QR = + \sqrt{4 \times 49}$
$= 2 \times 7 = 14 cm$

1154706_345944_ans_015c7fea23b14d5981a3bf04578ec4eb.jpg

Prove that the line segments joining the midpoints of the sides of a triangle from four triangles each of which is similar to the original triangle.

$\displaystyle \bigtriangleup ABC$ in which

$D,E,F$ are the mid-points of sides

$BC,CA,$ and

$AB$ , respectively. We shall prove that each of the triangles

$AFE , FBD, EDC, AND, DEF$ is similar to

$\displaystyle \bigtriangleup ABC$

$\displaystyle \bigtriangleup s$ AFE and ABC

$F$ and

$E$ are mid-points of

$AB$ and

$AC$ respectively

$\displaystyle FE\parallel BC$

$\displaystyle \Rightarrow \angle AFE = \angle B$ [ Corresponding angles]

$\displaystyle \bigtriangleup ABC$

$\displaystyle\angle AFC =\angle B$

AND

$\displaystyle \bigtriangleup AFE \sim \bigtriangleup ABC$ [By AA corollary ]

Similarly

$\displaystyle \bigtriangleup FBD \sim \bigtriangleup ABC$

and

$\displaystyle \bigtriangleup EDC \sim \bigtriangleup ABC$

Next we shall show that

$\displaystyle \bigtriangleup DEF \sim \bigtriangleup ABC$

clearly

$\displaystyle ED\parallel AF \parallel DF \parallel EA$

$\displaystyle \therefore AFDE$ is a parallelogram

$\displaystyle \Rightarrow \angle EDF = \angle A$

$\displaystyle \left [ \because opposite angles of a \parallel ^{gm} are equal \right ]$

Similarly BDEF is a parallelogram

$\displaystyle \Rightarrow \angle DEF = \angle B$

$\displaystyle \left [ \because opposite angles of a \parallel ^{gm} are equal \right ]$

thus in triangles DEF and ABC we have

$\displaystyle \angle EDF = \angle A$

and

$\displaystyle \angle DEF = \angle B$

Do by AA corollary

$\displaystyle \bigtriangleup DEF \sim \bigtriangleup ABC$

Hence each one of the triangle

$AFE, FBD, EDC$ and

$DEF$ is similar to

$\displaystyle \bigtriangleup ABC$

If three or more parallel lines are intersected by two transversals prove that the intercepts made by them on the transversals are proportional

Three parallel lines

$l,m,n$ which are cut by the transversals

$AB$ and

$CD$ in

$P, Q, R$ and

$E ,F,G$ respectively

We shall prove,

$\displaystyle \frac{PQ}{QR}=\frac{EF}{FG}$

Draw

$\displaystyle PL \parallel CD$ meeting the lines

$m$ and

$n$ in

$M$ and

$L$ respectively

PROOF: Since

$\displaystyle PE \parallel MF$ and

$\displaystyle PM \parallel EF$

$\displaystyle \therefore$

$PMFE$ is a parallelogram

$\displaystyle \Rightarrow PM = EF$ ....(i) (Opposite sides of a parallelogram are equal )

Also

$\displaystyle MF \parallel n$ and

$\displaystyle ML \parallel FG$

$\therefore MLGF$ is a parallelogram and

$ML = FG$ (ii) (Opposite side of a

$\displaystyle \parallel ^{gm}$ are equal )

Now in

$\displaystyle \bigtriangleup PRL$ ,

$QM$ is parallel to

$RL$

$\displaystyle \frac{PQ}{QR}=\frac{PM}{ML}$ [Using Basic proportionality Theorem]

Thus,

$\displaystyle \frac{PQ}{QR}=\frac{EF}{FG}$

In Fig., $DE\,||\,OQ$ and $DF\,||\,OR$ . Show that $EF\,||\,QR$ .

$\triangle POQ,$

$DE \parallel OQ$

$\therefore$ By basic proportionality theorem,

$\dfrac{PE}{EQ}=\dfrac{PD}{DO}............(1)$

Similarly,
In

$\triangle POR,$

$DF \parallel OR$

$\therefore$ By basic proportionality theorem,

$\dfrac{PD}{DO}=\dfrac{PF}{FR}............(2)$

$\therefore$ from

$(1)$ and

$(2)$ ,

$\dfrac{PE}{EQ}=\dfrac{PF}{FR}$

$\therefore$ By converse of Basic Proportionality Theorem,

$EF\parallel QR$

In given figure, $A, B$ and $C$ are points on $OP, OQ$ and $OR$ respectively such that $AB\,||\,PQ$ and $AC\,||\,PR$ . Show that $BC\,||\,QR$ .

$\triangle POR,$

$PR \parallel AC$ Given

$\therefore$ By basic proportionality theorem,

$\dfrac{PA}{AO}=\dfrac{RC}{CO}............(1)$

Similarly,

$\triangle POQ,$

$AB \parallel PQ$ Given

$\therefore$ By basic proportionality theorem,

$\dfrac{PA}{AO}=\dfrac{QB}{BO}............(2)$

$\therefore$ From

$(1)$ and

$(2)$ ,

$\dfrac{RC}{CO}=\dfrac{QB}{BO}$

$\therefore$ By converse of basic proportionality theorem,

$BC\,\parallel\,QR$

$[hence \, proved]$

Using Theorem $6.1,$ prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Theorem 6.1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given:

$\triangle\, ABC, D$ is midpoint of

$AB$ and

$DE$ is parallel to

$BC.$

$\therefore$

$AD=DB$

To prove:

$AE = EC$

Proof:

Since,

$DE \parallel BC$

$\therefore$ By Basic Proportionality Theorem,

$\dfrac{AD}{DB}=\dfrac{AE}{EC}$

Since,

$AD = DB$

$\therefore$

$\dfrac{AE}{EC}=1$

$\therefore$

$AE=EC$

In Fig., (i) and (ii), $DE \,||\, BC$ . Find $EC$ in (i) and $AD$ in (ii).

(i) Given :

$DE \parallel BC$ in

$\triangle$ ABC,

Using Basic proportionality theorem,

$\therefore \dfrac{AD}{DB} = \dfrac{AE}{EC}$

$\Rightarrow \dfrac{1.5}{3} =\dfrac{ 1}{EC}$

$\Rightarrow EC = \dfrac{3}{1.5}$

$EC = 3\times \dfrac{10}{15} = 2$ cm

$EC = 2$ cm.

(ii) In

$\triangle ABC, DE\parallel BC$ (Given)

Using Basic proportionality theorem,

$\therefore \dfrac{ AD}{DB} = \dfrac{AE}{EC}$

$\Rightarrow \dfrac{AD}{7.2} =\dfrac{ 1.8}{5.4}$

$\Rightarrow AD = 1.8\times \dfrac{7.2}{5.4} = \dfrac{18}{10} \times \dfrac{ 72}{10}\times \dfrac{ 10}{54 }= \dfrac{24}{10}$

$\Rightarrow AD = 2.4$ cm

So,

$AD = 2.4$ cm

In Fig., $DE\, ||\, AC$ and $DF\, ||\, AE$ . Prove that $\dfrac{BF}{FE}=\dfrac{BE}{EC}$ .

$\triangle ABC,$

$DE \parallel AC$

$\therefore$ By proportionality theorem,

$\dfrac{BD}{DA}=\dfrac{BE}{EC}............(1)$

Similarly,

$\triangle ABE,$

$DF \parallel AE$

$\therefore$ By proportionality theorem,

$\dfrac{BD}{DA}=\dfrac{BF}{FE}............(2)$

$\therefore$ from

$(1)$ and

$(2)$ ,

$\dfrac{BE}{EC}=\dfrac{BF}{FE}$ .

Hence Proved.

In Fig., if $LM \,||\, CB$ and $LN\, ||\, CD$ , prove that $\dfrac{AM}{AB}=\dfrac{AN}{AD}$ .

$\triangle$ ABC,

$LM \parallel BC$

$\therefore$ By proportionality theorem,

$\dfrac{AM}{AB}=\dfrac{AL}{AC}............(1)$

Similarly,
In

$\triangle$ ADC,

$LN \parallel CD$

$\therefore$ By proportionality theorem,

$\dfrac{AN}{AD}=\dfrac{AL}{AC}............(2)$

$\therefore$ from

$(1)$ and

$(2)$ ,

$\dfrac{AM}{AB}=\dfrac{AN}{AD}$

State which pairs of triangles in Fig. are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

( i )

$\triangle\,ABC$ and

$\triangle\,PQR$

$\angle\,A=\angle P$ ,

$\angle\,B=\angle Q$ ,

$\angle\,C=\angle R$ ,

$\therefore$ By AAA criterion of similarity,

$\triangle\,ABC \sim \triangle\,PQR$

( ii )

$\triangle\,ABC$ and

$\triangle\,QRP$

$\dfrac{AB}{QR}=\dfrac{BC}{RP}=\dfrac{AC}{QP}=\dfrac{1}{2}$

$\therefore$ By SSS criterion of similarity,

$\triangle\,ABC \sim \triangle\,QRP$

( iii )

$\triangle\,LMP$ and

$\triangle\,DEF$

$\dfrac{LM}{DE}=\dfrac{2.7}{4}, \dfrac{LP}{DF}=\dfrac{1}{2}$

The sides are not in the equal ratios, Hence the two triangles are not similar.

( iv )

$\triangle\,MNL$ and

$\triangle\,QPR$

$\angle M=\angle Q$ ,

$\dfrac{MN}{QP}=\dfrac{ML}{QR}=\dfrac{1}{2}$

$\therefore$ By SAS criterion of similarity,

$\triangle\,MNL \sim \triangle\,QPR$

( v )

$\triangle\,ABC$ and

$\triangle\,EFD$

$\angle A=\angle F$ ,

$\dfrac{AB}{FD}=\dfrac{BC}{FD}=\dfrac{1}{2}$

$\therefore$ By SAS criterion of similarity,

$\triangle\,ABC \sim \triangle\,EFD$

( vi )

$\triangle\,DEF$ and

$\triangle\,PQR$

Since, sum of angles of a triangle is

$180^o$ , Hence,

$\angle F=30^o$ and

$\angle P=70^o$

$\angle\,D=\angle P$ ,

$\angle\,E=\angle Q$ ,

$\angle\,F=\angle R$ ,

$\therefore$ By AAA criterion of similarity,

$\triangle\,DEF \sim \triangle\,PQR$

$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$ . For each of the following cases, state whether $EF \,|| \,QR$ :
(i) $PE = 3.9$ cm, $EQ = 3$ cm, $PF = 3.6$ cm and $FR = 2.4$ cm
(ii) $PE = 4$ cm, $QE = 4.5$ cm, $PF = 8$ cm and $RF = 9$ cm
(iii) $PQ = 1.28$ cm, $PR = 2.56$ cm, $PE = 0.18$ cm and $PF = 0.36$ cm

$E$ and

$F$ are two points on side

$PQ$ and

$PR$ in

$\triangle PQR.$

(i)

$PE = 3.9$ cm,

$EQ = 3$ cm and

$PF = 3.6$ cm,

$FR = 2.4$ cm

Using Basic proportionality theorem,

$\therefore \dfrac{PE}{EQ} = \dfrac{3.9}{3 }= \dfrac{39}{30} =\dfrac{13}{10} = 1.3$

$\dfrac{PF}{FR} = \dfrac{3.6}{2.4} =\dfrac{ 36}{24} =\dfrac{ 3}{2} = 1.5$

$\dfrac{PE}{EQ} \neq \dfrac{ PF}{FR}$

So,

$EF$ is not parallel to

$QR.$

(ii)

$PE = 4$ cm,

$QE = 4.5$ cm,

$PF = 8$ cm,

$RF = 9$ cm

Using Basic proportionality theorem,

$\therefore \dfrac{PE}{QE} = \dfrac{4}{4.5} =\dfrac{ 40}{45} =\dfrac{ 8}{9}$

$\dfrac{PF}{RF} = \dfrac{8}{9}$

$\dfrac{PE}{QE} = \dfrac{PF}{RF}$

So,

$EF$ is parallel to

$QR.$

(iii)

$PQ = 1.28$ cm,

$PR = 2.56$ cm,

$PE = 0.18$ cm,

$PF = 0.36$ cm

Using Basic proportionality theorem,

$EQ = PQ - PE = 1.28 - 0.18 = 1.10$ cm

$FR = PR - PF = 2.56 - 0.36 = 2.20$ cm

$\dfrac{PE}{EQ} = \dfrac{0.18}{1.10} = \dfrac{18}{110} = \dfrac{9}{55 }$ ... (i)

$\dfrac{PE}{FR} = \dfrac{0.36}{2.20} = \dfrac{36}{220} =\dfrac{ 9}{55}$ ... (ii)

$\therefore \dfrac{PE}{EQ} =\dfrac{ PF}{FR.}$

So,

$EF$ is parallel to

$QR.$

Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Theorem 6.2: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Given: In

$\triangle ABC, D$ and

$E$ are midpoints of

$AB$ and

$AC$ respectively,
i.e.,

$AD=DB$ and

$AE=EC$

To Prove:

$DE \parallel BC$

Proof:
Since,

$AD=DB$

$\therefore$

$\dfrac{AD}{DB}=1............(1)$

Also,

$AE=EC$

$\therefore$

$\dfrac{AE}{EC}=1............(2)$

From

$(1)$ and

$(2)$ ,

$\dfrac{AD}{DB}$

$=$

$\dfrac{AE}{EC}$

$=\,1$
i.e.,

$\dfrac{AD}{DB}$

$=$

$\dfrac{AE}{EC}$

$\therefore$ By converse of Basic Proportionality theorem,

$DE \parallel BC$

The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\dfrac{AO}{BO}=\dfrac{CO}{DO}$ . Show that $ABCD$ is a trapezium.

Given:
The diagonals of a quadrilateral ABCD intersect each other at the point O such that

$\dfrac{AO}{BO}=\dfrac{CO}{DO}$
i.e.,

$\dfrac{AO}{CO}=\dfrac{BO}{DO}$

To Prove:

$ABCD$ is a trapezium

Construction:
Draw

$OE\parallel DC$ such that

$E$ lies on

$BC.$

Proof:
In

$\triangle BDC$ ,
By Basic Proportionality Theorem,

$\dfrac{BO}{OD}=\dfrac{BE}{EC}\,............(1)$

But,

$\dfrac{AO}{CO}=\dfrac{BO}{DO}$ (Given)

$.........(2)$

$\therefore$ From

$(1)$ and

$(2)$

$\dfrac{AO}{CO}=\dfrac{BE}{EC}$
Hence, By Converse of Basic Proportionality Theorem,

$OE\parallel AB$

Now Since,

$AB\parallel OE\parallel DC$

$\therefore$

$AB\parallel DC$

Hence,

$ABCD$ is a trapezium.

Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB\,||\,DC$ intersect each other at the point $O.$ Using a similarity criterion for two triangles, show that $\dfrac{OA}{OC}=\dfrac{OB}{OD}$ .

Given:
ABCD is a trapezium with

$AB\parallel DC$ .
O is the point of intersection of two diagonals

To Prove:

$\dfrac{OA}{OC}=\dfrac{OB}{OD}$

Proof:
In

$\triangle\,AOB$ and

$\triangle\,DOC$

$\angle BAO=\angle OCD$ (Alternate Angles)

$\angle ABO=\angle ODC$ (Alternate Angles)

$\angle AOB=\angle DOC$ (Vertically opposite angles)

$\therefore$ By AAA criterion of similarity,

$\triangle\,AOB \sim \triangle\,DOC$

$\therefore$

$\dfrac{OA}{OC}=\dfrac{OB}{OD}$ (Corresponding Sides of Similar Triangles)

In Fig, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P.$ Show that:
(i) $\triangle AEP \sim \triangle CDP$
(ii)   $\triangle ABD \sim  \triangle CBE$
(iii)   $\triangle AEP \sim  \triangle ADB$
(iv)   $\triangle PDC \sim  \triangle BEC$

$\triangle\,AEP$ and

$\triangle\,CDP$ ,

$\angle APE=\angle CPD$ (Vertically opposite angle)

$\angle AEP=\angle CDP=90^o$

$\therefore$ By AA criterion of similarity,

$\triangle\,AEP$

$\sim$

$\triangle\,CDP$
In

$\triangle ABD$ and

$\triangle CBE$

$\angle ADB=\angle CEB=90^o$

$\angle B$ is common

$\therefore$ By AA criterion of similarity,

$\triangle ABD$

$\sim$

$\triangle CBE$
In

$\triangle AEP$ and

$\triangle ADB$

$\angle AEP=\angle ADB=90^o$

$\angle A$ is common

$\therefore$ By AA criterion of similarity,

$\triangle AEP$

$\sim$

$\triangle ADB$

$\triangle PDC$ and

$\triangle BEC$

$\angle PDC=\angle BEC=90^o$

$\angle C$ is common

$\therefore$ By AA criterion of similarity,

$\triangle PDC$

$\sim$

$\triangle BEC$

$E$ is a point on the side $AD$ produced of a parallelogram ABCD and $BE$ intersects $CD$ at $F.$ Show that $\triangle ABE \sim \triangle CFB$ .

$\triangle ABE$ and

$\triangle CFB,$

$\angle ABE=\angle CFB$ (Alternate angles)

$\angle BAE=\angle BCF$ (opposite angles of a parallelogram)

$\therefore$ By AA criterion of similarity,

$\triangle ABE$

$\sim$

$\triangle CFB$

A guy wire attached to a vertical pole of height $18\ m$ is $24\ m$ long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Let

$AC=18\ m$ be the pole.

$BC=24\ m$ is the length of a guy wire and is attached to stake

$B.$

$\therefore$ In

$\triangle ABC$

By pythagoras theorem,

$BC^2=AB^2+AC^2$

$\therefore$

$24^2=AB^2+18^2$

$\Rightarrow AB^2=576-324=252$

$\therefore$

$AB=6\sqrt7\,m$

Hence, the stake has to be

$6\sqrt7\,m$ from base

$A.$

In Fig., $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that:
(i) $\triangle ABC \sim \triangle AMP$
(ii) $\dfrac{CA}{PA} = \dfrac{BC}{MP}$

$\triangle ABC$ and

$\triangle AMP,$

$\angle ABC=\angle AMP=90^o$

$\angle A$ is common

$\therefore$ By AA criterion of similarity,

$\triangle ABC$

$\sim$

$\triangle AMP$

$\therefore$

$\dfrac{CA}{PA}=\dfrac{BC}{MP}$ (Corresponding Sides of Similar Triangles)

Hence Proved

In the given figure., if $\triangle ABE \cong \triangle ACD$ , show that $\triangle ADE \sim \triangle ABC.$

Given

$\triangle ABE\cong \triangle ACD$

$\therefore$

$AB=AC$

$.........(1)$

Also,

$AE=AD$

$.........(2)$

From

$(1)$ and

$(2)$ ,

$\dfrac{AB}{AD}=\dfrac{AC}{AE}$ and

$\angle A$ is Common

$\therefore$ By SAS Criterion of Similarity,

$\triangle ADE\sim\triangle ABC$

$[hence \, proved]$

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Given:

$ABCD$ is a rhombus.
Hence,

$AB=BC=CD=AD$
And, AC perpendicular to BD

$DO=\dfrac{1}{2}DB$ and

$AO=\dfrac{1}{2}AC$

To Prove:

$AB^2+BC^2+CD^2+AD^2=AC^2+BD^2$

Proof:
In

$\triangle AOD$ , By Pythagoras Theorem,

$AD^2=AO^2+OD^2\,.....(1)$
Similarly,

$DC^2=DO^2+OC^2\,.....(2)$

$BC^2=OB^2+OC^2\,.....(3)$

$AB^2=AO^2+OB^2\,.....(4)$

Adding 1, 2, 3, 4 we get,

$AB^2+BC^2+CD^2+AD^2=2AO^2+2BO^2+2CO^2+2DO^2\,....(5)$
Since,

$DO=OB=\dfrac{1}{2}DB$ and

$AO=OC=\dfrac{1}{2}AC\,....(6)$

From 5 and 6,

$AB^2+BC^2+CD^2+AD^2=\dfrac{AC^2}{2}+\dfrac{BD^2}{2}+\dfrac{AC^2}{2}+\dfrac{BD^2}{2}$

$\therefore$

$AB^2+BC^2+CD^2+AD^2=AC^2+BD^2$

Hence Proved

An aeroplane leaves an airport and flies due north at a speed of $1000\ km$ per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of $1200\ km$ per hour. How far apart will be the two planes after $1\dfrac{1}{2}$ hours?

The first aeroplane leaves an airport and flies due north at a speed of

$1000$ km per hr.

$\therefore$ Distance travelled in

$1.5$ hrs.

$=1000\times1.5=1500$ km
Similarly, Distance traveled by second aeroplane,

$=1200\times1.5=1800$ km
In

$\triangle ABC$ ,

$BC$ is distance travelled by first aeroplane and

$BA$ is the distance traveled by second aeroplane.
Hence, applying Pythagoras Theorem,

$AC^2=AB^2+BC^2$

$\therefore$

$AC^2=1500^2+1800^2$

$\therefore$

$AC=300\sqrt{61}$ km
Hence, two planes are

$300\sqrt{61}$ km apart.

In Fig., $\dfrac{QR}{QS}=\dfrac{QT}{PR}$ and $\angle 1 = \angle 2$ . Show that $\triangle PQS \sim \triangle TQR$ .

$\triangle\,PQR$ ,
Since,

$\angle 1=\angle 2$

$\therefore$

$PR=PQ$ (Opposite sides of equal angles are equal)

$.....(1)$

In

$\triangle PQS$ and

$\triangle TQR$ ,

$\dfrac{QR}{QS}=\dfrac{QT}{PR}\,....(Given)$
i.e.,

$\dfrac{QR}{QS}=\dfrac{QT}{PQ}\,....(From\,\,\,1)$
Also,

$\angle Q$ is common

$\therefore$ By SAS criterion of similarity,

$\triangle PQS \sim \triangle TQR.$

In given figure $E$ is a point on side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC.$ If $AD \bot BC$ and $EF \bot AC,$ prove that $\triangle ABD \sim \triangle ECF$ .

Given

$AB=AC$

$\therefore\, \angle B=\angle C\,...........(1)$

$\triangle ABD$ and

$\triangle ECF$

$\angle B=\angle C$ (From 1)

$\angle ADB=\angle EFC=90^o$ [

$\because AD \bot BC$ and

$EF \bot AC,$ ]

$\therefore$ By AA Criterion of Similarity,

$\triangle ABD$

$\sim$

$\triangle ECF$

$[hence \, proved]$

State and prove Pythagoras theorem.

Given triangle ABC is a right angle triangle at B

Draw BD perpendicular to AC

$\Delta ABC$ and

$\Delta ADB$

$\angle ABC=\angle ADB$

$\angle A=\angle A$

$\therefore \Delta ABC\sim \Delta ADB$

$\therefore \dfrac{AB}{A D}=\dfrac{BC}{DB}=\dfrac{AC}{AB}$

$\Rightarrow AB^{2}=AC\times AD$ .............................................(1)

$\Delta ABC$ and

$\Delta BDC$

$\angle ABC=\angle BDC$

$\angle C=\angle C$

$\therefore \Delta ABC\sim \Delta BDC$

$\therefore \dfrac{AB}{BD}=\dfrac{BC}{DC}=\dfrac{AC}{BC}$

$\Rightarrow BC^{2}=AC\times DC$

Add (1) and (2) we get

$AB^{2}+BC^{2}=AC\times AD+AC\times DC$

$=AC(AD+DC)=AC\times AC$

$=AC^{2}$

Hence proved.

$D$ and $E$ are points on the sides $CA$ and $CB$ respectively of a triangle $ABC$ right angled at $C.$ Prove that $AE^2 + BD^2 = AB^2 + DE^2$ .

Given:

$\triangle\,ACB$ is a right angles triangle at C.

Construction:
Join

$AE\,,BD$ and

$ED$ .

To Prove:

$AE^2+BD^2=AB^2+DE^2$

Proof:
In

$\triangle \,ACE$
By pythagoras theorem,

$AE^2=EC^2+AC^2\,.....(1)$

In

$\triangle \,BCD$
By pythagoras theorem,

$BD^2=BC^2+CD^2\,.....(2)$

In

$\triangle \,ECD$
By pythagoras theorem,

$ED^2=EC^2+DC^2\,.....(3)$

In

$\triangle \,ABC$
By pythagoras theorem,

$AB^2=BC^2+AC^2\,.....(4)$

Adding 1) and 2) we get,

$AE^2+BD^2=EC^2+AC^2+BC^2+CD^2\,......(5)$

From 3), 4) and 5) we get,

$AE^2+BD^2=AB^2+DE^2$

$[hence \, proved]$

In Fig., two chords AB and CD intersect each other at the point P. Prove that :
(i) $\triangle APC \sim \triangle DPB$
(ii) $AP . PB = CP . DP$

(i) Given : In

$\triangle APC$ and

$\triangle DPB$ ,

$\angle APC=\angle DPB$ ...[Vert. opp. ∠s]

$\angle CAP=\angle BDP$ ...[Angles subtended by the same arc of a circle are equal]

$\therefore$ By AA-condition of similarity,

$\triangle APC\sim \triangle DPB$

(ii)

$\triangle APC\sim\triangle DPB$

So, sides are proportional

$\Rightarrow \dfrac{AP}{DP}=\dfrac{CP}{PB}$

$\therefore AP\times PB=CP\times DP$

Nazima is fly fishing in a stream. The tip of her fishing rod is $1.8$ m above the surface of the water and the fly at the end of the string rests on the water $3.6$ m away and $2.4$ m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig.)? If she pulls in the string at the rate of $5$ cm per second, what will be the horizontal distance of the fly from her after $12$ seconds?

Let AB is the height of the tip of the fishing rod from the water surface .Let BC is the horizontal distance of the fly from the tip of the fishing rod.

Then AC is the length of the string.

Then according to the Pythagorean theorem-

$\Rightarrow AC^2=AB^2+BC^2$

$\Rightarrow AC^2=(1.8)^2+(2.4)^2$

$\Rightarrow AC^2=3.24+5.76$

$\Rightarrow AC^2=9.00$

$\Rightarrow AC=\sqrt{9 }m=3m$

$\therefore$ The length of the string out is

$3$ m.

She pulls the string at the rate of 5 cm per second .

$\therefore$ She pulls in 12 seconds=

$12\times 5=60 cm=0.6 m$

Let the fly be at a point of D after 12 seconds

Length of the string out of 12 second is AD.

AD=AC-string pull by Nazima after 12 sec.

$AD=3-.6=2.4 m$

$\triangle ADB$

$AD^2=AB^2+BD^2$

$\Rightarrow BD^2=AD^2-AB^2$

$\Rightarrow BD^2=(2.4)^2-(1.8)^2$

$\Rightarrow BD^2=5.76-3.24$

$\Rightarrow BD^2=2.52$

$\Rightarrow BD=\sqrt{2.52}=1.587 m$

Horizontal distance to fly

$=BD+1.2$

$\Rightarrow 1.587+1.2\Rightarrow 2.787$ m

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Given:

$\triangle ABC$ is an Equilateral Triangle.

$AB=BC=AC$

$CD$ Perpendicular to

$AB$
To Prove:

$3AC^2=4CD^2$
Proof:

$\triangle ADC$

By Pythagoras Theorem,

$AC^2=AD^2+DC^2\,.....(1)$

$\triangle BDC$

By Pythagoras Theorem,

$BC^2=DC^2+BD^2\,.....(2)$

Adding 1) and 2) We get,

$2AC^2=2DC^2+AD^2+BD^2$

$(Since\,\, AC=BC)$

Since,

$AD=BD=\dfrac{1}{2}AC$

$(Since\,\,AC=AB)$

$\therefore$

$2AC^2=2DC^2+\dfrac{1}{2}AC^2$

$\therefore$

$2AC^2-\dfrac{1}{2}AC^2=2DC^2$

$\therefore$

$3AC^2=4CD^2$

In fig., $PS$ is the bisector of $\angle QPR$ of $\triangle PQR$ . Prove that $\dfrac{QS}{SR}=\dfrac{PQ}{PR}$ .

Given:

$\angle QPS=\angle RPS$

To Prove:

$\dfrac{QS}{SR}=\dfrac{PQ}{PR}$

Construction:
Extend

$RP$ to

$T$ and
Join

$QT$ such that

$TQ\parallel PS$

Proof:
Since,

$QT\parallel PS$

$\therefore$

$\angle TQP=\angle QPS$ (Alternate Angles)
Also,

$\angle QTP=\angle RPS=\angle QPS$ (Corresponding Angles and

$PS$ is the bisector of

$\angle QPR$ of

$\triangle PQR$ )

$\therefore$

$\angle TQP=\angle QTP$

$\therefore\,TP=QP\,.......(1)$

Since,

$QT\parallel PS,$ by basic proportionality theorem,

$\therefore$

$\dfrac{QS}{SR}=\dfrac{TP}{PR}$

$\therefore$

$\dfrac{QS}{SR}=\dfrac{PQ}{PR}$ (From 1)

Hence proved

In Fig., two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:
(i) $\triangle PAC \sim \triangle PDB$
(ii) $PA . PB = PC . PD$

(i) In

$\triangle PAC$ and

$\triangle PDB$ ,

$\angle BAC=180^{\circ}−\angle PAC$ (linear pairs)

$\angle PDB=\angle CDB=180^{\circ}−\angle BAC$

$=180^{\circ}−(180^{\circ}−\angle PAC)=\angle PAC$ )

$\angle PAC=\angle PDB$ ... (1)

$\angle APC=\angle BPD$ [Common] ... (2)

We know that in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
Therefore, for cyclic quadrilateral ABCD.
consider ABCD

$\Delta PAC$ and

$\Delta PBD.$

$\angle PCA = \angle PBD$ .... (3)

$\therefore$ By AAA-criterion of similarity,

$\triangle PAC\sim \triangle DPB$

(ii)

$\triangle PAC\sim \triangle DPB$

So, the corresponding sides of the similar triangles are proportional

$\dfrac{PA}{PD}=\dfrac{PC}{PB}$

$\Rightarrow PA.PB = PC.PD$

In fig., $ABC$ is a triangle in which $\angle ABC > 90^o$ and $AD \bot CB$ produced. Prove that $AC^2 = AB^2 + BC^2 + 2 BC . BD$ .

To Prove:

$AC^2 = AB^2 + BC^2 + 2 BC . BD$ .

Proof:

$\triangle ADC$

By Pythagoras Theorem,

$AC^2=AD^2+DC^2\,.....(1)$

$\triangle ADB$

By Pythagoras Theorem,

$AB^2=AD^2+BD^2\,.......(2)$

Subtracting 1) and 2) we get,

$AC^2-AB^2=DC^2-(DC-BC)^2$

$\therefore$

$AC^2-AB^2=DC^2-DC^2+2DC.BC-BC^2$

$\therefore$

$AC^2-AB^2=2(DB+BC)BC-BC^2$

$\therefore$

$AC^2-AB^2=2DB.BC+2BC^2-BC^2$

$\therefore$

$AC^2=AB^2+2DB.BC+BC^2$

$[hence \, proved]$

In Fig., $D$ is a point on side $BC$ of $\triangle ABC$ such that $\dfrac{BD}{CD}=\dfrac{AB}{AC}.$ Prove that $AD$ is the bisector of $\angle BAC$ .

$D$ is a point on

$BC$ of

$ABC.$

and

$\dfrac{BD}{CD}=\dfrac{AB}{AC}$

Let us construct

$BA$ to

$E$ such that

$AE = AC.$ Join

$CE.$

Now, as

$AE = AC$ ,

$\dfrac{BD}{CD}=\dfrac{AB}{AC}$

$\Rightarrow \dfrac{BD}{CD}=\dfrac{AB}{AE}$

Also,

$\angle AEC=\angle ACE$ (angles opp. to equal sides of a triangle are equal) ....... (i)

By converse of Basic Proportionality Theorem,

$DA\parallel CE$

$\angle DAC=\angle ACE$ ...... (ii) ...[Alternate angles]

$\angle BAD=\angle AEC$ ......... (iii) ...[Corresponding ∠s]

Also,

$\angle AEC=\angle ACE$ ...[From (i)]

and

$\angle BAD=\angle DAC$ ...[From (ii) and (iii)]

So,

$AD$ is the bisector

$\angle BAC$ .

Two triangles are similar but not congruent and the lengths of the sides of the first are $6$ cm, $11$ cm and $12$ cm. The sides of the second also have integral lengths, and one of them is congruent to a side of the first. What is the perimeter of the second triangle?

Let

$x,y$ and

$z$ be integral lengths of second triangle.

Since, triangles are similar,

$\cfrac{x}{6}=\cfrac{y}{11}=\cfrac{z}{12}$

Case (i): If

$x=6$ then

$y=11$ and

$z=12$ , it is not possible because triangles are not congruent.

Case (ii): If

$x=11$ then

$y=\cfrac{121}{6}=20.1$ and

$z=22$ , since

$y$ is not integral

$x$ cannot be equal to

$11.$

Case (iii): If

$x=12$ then

$y=22$ and

$z=24$

Hence, the sides of second triangle are

$12$ cm,

$22$ cm and

$24$ cm.

Thus perimeter of the second triangle

$=12+22+24=58$ cm.

In $\triangle$ ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 8 cm, AB = 12 cm, AE = 12 cm, find CE.

The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.

It is given that

$AD=8$ cm,

$AB=12$ cm and

$AE=12$ cm.

Using the basic proportionality theorem, we have

$\frac { AD }{ AB } =\frac { AE }{ AC } =\frac { DE }{ BC } \\ \Rightarrow \frac { AD }{ AB } =\frac { AE }{ AC } \\ \Rightarrow \frac { 8 }{ 12 } =\frac { 12 }{ AC } \\ \Rightarrow 8AC=12\times 12\\ \Rightarrow 8AC=144\\ \Rightarrow AC=\frac { 144 }{ 8 } =18$

Now,

$CE=AC-AE=18-12=6$ cm

Hence,

$CE=6$ cm.

In $\triangle$ PQR 2PM = 3 PN and 2PQ = 3PR. Prove that MQRN is a trapezium

(i)

$\dfrac{2PM}{2PQ} = \dfrac{3 PN}{3 PR}$ [

$\because$ Data]

$\dfrac{PM}{PQ} = \dfrac{PN}{PR}$

$\therefore MN || QR$ [

$\because$ Converse of corollary 2 of B.P.T.]
(ii) Since

$QP$ and

$RP$ intersects at '

$P$ ',

$MQ$ cannot be parallel to

$RN$

$\Rightarrow MQRN$ is a trapezium
[

$\because$ If only one pair of opposite sides of a quadrilateral is parallel then it is a trapezium]

If AD = $4x-3$ , BD = $3x-1$ , AE = $8x-7$ and CE = $5x-3$ find the value $x$ .

It is given that

$AD=4x-38$ ,

$BD=3x-1$ ,

$AE=8x-7$ and

$CE=5x-3$ . Let

$AC=x$

Using the basic proportionality theorem, we have

$\frac { AD }{ BD } =\frac { AE }{ AC } \\ \Rightarrow \frac { 4x-3 }{ 3x-1 } =\frac { 8x-5 }{ x } \\ \Rightarrow x(4x-3)=(3x-1)(8x-5)\\ \Rightarrow 4x^{ 2 }-3x=3x(8x-5)-1(8x-5)\\ \Rightarrow 4x^{ 2 }-3x=24x^{ 2 }-15x-8x+5\\ \Rightarrow 4x^{ 2 }-3x=24x^{ 2 }-23x+5\\ \Rightarrow 24x^{ 2 }-23x+5-4x^{ 2 }+3x=0\\ \Rightarrow 20x^{ 2 }-20x+5=0\\ \Rightarrow 5(4x^{ 2 }-4x+1)=0\\ \Rightarrow 4x^{ 2 }-4x+1=0\\ \Rightarrow (2x)^{ 2 }-(2\times 2x\times 1)x+1^{ 2 }=0\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (∵(a-b)^{ 2 }=a^{ 2 }+b^{ 2 }-2ab)\\ \Rightarrow (2x-1)^{ 2 }=0\\ \Rightarrow (2x-1)=0\\ \Rightarrow 2x=1\\ \Rightarrow x=\frac { 1 }{ 2 }$

Hence,

$x=\frac { 1 }{ 2 }$ .

Prove that: In a trapezium, the line joining the midpoints of non-parallel sides is
(i) Parallel to the parallel sides and
(ii) Half of the sum of the parallel sides.

Data: In the trapezium

$ABCD$ ,

$AD \;|\;|\; BC$ ,

$AX = XB$ and

$DY = YC$ .

To Prove: (i)

$XY \;|\;|\; AD$ or

$XY || BC$
(ii)

$XY = \dfrac{1}{2} (AD + BC)$

Construction: Extend

$BA$ and

$CD$ to meet at

$Z$ . Join

$A$ and

$C$ . Let in cut

$XY$ at

$P.$

Proof:

$(i)$ In

$\triangle ZBC,$

$AD \;|\;|\; BC$

$\dfrac{ZA}{AB} = \dfrac{ZD}{DC}$

$\dfrac{ZA}{2AX} = \dfrac{ZD}{2DY}$ [

$\because$

$X$ and

$Y$ are mid points of

$AB$ and

$DC$ ]

$\dfrac{ZA}{AX} = \dfrac{ZD}{DY}$

$\Rightarrow XY || AD$ [

$\because$ Converse of B.P.T.]

(ii) In

$\triangle ABC, AX = XB$ [

$\because$ Data]

$XP || BC$ [Already Proved]

$AP = PC$ [

$\because$ Converse of mid point theorem]

$XP = \dfrac{1}{2} BC$ [

$\because$ Midpoint Theorem]

$\triangle ADC,$

$PY = \dfrac{1}{2}AD$ .

By adding, we get,

$XP + PY = \dfrac{1}{2} BC + \dfrac{1}{2} AD$

$\therefore XY = \dfrac{1}{2} (BC + AD)$ [Hence Proved]

In $\triangle$ PQR, E and F are points on the sides PQ and PR respectively. verify EF || QR. PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm.

It is given that

$PE=3.9$ cm,

$EQ=3$ cm,

$PF=3.6$ cm and

$FR=2.4$ cm.

Using the basic proportionality theorem, we have

$\frac { PE }{ EQ } =\frac { 3.9 }{ 3 } =1.3\\ \frac { PF }{ FR } =\frac { 3.6 }{ 2.4 } =1.5$

Since

$\frac { PE }{ EQ } ≠\frac { PF }{ FR }$

Hence,

$EF$ is not parallel to

$QR$ .

In $\triangle$ PQR, E and F are points on the sides PQ and PR respectively. Then verify that EF || QR if PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm.

It is given that

$PQ=1.28$ cm,

$PR=2.56$ cm,

$PE=0.18$ cm and

$PF=0.36$ cm.

We find the following:

$EQ=PQ-PE=1.28-0.18=1.10$ cm

$FR=PR-PF=2.56-0.36=2.20$ cm

Using the basic proportionality theorem, we have

$\frac { PE }{ EQ } =\frac { 0.18 }{ 1.10 } =0.1636\\ \frac { PF }{ FR } =\frac { 0.36 }{ 2.20 } =0.1636$

Since

$\frac { PE }{ EQ } =\frac { PF }{ FR }$

Hence,

$EF$ is parallel to

$QR$ .

At a certain time of the day, a man 6 feet tall, casts his shadow 8 feet long. Find the length of the shadow cast by a building 45 feet high, at the same time which is next to the man.

Let the length of the shadow of the building

$= x$ .
In

$\triangle$ ABC, AB || DE (both are standing vertically on the ground)

$\dfrac{AB}{DE} = \dfrac{BC}{EC}$

$\dfrac{45}{6} = \dfrac{x}{8} \Rightarrow x = \dfrac{45 \times 8}{6} = 60\ ft$

$\therefore$ Length of shadow cast by the building is

$60\ ft$ .

In $\triangle$ PQR, E and F are points on the sides PQ and PR respectively. Verify EF || QR. Given that PE = 4 cm, QE = 4.5 cm, PF = 8 cm and FR = 9 cm.

It is given that

$PE=4\ \text{cm}$ ,

$QE=4.5\ \text{cm}$ ,

$PF=8\ \text{cm}$ and

$FR=9\ \text{cm}$ .

Using the basic proportionality theorem, we have

$\dfrac { PE }{ EQ } =\dfrac { 4 }{ 4.5 } =0.88\\ \dfrac { PF }{ FR } =\dfrac { 8 }{ 9 } =0.88$

Since

$\dfrac { PE }{ EQ } =\dfrac { PF }{ FR }$

Hence,

$EF$ is parallel to

$QR$ .

In $\triangle$ ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 6 cm, DB = 9 cm and AE = 8 cm, find AC.

It is given that

$AD=6$ cm,

$DB=9$ cm and

$AE=8$ cm.

Using the basic proportionality theorem, we have

$\frac { AD }{ AB } =\frac { AE }{ AC } =\frac { DE }{ BC } \\ \Rightarrow \frac { AD }{ AB } =\frac { AE }{ AC } \\ \Rightarrow \frac { 6 }{ 15 } =\frac { 8 }{ AC } \\ \Rightarrow 6AC=15\times 8\\ \Rightarrow 6AC=120\\ \Rightarrow AC=\frac { 120 }{ 6 } =20$

Hence,

$AC=20$ cm.

Study the adjoining figure. Write the ratios in relation to basic proportionality theorem, in terms of $a, b, c$ and $d$ .

Consider two triangles

$PXY$ and

$PQR$

$PQ=PX+XQ=a+b$

$PR=PY+YR=c+d$

$\therefore \cfrac {PX}{PQ}=\cfrac a{a+b}$

$\cfrac {PY}{PR}=\cfrac c{c+d}$

$\therefore \cfrac {PX}{XQ}=\cfrac ab; \cfrac {PY}{YR}=\cfrac cd$

By Basic Proportionality Theorem, if a line is parallel to a triangle which intersects the other sides into two distinct points, then the line divides the sides in proportion.

Here,

$XY||QR$

In the figure, $PR \parallel BC$ and $QR \parallel BD$ . Prove that $PQ || CD$ .

The converse of basic proportionality theorem states that if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

In the given figure,

$PR||BC$ and

$QR||BD$

$△ABC$ ,

$PR||BC$

Using the basic proportionality theorem, we have

$\dfrac { AP }{ PC } =\dfrac { AR }{ RB } \quad \dots(1)$

Now, in

$△ABD$ ,

$QR||BD$

Again using the basic proportionality theorem, we have

$\frac { AQ }{ QD } =\frac { AR }{ RB } \quad \dots(2)$

Comparing equations

$(1)$ and

$(2)$ we get,

$\dfrac { AP }{ PC } =\dfrac { AQ }{ QD }$

Now on applying this relation to

$△ADC$ , then using the converse of basic proportionality theorem we get

$PQ||DC$ .

Hence,

$PQ||CD$ .

In $\triangle ABC, DE \parallel BC$ and $CD \parallel EF$ . Prove that $AD^2 = AF \times AB$ .

We have to prove that

$AD^2=AF \times AB$

Consider

$\triangle ABC, DE\parallel BC$ by basic proportionality theorem.

If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides the sides in proportion.

i.e,

$\cfrac {AD}{DB}=\cfrac {AE}{EC}$ .....(i)

From

$\triangle ADC, EF||DC$

$\therefore,$ By Basic Proportionality Theorem,

$\cfrac {AF}{FD}=\cfrac {AE}{EC}$ .....(ii)

From (i) and (ii),

$\cfrac {AD}{DB}=\cfrac {AF}{FD}$

$\Rightarrow \cfrac {DB}{AD}=\cfrac {FD}{AF}$ [Taking Reciprocal]....(iii)

Adding 1 on both sides of equation (iii),

$\cfrac {DB+AD}{AD}=\cfrac {FD+AF}{AF}$

$\cfrac {AB}{AD}=\cfrac {AD}{AF}$

$\therefore (AD)^2=AF \times AB$

The diagonal BD of parallelogram ABCD intersects AE at 'F'. 'E' is any point on BC.
Prove that DE.EF = FB. FA

$ABCD$ is a parallelogram.

$△ADF$ and

$△EFB$ ,,

$∠ADF = ∠EBF$ (Alternate angles)

$∠DAF = ∠BEF$ (Alternate angles)

$∠AFD = ∠BFE$ (Vertically opposite angles)

Therefore,

$△ADF∼△EFB$ and thus,

$\frac { AF }{ EF } =\frac { DF }{ FB } \\ \Rightarrow AF\times FB=DF\times EF$

Hence

$AF\times FB=DF\times EF$ .

X is any point inside $\triangle$ ABC. XA, XB and XC are joined. 'E' is any point on AX. If EF || AB, FG || BC. Prove that EG || AC.

The converse of basic proportionality theorem states that if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

$△XAB$ ,

$EF||AB$ , therefore,

$\frac { XE }{ EA } =\frac { XF }{ FB } \quad .......(1)$

$△XBC$ ,

$FG||BC$ , therefore,

$\frac { XF }{ FB } =\frac { XG }{ GC } \quad .......(2)$

Comparing equations 1 and 2, we get

$\frac { XE }{ EA } =\frac { XG }{ GC }$

Now in

$△XCA$ , if we use the converse of the basic proportionality theorem then we get that

$EG||AC$ .

Hence,

$EG||AC$ .

A girl of height 90 cm is walking away from the base of a lamp - post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

Let the girl of height be at point

$D$ on the ground from the lamp post after

$4$ seconds. Therefore,

$AD = 1.2$ m/sec

$× 4$ sec

$= 4.8$ m

$= 480$ cm

Suppose the length of the shadow of the girl be

$x$ cm when she at position

$D$ . Therefore,

$BD = x$ cm

$∆BDE$ and

$∆BAC$ ,

$∠BDE = ∠BAC$

$(90°)$

$∠DBE = ∠ABC$ (Common)

Thus,

$∆BDE ∼ ∆BAC$ (AA similarity)

$\dfrac { BE }{ BC } =\dfrac { DE }{ AC } =\dfrac { BD }{ AB }$

( Corresponding sides are proportional )

$\dfrac { 90 }{ 360 } =\dfrac { x }{ 480+x } \\ \Rightarrow \dfrac { 1 }{ 4 } =\dfrac { x }{ 480+x } \\ \Rightarrow 4x=480+x\\ \Rightarrow 4x-x=480\\ \Rightarrow 3x=480\\ \Rightarrow x=\dfrac { 480 }{ 3 } =160$

Hence length of her shadow after

$4$ seconds is

$160$ cm.

In the figure
$\angle QPR = \angle UTS = 90^o$ and PR || TS
Prove that $\triangle PQR \sim \triangle TUS$

$∆PQR$ and

$∆TUS$ , it is given that

$PR||TS$ . Also,

$∠QPR=∠UTS=90^{ 0 }$ (Given)

$∠PRQ = ∠TSU$ (Alternate angles)

$∠PQR = ∠STU$ (Alternate angles)

Therefore,

$△PQR∼△TUS$ (AA criterion)

Hence proved.

In $\triangle$ ABC, $\angle B = \angle C$ , D and E are the points on AB and AC such that BD = CE, prove that DE || BC.

Given is a

$\triangle ABC$ such that

$\angle B=\angle C \ and \ BD=CE$ .

To prove:

$DE\parallel BC$

We know that, sides opposite to equal angles of a triangle are equal.

$AB=AC\quad \quad [\because \ \angle B=\angle C]$

$\therefore \ AD+DB=AE+EC$

Also,

$BD=CE$

$\therefore \ AD=AE$

Hence,

$\dfrac{AD}{DB}=\dfrac{AE}{EC}$

According to the converse of basic proportionality theorem, if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Therefore, by using the converse of the basic proportionality theorem, we get

$DE||BC$ .

$[Hence\ proved]$

In the adjoining figure $\angle ABC = 90^o$ and $\angle AMP = 90^o$
Prove that
(i) $\triangle ABC \sim \triangle AMP$
(ii) $\dfrac{CA}{PA} = \dfrac{BC}{MP}$

In the given figure,

$△ABC$ and

$△AMP$ are right angled triangles.

(i)

$∠ABC = ∠AMP=90^0$ (Given)

$∠CAB = ∠MAP$ (Common angles)

$∠ACB = ∠APM$ (Remaining angles)

Therefore,

$△ABC∼△AMP$

(ii) When the triangles are equiangular then their corresponding sides are proportional, therefore,

$\frac { CA }{ PA } =\frac { BC }{ MP }$

Hence proved.

If the diagonals of a quadrilateral divide each other proportionally, then prove that the quadrilateral is a trapezium.

Let

$ABCD$ be a quadrilateral as shown in the above figure:

Diagonal

$AC$ and

$BD$ divide each other proportionately that is,

$\dfrac { AO }{ OC } =\dfrac { BO }{ OD }$

Therefore, the triangles containing these sides are equiangular that is

$△AOB$ and

$△COD$ and thus,

$∠OAB = ∠OCD$

$∠OBA = ∠ODC$ (Alternate angles)

Therefore,

$DC||AB$

Hence,

$ABCD$ is a trapezium.

A ladder resting against a vertical wall has its foot on the ground at a distance of 6ft. from the wall. A man on the ground climbs two - thirds of the ladder. What will be his distance from the wall now?

Man climb

$\cfrac{2}{3}$ of the ladder.

So, in

$\triangle ADE$ &

$\triangle ABC$

$\angle A=\angle A$ (common)

$\angle ADE=\angle ABC$ (corresponding Angle)

$\angle AED=\angle ACB$ (corresponding Angle)

$\triangle ADE\sim\triangle ABC$

$\Longrightarrow \cfrac{AD}{AB}=\cfrac{DE}{BC}$

$\Longrightarrow \cfrac{3}{5}=\cfrac{DE}{6}\Longrightarrow DE=\cfrac{18}{5}=3.6ft$

1015491_564258_ans_7221339d3a5b402381a302d806f6fe54.png

In the trapezium $ABCD$
$AB \parallel DC, EF \parallel AB, DC = 2AB$ and $\dfrac{BE}{EC} = \dfrac{3}{4}$ . Prove that $7EF = 10AB$ .

Let us consider that

$\cfrac {BE}{EC}=\cfrac {AF}{FD}=\cfrac 34$

From

$\triangle BDC$ ,

$\cfrac {GE}{CD}=\cfrac 37$

$\Rightarrow \cfrac {GE}{2AB}=\cfrac 37$

$\therefore GE=\cfrac {6AB}7$ [Since

$CD=2AB$ ]....(i)

From

$\triangle DAB$

$\Rightarrow \cfrac {DF}{AF}=\cfrac 43$

$\therefore \cfrac {FG}{AB}=\cfrac 47$

$\therefore GF=\cfrac {4AB}7$ ....(ii)

Adding (i) and (ii),

$GE+FG=\cfrac {6AB}7+\cfrac {4AB}7$

$\Rightarrow FE=\cfrac {10AB}7$

$\Rightarrow 7FE=10AB$

$AM = 8x^2, MC = 2x^2$ , then find BM and AB.

It is given that

$∠ABC=90^{ 0 }$ and

$BM⊥AC$ as shown in the above figure:

Now with

$AM=8x^2$ and

$MC=2x^2$ , consider,

$BM^{ 2 }=AM\times MC=8x^{ 2 }\times 2x^{ 2 }=16x^{ 4 }\\ \Rightarrow BM=\sqrt { 16x^{ 4 } } =4x^{ 2 }$

Similarly, with

$AM=8x^2$ and

$MC=2x^2$ , consider,

$AB^{ 2 }=AC\times AM\\ \Rightarrow AB^{ 2 }=(AM+MC)\times AM\\ \Rightarrow AB^{ 2 }=(8x^{ 2 }+2x^{ 2 })\times 8x^{ 2 }\\ \Rightarrow AB^{ 2 }=10x^{ 2 }\times 8x^{ 2 }\\ \Rightarrow AB^{ 2 }=80x^{ 4 }\\ \Rightarrow AB=\sqrt { 80x^{ 4 } } \\ \Rightarrow AB=4x^{ 2 }\sqrt { 5 }$

Hence,

$BM=4x^2$ and

$AB=4x^{ 2 }\sqrt { 5 }$ .

In $\triangle ABC, \angle ABC = 90^o, BD \perp AC$ If $BD = 8$ cm, $AD = 4$ cm, find $CD$

$△ABC$ , it is given that

$∠ABC =90^0$ ,

$BD⊥AC$ ,

$BD=8$ cm and

$AD=4$ cm, therefore,

$DB^{ 2 }=AD\times CD\\ \Rightarrow 8^{ 2 }=4\times CD\\ \Rightarrow 64=4\times CD\\ \Rightarrow CD=\frac { 64 }{ 4 } =16$

Hence

$CD=16$ cm.

If $AB = 5.7$ cm, $BD = 3.8$ cm, $CD = 5.4$ cm, find $BC$

$△DAB$ and

$△DBC$ , it is given that

$∠ABC =90^0$ ,

$BD⊥AC$ ,

$AB=5.7$ cm,

$BD=3.8$ cm and

$CD=5.4$ cm, therefore,

$BD^{ 2 }=AD\times DC\\ \Rightarrow 3.8^{ 2 }=AD\times 5.4\\ \Rightarrow AD=\frac { 3.8\times 3.8 }{ 5.4 } =2.67=2.7$

Now,

$AC=AD+DC=2.4+5.4=8.1$ cm

Thus,

$BC^{ 2 }=AC\times DC\\ \Rightarrow BC^{ 2 }=8.1\times 5.4\\ \Rightarrow BC^{ 2 }=47.74\\ \Rightarrow BC=\sqrt { 47.74 }$

Hence

$BC=\sqrt { 47.74 }$ cm.

In given figure $\triangle$ ABC and $\triangle$ BDC are on the same base BC

Prove that $\dfrac{area(\triangle ABC)}{area(\triangle DBC)} = \dfrac{AO}{DO}$

In the given figure, construct

$AM$ and

$DN$ such that

$△ABC$ and

$△BDC$ are on the same base

$BC$ with

$AM⊥BC$ and

$DN⊥BC$ as shown in the above figure:

We know that the area of a triangle is

$A=\dfrac {1}{2}\times b\times h$ where

$b$ is the base and

$h$ is the height of the triangle.

Here, area of

$△ABC$ is

$A_1=\dfrac {1}{2}\times BC\times AM$ and area of

$△BDC$ is

$A_2=\dfrac {1}{2}\times BC\times DN$

Dividing both the areas, we get

$\dfrac { { A }_{ 1 } }{ { A }_{ 2 } } =\dfrac { \dfrac { 1 }{ 2 } \times BC\times AM }{ \dfrac { 1 }{ 2 } \times BC\times DN } =\dfrac { AM }{ DN } \quad .......(1)$

Now, consider

$△AOM$ and

$△DON$ ,

$∠AMO=∠DNO=90^{ 0 }$ (Construction)

$∠AOM=∠DON$ (Vertically opposite angles)

Therefore,

$△AOM∼△DON$ (A similarity criterion)

Thus,

$\dfrac { AM }{ DN } =\dfrac { AO }{ DO } \quad .......(2)$

Substitute equation

$(2)$ in equation

$(1)$ as follows:

$\dfrac { { A }_{ 1 } }{ { A }_{ 2 } } =\dfrac { AM }{ DN } =\dfrac { AO }{ DO }$

$\Rightarrow \dfrac { { A }rea(△ABC) }{ { A }rea(△BDC) } =\dfrac { AO }{ DO }$

$[hence \quad proved]$

Rhombus $PQRB$ is inscribed in $\triangle ABC$ such that $\angle B$ is one of the its angle, $P$ , $Q$ and $R$ lie on $AB$ , $AC$ and $BC$ respectively. If $AB = 12 \text{ cm}$ and $BC = 6 \text{ cm}$ find the sides of rhombus $PQRB$ .

$\triangle ABC, PQ || BC$ ......... (

$\because$ In a rhombus, opposite sides are parallel)

$\therefore \dfrac{AP}{AB} = \dfrac{PQ}{BC}$ ......... (Using Corollary of Thales theorem)

$\implies \dfrac{12 - x}{12} = \dfrac{x}{6}$

$\implies 12 x = 72 - 6x$

$\implies 18 x = 72$

$\therefore x = 4\ cm$
The sides of the rhombus is

$4$ cm each.

In △ABC,∠ABC=90o,BD⊥AC. If $$AB=75

$△ABC$ and

$△BDA$ , it is given that

$∠ABC =90^0$ ,

$BD⊥AC$ ,

$AB=75$ cm,

$AC=1.25$ m or

$AC=125$ cm and

$BC=1$ m or

$BC=100$ cm. ( as we know that

$1$ m

$=100$ cm)

Therefore,

$AB^{ 2 }=AC\times AD\\ \Rightarrow 75^{ 2 }=125\times AD\\ \Rightarrow AD=\frac { 75\times 75 }{ 125 } =45$

Also,

$△ABC$ and

$△CDB$ are similar right angled triangle, thus we have,

$BC^{ 2 }=AC\times DC\\ \Rightarrow 100^{ 2 }=125\times DC\\ \Rightarrow DC=\frac { 100\times 100 }{ 125 } =80$

Finally,

$△ABD$ and

$△CBD$ are similar right angled triangle, therefore,

$BD^{ 2 }=AD\times DC=45\times 80=3600\\ \Rightarrow BD=\sqrt { 3600 } =60$

Hence

$BD=60$ cm.

In $\triangle PQR, \angle PQR = 90^o, QD \perp PR$ .
If $PD = 4DR$ . Prove that $PQ = 2QR$ .

It is given that

$∠PQR=90^{ 0 }$ ,

$QS⊥PR$ ,

$PD=4DR$

Consider,

$PQ^{ 2 }=PR\times PD\\ PR=\frac { PQ^{ 2 } }{ PD } \quad .......(1)$

Similarly, we have,

$QR^{ 2 }=PR\times RD\\ PR=\frac { QR^{ 2 } }{ RD } \quad .......(2)$

Comparing equations 1 and 2 as follows:

$\frac { PQ^{ 2 } }{ PD } =\frac { QR^{ 2 } }{ RD } \\ \Rightarrow \frac { PQ^{ 2 } }{ 4DR } =\frac { QR^{ 2 } }{ RD } \\ \Rightarrow \frac { PQ^{ 2 } }{ 4 } =QR^{ 2 }\\ \Rightarrow PQ^{ 2 }=4QR^{ 2 }\\ \Rightarrow \sqrt { PQ^{ 2 } } =\sqrt { 4QR^{ 2 } } \\ \Rightarrow PQ=2QR$

Hence,

$PQ=2QR$ .

In the adjacent figure $\triangle ABC \sim \triangle PQR$ , and $\angle C = 53^{\circ}$ . Find the side $PR$ and $\angle P$

$\triangle ABC \sim \triangle PQR$
When two triangles are similar their corresponding angles are equal and corresponding angles are equal and corresponding sides are in proportion.

$\dfrac {PR}{AC} = \dfrac {PQ}{AB}\Rightarrow \dfrac {PR}{5} = \dfrac {2}{4}$

$PR = \dfrac {2}{4} \times 5 = 2.5$
Again

$\angle R = \angle C = 53^{\circ}$
Sum of all three angles in a triangle as

$180^{\circ}$
i.e.

$\angle P + \angle Q + \angle R = 180^{\circ}$

$\angle P = 90^{\circ} + 53^{\circ} = 180^{\circ}$

$\angle P = 180^{\circ} - 143^{\circ} = 37^{\circ}$

In the above figure find the ratio between the $\triangle$ AOB and $\triangle$ COD, if $AB = 3CD$ .

Let

$△AOB∼△COD$ .

We know that the arc of similar triangles are proportional to squares of their corresponding altitude, therefore with

$AB=3CD$ , we have,

$\frac { { A }r(△AOB) }{ { A }r(△COD) } =\frac { AB^{ 2 } }{ CD^{ 2 } } =\frac { (3CD)^{ 2 } }{ CD^{ 2 } } =\frac { 9CD^{ 2 } }{ CD^{ 2 } } =\frac { 9 }{ 1 }$

Hence, the ratio between the

$△AOB$ and

$△COD$ is

$9:1$ .

$ABC$ is a right angled triangle with $\angle ABC={ 90 }^{ o }$ . $D$ is any point on $AB$ and $DE$ is perpendicular to $AC$ . Prove that :
(i) $\triangle ADE\sim \triangle ACB$ .
(ii) If $AC=13 cm$ , $BC=5 cm$ and $AE=4 cm$ . Find $DE$ and $AD$ .
(iii) Find, area of $\triangle ADE :$ area of quadrilateral $BCED$ .

To Prove :

$\triangle ADE \sim \triangle ACB$
Proof :
(i) In

$\triangle ADE$ and

$\triangle ACB$
(1)

$\angle A=\angle A$ [common]
(2)

$\angle AED=\angle ABC={ 90 }^{ o }$ (given)

$\therefore$

$\triangle ADE\sim \triangle ACB$ [AA axiom]
(ii)

${ \left( AC \right) }^{ 2 }={ \left( AB \right) }^{ 2 }+{ \left( BC \right) }^{ 2 }$

$169={ \left( AB \right) }^{ 2 }+25$

$AB=12 cm$

$\because$

$\triangle ADE \sim \triangle ACB$

$\therefore$

$\dfrac { DE }{ BC } =\dfrac { AD }{ AC } =\dfrac { AE }{ AB }$

$\therefore$

$\dfrac { DE }{ BC } =\dfrac { AE }{ AB }$

$\dfrac { DE }{ 5 } =\dfrac { 4 }{ 12 }$

$DE=\dfrac { 20 }{ 12 } =\dfrac { 5 }{ 3 } =1\dfrac { 2 }{ 3 } cm$
Now,

$\dfrac { AD }{ AC } =\dfrac { AE }{ AB }$

$\dfrac { AD }{ 13 } =\dfrac { 4 }{ 12 }$

$AD=\dfrac { 13\times 4 }{ 12 } =\dfrac { 13 }{ 3 } =4\dfrac { 1 }{ 3 } cm$ .
(iii)

$\dfrac { Ar.\quad of\left( \triangle ABC \right) }{ Ar.\quad of\left( \triangle ADE \right) } =\dfrac { { AB }^{ 2 } }{ { AE }^{ 2 } } =\dfrac { 144 }{ 16 } =\dfrac { 9 }{ 1 }$

$\dfrac { Ar.\quad of\left( \triangle ADE \right) +Ar.\quad of\left( BCED \right) }{ Ar.\quad of\left( \triangle ADE \right) } =9$

$1+\dfrac { Ar.\quad of\left( BCED \right) }{ Ar.\quad of\left( \triangle ADE \right) } =9$

$\dfrac { Ar.\quad of\left( \triangle ADE \right) }{ Ar.\quad of\left( BCED \right) } =\dfrac { 1 }{ 8 }$

In the given figure, $AB$ and $DE$ are perpendicular to $BC$ .
(i) Prove that $\triangle ABC \sim \triangle DEC$
(ii) If $AB = 6\ cm: DE = 4\ cm$ and $AC = 15\ cm$ , calculate $CD$ .
(iii) Find the ratio of the area of $\triangle ABC$ : area of $\triangle DEC$ .

(i) From

$\Delta ABC$ and

$\Delta DEC$ ,

$\angle ABC=\angle DEC =90^\circ$ (Given)
and

$\angle ACB =\angle DCE =$ Common

$\therefore$

$\Delta ABC \sim \Delta DEC$ (By

$A-A$ similarity)
(ii) In

$\Delta$ ABC and

$\Delta$ DEC,

$\Delta ABC \sim \Delta DEC$ (proved in (i) part)

$\therefore \displaystyle\frac{AB}{DE}=\frac{AC}{CD}$
Given:

$AB=6\ cm, DE=4\ cm, AC=15\ cm$ ,

$\therefore \displaystyle\frac{6}{4}=\frac{15}{CD}$

$\Rightarrow 6\times CD=15\times 4$

$\Rightarrow CD=\displaystyle\frac{60}{6}$

$\Rightarrow CD=10\ cm$ .
(iii)

$\displaystyle\frac{Area\ of\ \Delta ABC}{Area\ of\ \Delta DEC}=\frac{AB^2}{DE^2}$

$(\because \Delta ABC \sim \Delta DEC)$

$=\displaystyle\frac{(6)^2}{(4)^2}=\frac{36}{16}=\frac{9}{4}$

$\therefore$ Area of

$\Delta ABC :$ Area of

$\Delta DEC =9:4$ .

In the given figure drawn, seg $BE \perp$ set $AB$ and seg $BA \perp$ seg $AD$ .
If $BE=6$ and $AD=9$ , find $\dfrac { A\left( \Delta ABE \right) }{ A\left( \Delta BAD \right) }$

Given,

$\Delta ABE$ and

$\Delta BAD$ have the same base and

$BE=6$ and

$AD=9$
Now,

$\dfrac { A\left( \Delta ABE \right) }{ A\left( \Delta BAD \right) } =\dfrac { BE }{ AD }$ (

$\because$ Triangle having equal base)

$\therefore$

$\dfrac { A\left( \Delta ABE \right) }{ A\left( \Delta BAD \right) } =\dfrac { 6 }{ 9 }$

$\therefore$

$\dfrac { A\left( \Delta ABE \right) }{ A\left( \Delta BAD \right) } =\dfrac { 2 }{ 3 }$ .

Draw two congruent figures. Are they similar? Explain

The congruent figures are similar as they are same in shape when compared.

Congruent figures are identical in size, shape and measure. The corresponding sizes between two triangles are sides in the same relative position. Two figures are similar if they have the same shape, but not necessarily the same size.

In the trapezium $ABCD,\; AB \parallel DC$ and $\triangle AED \sim \triangle BEC$ . Prove that $AD = BC.$

To Prove:

$AD = BC$

Proof: Compare

$\triangle EDC$ and

$\triangle EBA$

$\angle EDC = \angle EBA$ [Alternate Angles]

$\angle ECD = \angle EAB$ [Alternate Angles]

$AA$ criterion of similarity,

$\triangle EDC \sim \triangle EBA$

$\Rightarrow \dfrac{ED}{EB} = \dfrac{EC}{EA}$ [By CPST]

$\Rightarrow \dfrac{ED}{EC} = \dfrac{EB}{EA}\quad\quad\quad\dots(i)$

But,

$\triangle AED \sim \triangle BEC$ [Given]

$\Rightarrow \dfrac{ED}{EC} = \dfrac{EA}{EB} = \dfrac{AD}{BC}$

$\dots(ii)$ [By CPST]

$\dfrac{EB}{EA} = \dfrac{EA}{EB}$ [From

$(i)$ and

$(ii)$ ]

$\Rightarrow EA^2 = EB^2$

$\Rightarrow EA = EB$

Hence, from equation

$(ii),$

$\dfrac{AD}{BC} = 1$

$\Rightarrow AD = BC$

Hence proved.

In $\triangle ABC, \angle ABC = \angle DAC, AB = 8\ cm, AC = 4\ cm, AD = 5\ cm$ . Prove that $\triangle ACD$ is similar to $\triangle BCA$

Given that, in

$\Delta ABC,\ \angle ABC=\angle DAC,\ AB=8\ cm,\ AC=4\ cm\ and\ AD=5\ cm$ .

To prove:

$\Delta ACD \sim \Delta BCA$

Proof:

$\Delta ACD$ and

$\Delta BCA$ ,

$\angle ACD=\angle BCA\quad [\text{Common angle}]$

$\angle ABC=\angle DAC\quad [\text{Given}]$

Hence, by

$AA$ similarity criterion, we can say that,

$\Delta ACD \sim \Delta BCA \quad \quad [Hence\ proved]$

$\Delta DEF \sim \Delta MNK$ . If $DE=2, MN=5$ , then find the value of $\cfrac{Area(\Delta DEF)}{Area(MNK)}$

If we know that

$\Delta DEF \sim \Delta MNK$

Then

$\cfrac{DE}{MN} = \cfrac{EF}{NK} = \cfrac{FD}{KM} =$

$\sqrt{\dfrac{Area(\Delta DEF)}{Area(\Delta MNK)}}$

It is given that

$DE = 2, MN = 5$

Therefore,

$\cfrac{DE}{MN} =\sqrt{\dfrac{Area(\Delta DEF)}{Area(\Delta MNK)}} = \dfrac25$

$\Rightarrow \dfrac{Area(\Delta DEF)}{Area(\Delta MNK)} = \bigg(\dfrac25\bigg)^2 = \dfrac4{25}$

In $\triangle ABC$ , $AL, BM$ and $CN$ are the altitudes which intersect at 'O'.
Prove that (i) $\triangle AMB \sim \triangle ANC$ (ii) $\dfrac{AN}{BN}. \dfrac{BL}{CL}. \dfrac{CM}{AM} = 1$

$\triangle AMB$ and

$\triangle ANC$

$\angle AMB = \angle ANC = 90^o$ .......... (

$\because$ Data)

$\angle BAM = \angle NAC$

$\angle ABM = \angle ACN$

$\therefore \triangle AMB \sim \triangle ANC$ ......... [By AAA rule]

Also,

$\triangle BLA \sim \triangle BNC$
and

$\triangle BMC \sim \triangle ALC$
Now,

$\triangle AMB \sim \triangle ANC$

$\therefore \dfrac{AN}{AM} = \dfrac{AC}{AB}$

$\triangle BLA \sim \triangle BNC$

$\therefore \dfrac{BL}{BN} = \dfrac{AB}{BC}$

$\triangle BMC \sim \triangle ALC$

$\therefore \dfrac{CM}{CL} = \dfrac{BC}{AC}$

$\dfrac{AN}{AM} \times \dfrac{BL}{BN} \times \dfrac{CM}{CL} = \dfrac{AC}{AB} \times \dfrac{AB}{BC} \times \dfrac{BC}{AC}$

$\therefore \dfrac{AN}{BN} \times \dfrac{BL}{LC} \times \dfrac{CM}{AM} = 1$

$AN \times BL \times CM = BN \times LC \times AM$

Take two similar shapes. If you slide rotate or flip one of them, does the similarity remain.

The similarity will still remain even when the figure is rotated or flipped. The position of the proportional lengths might change but they will still be similar. For Example:

When a parallelogram is flipped, its similarity will still remain as the sides are still proportional even though the position is changed.

When a circle is flipped, similarity is retained as circle is similar from all point of views.

In the figure D, E are points on the sides AB and AC respectively of $\triangle$ ABC such that $area(\triangle DBC) = area (\triangle EBC)$ . Prove that DE || BC.

Given D and E are point on sides AB and AC respectively of triangle ABC such that

$Area(\Delta DBC)$

$=$

$Area(\Delta EBC)$ .

Given that

$Area(\Delta DBC)$

$=$

$Area(\Delta EBC)$ .

Then the height of triangle DBC and Triangle EBC are the same because both triangles are the same

Then it possible that both triangle on the same base and between parallel line BC and DE

Therefore, they will lie between the same parallel lines.

$\therefore BC \parallel DE$

$[hence \, proved]$

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then prove that the two sides are divided in the same ratio.

Given:

$DE || BC$

To prove that:

$\cfrac{EC}{AE} = \cfrac{BD}{AD}$

Proof:

$\angle AED = \angle ACB$ Corresponding angles

$\angle ADE = \angle ABC$ Corresponding angles

$\angle EAD$ is common to both the triangles

$\Rightarrow \Delta AED \sim \Delta ACB$ by $AAA$ similarity

$\Rightarrow \cfrac{AC}{AE} = \cfrac{AB}{AD}$

$\Rightarrow \cfrac{AE+EC}{AE} = \cfrac{AD + BD}{AD}$

$\Rightarrow \cfrac{EC}{AE} = \cfrac{BD}{AD}$

Hence proved

Prove that, if a line parallel to a side of a triangle intersect the other sides in two distinct points, then the line divides those sides in proportion.

Given: In a

$\triangle PQR$ , line

$l\parallel$ side

$QR$ , line

$l$ intersect the sides

$PQ$ and

$PR$ in two distinct points

$M$ and

$N$ respectively.

To prove:

$\dfrac {PM}{MQ} = \dfrac {PN}{NR}$ ... (i)

Construction:

$seg QN$ and

$seg RM$ are drawn.

Proof:

$\dfrac {A(\triangle PMN)}{A(\triangle QMN)} = \dfrac {PM}{MQ}$
(Both triangles have equal height with common vertex

$M$ )

$\therefore \dfrac {A(\triangle PMN)}{A(\triangle RMN)} = \dfrac {PN}{NR}$ ... (ii)

But

$A(\triangle QMN) = A(\triangle RMN)$ , because they are between parallel lines

$MN$ and

$QR$ and have equal height corresponding to their common base

$MN$ ..... (iii)

From (i), (ii) and (iii), we get

$\dfrac {A(\triangle PMN)}{A(\triangle QMN)} = \dfrac {A(\triangle PMN)}{A(\triangle RMN)}$

$\therefore \dfrac {PM}{MQ} = \dfrac {PN}{NR}$

$[hence \, proved]$

In the given figure, line $PY \parallel$ side $BC$ , $AP=4$ , $PB=8$ , $AY=5$ and $YC=x$ . Find $x$ .

$\Delta ABC$ , line

$PY \parallel$ side

$BC$ .
By the theorem of B.P.T.,
Thus

$\dfrac { AY }{ YC } =\cfrac { AP }{ PB }$

$\Rightarrow \cfrac { 5 }{ x } =\dfrac { 4 }{ 8 }$

$\Rightarrow x\times 4=5\times 8$

$\Rightarrow x=\cfrac { 5\times 8 }{ 4 }$

$\Rightarrow x=5\times 2$
Therefore,

$x=10$ units

$\Delta ABC \sim \Delta PQR$ and $A(\Delta ABC) = 144 cm^2$ and $A(\Delta PQR) = 100 cm^2$ . If BC = 12 cm, then find the value of QR.

Given,

$\Delta ABC \sim \Delta PQR, A(\Delta ABC) = 144 cm^2, A(\Delta PQR) = 100 cm^2$ and

$BC = 12 cm$ .
We know that,

$\dfrac{A(\Delta ABC)}{A(\Delta PQR)} = \dfrac{BC^2}{QR^2}$ ....... (i)
Now,

$\dfrac{A(\Delta ABC)}{A(\Delta PQR)} = \dfrac{144 cm^2}{100 cm^2}$
From (i),

$\left ( \dfrac{BC}{QR} \right )^2 = \dfrac{144}{100}$

$\therefore \dfrac{12}{QR} = \dfrac{12}{10}$

$\therefore QR = 10 cm$ .

Prove that "If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side".

Given : The line

$l$ intersects the sides

$PQ$ and side

$PR$ of

$\Delta PQR$ in the points

$M$ and

$N$ respectively such that

$\dfrac{PM}{MQ} = \dfrac{PN}{NR}$ and

$P-M-Q$ ,

$P-N-R$ .
To Prove : Line

$l$

$\parallel$ Side

$QR$
Proof : Let us consider that line

$l$ is not parallel to the side

$QR$ . Then there must be another line passing through

$M$ which is parallel to the side

$QR$ .
Let line

$MK$ be that line.
Line

$MK$ intersects the side

$PR$ at

$K$ ,

$\left(P-K-R\right)$
In

$\Delta PQR$ , line

$MK \parallel$ side

$QR$

$\therefore$

$\dfrac{PM}{MQ} = \dfrac{PK}{KR}$ ....(1) (B.P.T.)
But

$\dfrac{PM}{MQ}=\dfrac{PN}{NR}$ ....(2) (Given)

$\therefore$

$\dfrac{PK}{KR}=\dfrac{PN}{NR}$ [From (1) and (2)]

$\therefore$

$\dfrac{PK+KR}{KR} = \dfrac{PN+NR}{NR}$ (

$P-K-R$ and

$P-N-R$ )

$\therefore$ the points

$K$ and

$N$ are not different.

$\therefore$ line

$MK$ and line

$MN$ coincide

$\therefore$ line

$MN \parallel$ Side

$QR$
Hence, the converse of B.P.T. is proved.

In $\triangle RST$ , line $PQ\parallel seg ST$ , $R-P-S$ and $R-Q-T$ . If $RP=4,PS=8,RQ=3$ , then find $QT$ .

$\triangle RST$ , line

$PQ\parallel ST$

$\cfrac { RP }{ PS } =\cfrac { RQ }{ QT }$ ....(i) [By B.P.T]
Now,

$RP=4,PS=8,RQ=3$

$\therefore \cfrac { 4 }{ 8 } =\cfrac { 3 }{ QT }$ ....[from (i)]

$\therefore 4\times QT = 8\times 3$

$\therefore QT=\cfrac{8\times 3}{4}$

$\therefore QT=6$

In the following figure drawn, line $PY$ $\parallel$ side $BC$ , $AP=6$ , $PB=12$ , $AY=5$ , $YC=x$ , then find $x$ .

$\Delta ABC$ , line

$l$

$\parallel$ side

$BC$ ,
By the Theorem of Basic Proportionality,

$\dfrac { AY }{ YC } =\dfrac { AP }{ PB }$

$\Rightarrow \dfrac { 5 }{ x }= \dfrac { 6 }{ 12 }$

$\Rightarrow x\times 6=5\times 12$

$\Rightarrow x=\dfrac { 5\times 12 }{ 6 }$

$\Rightarrow x=5\times 2$
Therefore,

$x=10$ units.

If $\triangle LMN\sim \triangle RST$ and $Ar\left( \triangle LMN \right) =100\ sq.cm,\ Ar\left( \triangle RST \right) =144\ sq.cm$ and $LM=5\ cm,$ then find $RS.$

Given,

$\triangle LMN\sim \triangle RST$ ,

$Ar\left( \triangle LMN \right) =100\ sq.cm$ ,

$Ar\left( \triangle RST \right) =144\ sq.cm$ , and

$LM=5\ cm$

By the theorem of areas of similar triangles,

$\dfrac { Ar\left( \triangle LMN \right) }{ Ar\left( \triangle RST \right) } =\dfrac { { LM }^{ 2 } }{ { RS }^{ 2 } }$

$\Rightarrow \dfrac { 100 }{ 144 } =\dfrac { 5^2 }{ { RS }^{ 2 } }$

$\Rightarrow { RS }^{ 2 }=25\times \dfrac { 144 }{ 100 } =36$

$\Rightarrow RS=6\ cm$

In the given figure ABC is a triangle. If $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ , then prove that DE || BC.

Given:

$\cfrac{AD}{AB} = \cfrac{AE}{AC}$

To prove that:

$DE || BC$

Proof:

$\cfrac{AD}{AB} = \cfrac{AE}{AC}$

$\angle A$ is common angle in

$\Delta ADE$ and

$\Delta ABC$ .

$\implies \Delta ADE \sim \Delta ABC$ by

$SAS$ similarity

$\implies \angle ADE = \angle ABC$ by properties of similar triangles

$\therefore DE || BC$ by property of corresponding angles for parallel lines.

$[hence \, proved]$

Prove that in a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.

${\textbf{Step - 1: Apply pythagoras theorem}}$

${\text{Given: - ABC is a triangle}}$

${\text{A}}{{\text{C}}^2} = {\text{A}}{{\text{B}}^2} + {\text{B}}{{\text{C}}^2}$

${\text{In }}\Delta{\text{PQR}}$

${\text{P}}{{\text{R}}^2} = {\text{P}}{{\text{Q}}^2} + {\text{Q}}{{\text{R}}^2}\;\;{\text{(by pythagoras theorem)}}$

$\Rightarrow {\text{P}}{{\text{R}}^2} = {\text{A}}{{\text{B}}^2} + {\text{B}}{{\text{C}}^2}\;......(1)\;(\because AB = PQ\;{\text{and QR = BC}})$

${\text{given}}$

${\text{A}}{{\text{C}}^{\text{2}}}{\text{ = A}}{{\text{B}}^{\text{2}}}{\text{ + B}}{{\text{C}}^2}\;\;.....(2)$

${\text{From equation 1 and 2 we have}}$

${\text{A}}{{\text{C}}^2} = P{R^2}$

$\Rightarrow {\text{AC = PR}}\;\;....{\text{(3)}}$

${\textbf{Step - 2: Find congruency }}$

${\text{Now}} {\text{in }}{{\Delta }}{\text{ABC and } \Delta PQR}$

${\text{i)AB = PQ}}$

${\text{ii)BC = QR}}$

${\text{iii)AC = PR}}\;{\text{(from}}\;{\text{3)}}$

$\therefore {\Delta \text{ ABC}} \cong \Delta{\text{PQR(by sss congruency)}}$

${\text{Therefore, by CPCT}}$

$\angle {\text{B = }}\angle {\text{Q}}$

$\because \angle {\text{Q}} = {90^ \circ }$

$\therefore \angle {\text{B}} = {90^ \circ }$

${\textbf{Hence,Proved}}$

2167959_604031_ans_9718e430b69d4a6ba21a8e85fa369d3b.png

The points $D$ and $E$ are on the sides $AB$ and $AC$ of $\triangle ABC$ respectively, such that $DE\parallel BC$ . If $AB = 3\ AD$ and the area of $\triangle ABC$ is $72\ \text{cm}^{2}$ , then find the area of the quadrilateral $DBCE$

1307342_622162_ans_5297d32bf2404f8faf43975c117f9931.jpeg

State and prove the Pythagoras theorem.

Statement: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given: ABC is a triangle in which

$\angle ABC={90}^{\circ}$

Construction: Draw

$BD\bot AC$ .

Proof:

$\triangle ADB$ and

$\triangle ABC$

$\angle A=\angle A$ [Common angle]

$\angle ADB=\angle ABC$ [Each

${90}^{\circ}$ ]

$\triangle ADB \sim \triangle ABC$ [

$A-A$ Criteria]

So,

$\dfrac {AD}{AB}=\dfrac{AB}{AC}$

Now,

${AB}^{2}=AD\times AC$ ..........(1)

Similarly,

${BC}^{2}=CD\times AC$ ..........(2)

Adding equations (1) and (2) we get,

${AB}^{2}+{BC}^{2}=AD\times AC+CD\times AC$

$=AC(AD+CD)$

$=AC\times AC$

$\therefore{AB}^{2}+{BC}^{2}={AC}^{2}$

$[hence \, proved]$

State and prove Basic Proportionality theorem.

Basic Proportionality Theorem states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points,then the line divides those sides of the triangle in proportion.

Let

$ABC$ be the triangle.

The line

$l$ parallel to

$BC$ intersect

$AB$ at

$D$ and

$AC$ at

$E$ .

To prove

$\dfrac{AD}{DB}=\dfrac{AE}{EC}$

Join

$BE,CD$

Draw

$EF \perp AB$ ,

$DG \perp CA$

Since

$EF \perp AB$ ,

$EF$ is the height of triangles

$ADE$ and

$DBE$

Area of

$\triangle ADE = \dfrac{1}{2} \times$ base

$\times$ height

$=\dfrac{1}{2}AD \times EF$

Area of

$\triangle DBE = \dfrac{1}{2} \times DB \times EF$

$\dfrac { area\quad of\quad \Delta ADE }{ area\quad of\quad \Delta DBE } =\dfrac { \dfrac { 1 }{ 2 } \times AD\times EF }{ \dfrac { 1 }{ 2 } \times DB\times EF } =\dfrac { AD }{ DB }$ ........(1)
Similarly,

$\dfrac { area\quad of\quad \Delta ADE }{ area\quad of\quad \Delta DCE } =\dfrac { \dfrac { 1 }{ 2 } \times AE\times DG }{ \dfrac { 1 }{ 2 } \times EC\times DG } =\dfrac { AE }{ EC }$ ......(2)

But

$\Delta DBE$ and

$\Delta DCE$ are the same base

$DE$ and between the same parallel straight line

$BC$ and

$DE$ .

Area of

$\Delta DBE=$ area of

$\Delta DCE$ ....(3)

From (1), (2) and (3), we have

$\dfrac{AD}{DB}=\dfrac{AE}{EC}$

Hence proved.

The points $D, E$ and $F$ are taken on the sides $AB, BC$ and $CA$ of a $\triangle ABC$ respectively, such that $DE\parallel AC$ and $FE \parallel AB$ .
Prove that $\dfrac {AB}{AD} = \dfrac {AC}{FC}$

Given that in

$\triangle ABC, DE\parallel AC$ .

$\dfrac {BD}{DA} = \dfrac {BE}{EC}$ (Thales theorem) ....

$(1)$
Also, given that

$FE\parallel AB$ .

$\therefore \dfrac {BE}{EC} = \dfrac {AF}{FC}$ (Thales theorem) ....

$(2)$
From

$(1)$ and

$(2)$ , we get

$\dfrac {BD}{AD} = \dfrac {AF}{FC}$

$\Rightarrow \dfrac {BD + AD}{AD} = \dfrac {AF + FC}{FC}$ (componendo rule)
Thus,

$\dfrac {AB}{AD} = \dfrac {AC}{FC}$ .

A man of height $1.8\ m$ is standing near a Pyramid. If the shadow of the man is of length $2.7\ m$ and the shadow of the Pyramid is $210\ m$ long at that instant, find the height of the Pyramid

Let

$AB$ and

$DE$ be the heights of the Pyramid and the man respectively.
Let

$BC$ and

$EF$ be the lengths of the shadows of the Pyramid and the man respectively.
In

$\triangle ABC$ and

$\triangle DEF$ , we have

$\angle ABC = \angle DEF = 90^{\circ}$

$\angle BCA = \angle EFD$ (angular elevation is same at the same instant)

$\therefore \triangle ABC\sim \triangle DEF$ (

$AA$ similarity criterion)
Thus,

$\dfrac {AB}{DE} = \dfrac {BC}{EF}$

$\Rightarrow \dfrac {AB}{1.8} = \dfrac {210}{2.7} \Rightarrow = \dfrac {210}{2.7}\times 1.8 = 140$ .
Hence, the height of the Pyramid is

$140\ m$ .
1034604_622314_ans_ec858ab52d4e42d582ccb6a0c911dfc6.jpg

Prove that the areas of two similar acute triangles are proportional to the squares of the corresponding sides.

Statement: Ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: Two triangles

$ABC$ and

$PQR$ such that

$\triangle ABC\sim \triangle PQR$

To prove:

$\cfrac { area\quad (ABC) }{ area\quad (PQR) } ={ \left( \cfrac { { AB }^{ 2 } }{ PQ } \right) }^{ }={ \left( \cfrac { { BC }^{ } }{ PQ } \right) }^{ 2 }={ \left( \cfrac { { CA }^{ } }{ RP } \right) }^{ 2 }\quad$

Proof: For finding the areas of the two triangles,we draw altitudes

$AM$ and

$PN$ of the triangles.

Now,

$area\quad (ABC)=\cfrac { 1 }{ 2 } BC\times AM$

And

$area\quad (PQR)=\cfrac { 1 }{ 2 } QR\times PN$

So,

$\cfrac { area\quad (ABC) }{ area\quad (PQR) } =\cfrac { \cfrac { 1 }{ 2 } \times BC\times AM }{ \cfrac { 1 }{ 2 } \times QR\times PN } =\cfrac { BC\times AM }{ QR\times PN } .....(1)\quad$

Now, in

$\triangle ABC$ and

$\triangle PQN$

$\angle B=\angle Q$ (As

$\triangle ABC\sim \triangle PQR$ )

and

$\angle M=\angle N$ (Each

$={90}^{o}$ )

so,

$\triangle ABM\sim \triangle PQN$ (AA similarity criterion)

Therefore

$\cfrac { AM }{ PN } =\cfrac { AB }{ PQ } .....(2)$

Also,

$\triangle ABC\sim \triangle PQR$

$\cfrac { AB }{ PQ } =\cfrac { BC }{ QR } =\cfrac { CA }{ RP } .......(3)$

$\cfrac { area\quad (ABC) }{ area\quad (PQR) } =\cfrac { AB }{ PQ } \times \cfrac { AM }{ PN }$ (from (1) and (3))

Therefore

$=\cfrac { AB }{ PQ } \times \cfrac { AB }{ PQ } ={ \left( \cfrac { AB }{ PQ } \right) }^{ 2 }$ (from (2))

Now using (3) we get

The perimeter of two similar triangles are $30\ cm$ and $20\ cm$ respectively. If one side of the first triangle is $12\ cm$ , determine the corresponding side of the second triangle.

Let the sides of one triangle be

$a ,b ,c$

Let the sides of another triangle be

$A, B,C$

as the triangle are similiar means that

$\dfrac { a }{ A } =\dfrac { b }{ B } =\dfrac { c }{ C } =K$

here

$k = constant$

$a = KA$

$b= KB$

$c = KC$

Ratio of perimeters is

$\dfrac { a+b+c }{ A+B+C } =\dfrac { 30 }{ 20 }$

$\dfrac { KA+KB+KC }{ A+B+C } =\dfrac { 30 }{ 20 }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1.5$

$\Rightarrow K=1.5$

According to the question

$12 = 1.5A$

$A=8$ cm

In a $\triangle ABC, D$ and $E$ are points on $AB$ and $AC$ respectively such that $\dfrac {AD}{DB} = \dfrac {AE}{EC}$ and $\angle ADE = \angle DEA$ . Prove that $\triangle ABC$ is isosceles

Since

$\dfrac {AD}{DB} = \dfrac {AE}{EC}$ , by converse of Thales theorem,

$DE\parallel BC$

$\therefore \angle ADE = \angle ABC and ... (1)$ and

$\angle DEA = \angle BCA .... (2)$
But, given that

$\angle ADE = \angle DEA ... (3)$
From

$(1), (2)$ and

$(3)$ , we get

$\angle ABC = \angle BCA$

$\therefore AC = AB$ (If opposite angles are equal, then opposite sides are equal).
Thus,

$\triangle ABC$ is isosceles.

In $\triangle PQR, AB\parallel QB$ . If $AB$ is $3\ cm, PB$ is $2\ cm$ and $PR$ is $6\, cm$ , then find the length $QR$

Given

$AB$ is

$3\ cm, PR$ is

$2\ cm\ PR$ is

$6\ cm$
In

$\triangle PAB$ and

$\triangle PQR$

$\angle PAB = \angle PQR$ (corresponding angles)
and

$\angle P$ is common.

$\therefore \triangle PAB \sim \triangle PQR$ (

$AA$ similarity criterion)
Since corresponding sides are proportional,

$\dfrac {AB}{QR} = \dfrac {PB}{PR}$

$QR = \dfrac {AB \times PR}{PB}$

$= \dfrac {3\times 6}{2}$
Thus,

$QR = 9\ cm$ .

In $\triangle ABC$ the points $M$ and $N$ are on $AB$ and $AC$ . If $AB = 1.3\ cm, AC = 2.6\ cm, AM =0.5\ cm, AN = 1.0\ cm$ then show that $MN \parallel BC$ or not?

$\triangle ABC$ ,

$\cfrac {AB}{AM} = \cfrac {1.3}{0.55}$
and

$\cfrac {AC}{AN} = \cfrac {2.6}{1.0} \neq \cfrac {1.3}{0.55}$
Since,

$\cfrac {AB}{AM} \neq \cfrac {AC}{AN}$

$\therefore$ By basic proprtionality theorem,

$MN$ is not parallel to

$BC$ .

If for acute angled $\Delta ABC$ and $\Delta PQR$ are similar by $ABC \leftrightarrow PQR$ correspondance, then prove that $\dfrac{A(ABC)}{A(PQR)} = \dfrac{AB^2}{PQ^2} = \dfrac{BC^2}{QR^2} = \dfrac{AC^2}{PR^2}$

We are given two triangles ABC and PQR such that

$\Delta ABC \leftrightarrow \Delta PQR$ .
Draw altitudes AM and PN of the triangles.
Now,

$A(ABC) = \dfrac{1}{2} \times BC \times AM$
and

$A(PQR) = \dfrac{1}{2} \times QR \times PN$
So,

$\dfrac{A(ABC)}{A(PQR)} = \dfrac{\dfrac{1}{2} \times BC \times AM}{\dfrac{1}{2} \times QR \times PN} = \dfrac{BC \times AM}{QR \times PN}$ ............. (1)
In

$\Delta ABM$ and

$\Delta PQN$ ,

$\angle B = \angle Q$ (As

$\Delta ABC \leftrightarrow \Delta PQR$ )
and

$\angle M = \angle N$ (Each of

$90^o$ )
So,

$\Delta ABM \leftrightarrow \Delta PQN$ (AA similarity criterion)

$\therefore \dfrac{AM}{PN} = \dfrac{AB}{PQ}$ .......... (2)
Also

$\Delta ABC \leftrightarrow \Delta PQR$
So,

$\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{CA}{RP}$ ......... (3)

$\therefore \dfrac{ABC}{PQR} = \dfrac{AB}{PQ} \times \dfrac{AM}{PN}$ [From (1) and (3)]

$= \dfrac{AB}{PQ} \times \dfrac{AB}{PQ}$ [From (2)]

$= \left( \dfrac{AB}{PQ}\right )^2$
Now using (3), we get

$\dfrac{A(ABC)}{A(PQR)} = \left( \dfrac{AB}{PQ} \right )^2 = \left( \dfrac{BC}{QR} \right )^2 = \left( \dfrac{CA}{RP} \right )^2$

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

$\textbf{Step 1: Apply property of congruent triangles}$

$Construction- Draw$

$DB'$

$equal\ to$

$AB$

$and$

$DC'$

$equal\ to$

$AC$

$in$

$\triangle DEF$

$and\ join$

$B'C'$ .

$AB = BD'$ By construction

$AC = DC'$ by construction

$\angle A = \angle D$ Given

$\triangle ABC \cong \triangle DB'C'$

$...(S.A.S\ test\ of\ Congruence)$

$\angle B = \angle DB'C'$ ...C.A.C.T

$\angle C = \angle DC'B'$ ....C.A.C.T

$\therefore \triangle ABC\sim \triangle DB'C'$

$....(A.A.A\ test\ of\ similarity)$

$\textbf{Step 2: Apply property of congruent triangles}$

$\cfrac {AB}{DB'} = \dfrac {AC}{DC'} = \dfrac {BC}{B'C'}$

$....(C.S.S.T)$

$But$

$\cfrac {AB}{DE} = \cfrac {AC}{DF}$ ...Given

$Or$

$\cfrac {DB'}{DE} = \cfrac {DC'}{DF}$

$\left \{AB = DB', AC = DC'\right \}$

$\Rightarrow B'C'\parallel EF$ .

$\text{(Side divides the two side in the same ratio then it is parallel to third side).}$

$\therefore \angle DB'C' = \angle E = \angle B$

$\angle DC'B' = \angle F = \angle C$

$\therefore \triangle ABC \sim \triangle DEF$

$\text{(S.A.S.test of similarity)}$

$\textbf{Hence Proved}$

In trapezium $ABCD$ , side $AB\parallel CD$ and diagonals $AC$ and $BD$ intersect each other at $O$ .
Prove that: $\cfrac { OA }{ OC } =\cfrac { OD }{ OB }$ .

In trapezium

$ABCD$ , side

$AB\parallel CD$

Diagonals

$AC$ and

$BD$ intersect each other at point

$O$ .

To prove:

$\cfrac { OA }{ OC } =\cfrac { OD }{ OB }$

Proof:

$\triangle AOB$ and

$\triangle COD$

$\angle OAB=\angle OCD$ ..........(Alternate angles)

and

$\angle AOB=\angle COD$ ........(Vertically opposite angles)

Hence,

$\triangle AOB \sim$

$\triangle COD$ .......(By AA test of similarity)

$\Rightarrow$

$\cfrac { OA }{ OC } =\cfrac { OD }{ OB }$ ........[C.S.S.T]

$[hence \, proved]$

Write property of side-angle-side similarity and side-side-side similarity.

Side Angle Side Similarity:

If in two triangles two corresponding sides are proportional and the corresponding included angles are equal, then the two said triangles are similar.

Side-Side-Side Similarity:

If the ratios of the all the corresponding sides of two triangles are equal, then the said triangles are similar.

The areas of two similar triangles are $250sq.cm$ and $360sq.cm$ , then determine the ratio of their corresponding sides.

$\cfrac { Area\quad of\quad 1st\quad triangle }{ Area\quad of\quad 2nd\quad triangle } ={ \left( \cfrac { Side\quad of\quad 1st\quad triangle }{ Side\quad of\quad 2nd\quad triangle } \right) }^{ 2 }\quad$

$\therefore \cfrac { 250 }{ 360 } =\left( \cfrac { Side\quad of\quad 1st\quad triangle }{ Side\quad of\quad 2nd\quad triangle } \right)^2$

$\Rightarrow \sqrt { \cfrac { 250 }{ 360 } } =\cfrac { Side\quad of\quad 1st\quad triangle }{ Side\quad of\quad 2nd\quad triangle }$

$\Rightarrow \sqrt { \cfrac { 25 }{ 36 } } =\cfrac { Side\quad of\quad 1st\quad triangle }{ Side\quad of\quad 2nd\quad triangle }$

$\Rightarrow \cfrac { Side\quad of\quad 1st\quad triangle }{ Side\quad of\quad 2nd\quad triangle } =\cfrac { 5 }{ 6 }$

Write converse of Pythagoras theorem and prove it.

Converse of Pythagoras Theorem.

Statement:
In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Proof:

Here, we are given a triangle ABC in which

$Ac^2 = AB^2 + BC^2$ .

We need to prove that

$\angle B = 90^o$ .

To start with, we construct a

$\Delta$ PQR right-angled at Q such that

$PQ = AB$ and

$QR = BC.$

Now, from

$\Delta$ PQR, we have:

$Pr^2 = PQ^2 + QR^2$ (Pythagoras Theorem, as

$\angle Q = 90^o$ )

$PR^2 = AB^2 + BC^2$ (By construction)........ (1)

But

$AC^2 = AB^2 + BC^2$ (Given).......... (2)

So,

$AC = PR$ (From (1) and (2)).............. (3)

Now, in

$\Delta ABC$ and

$\Delta PQR$ ,

$AB = PQ$ (By construction)

$BC = QR$ (By construction)

$AC = PR$ (Proved in (3))

So,

$\Delta ABC \simeq \Delta PQR$ (SSS congruence)

$\angle B = \angle Q$ (Corresponding angles of congruent triangles)

$\angle Q = 90^o$ (By construction)

$\angle B = 90^o$

In $\triangle ABC,DE\parallel BC$ , point $D$ lies $AB$ and $E$ lies on $AC$ . If $AD=3.4cm, AB=10cm, AE=5.1cm$ then find the value of $AC$ .

$In:\triangle ABC$ using the Thales's Theorem ("If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divided the two sides in the same ratio. ")

$In:\triangle ABC$

$DE||BC$ (given)

Therefore,

$\dfrac { AD }{ AB } =\dfrac { AE }{ AC }$

$\dfrac { 3.4 }{ 10 } =\dfrac { 5.1 }{ AC }$

${ AC= }\dfrac { 10\times 5.1 }{ 3.4 } =15cm$

The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

If two triangles are similar, then the ratio of the area of both triangles is proportional to square of the ratio of their corresponding sides.

To prove this theorem, consider two similar triangles $ΔABC$ and $ΔPQR$

According to the stated theorem

$\dfrac { ar\triangle ABC }{ ar\triangle PQR } ={ \left( \dfrac { AB }{ PQ } \right) }^{ 2 }={ \left( \dfrac { BC }{ QR } \right) }^{ 2 }={ \left( \dfrac { CA }{ RP } \right) }^{ 2 }$

$=\dfrac { 1 }{ 2 } \times base\times altitude$

To find the area of $ΔABC$ and $ΔPQR$ draw the altitudes $AD$ and $PE$ from the vertex $A$ and $P$ of $ΔABC$ and $ΔPQR$

Now, area of $ΔABC$ $=\dfrac { 1 }{ 2 } \times BC\times AD$

$=\dfrac { 1 }{ 2 } \times QR\times PE$

The ratio of the areas of both the triangles can now be given as:

$\dfrac { ar\triangle ABC }{ ar\triangle PQR } =\dfrac { \dfrac { 1 }{ 2 } \times BC\times AD }{ \dfrac { 1 }{ 2 } \times QR\times PE }$

$\dfrac { ar\triangle ABC }{ ar\triangle PQR } =\dfrac { BC\times AD }{ QR\times PE }$

Now in $\triangle ABD$ and $\triangle PQE$ it can be seen

$ΔABC\sim ΔPQR$ )

$∠ADB = ∠PEQ$ (Since both the angles are $90°$ )

From $AA$ criterion of similarity $ΔADB\sim ΔPEQ$

$\dfrac { AD }{ PE } =\dfrac { AB }{ PQ }$

Since it is known that $ΔABC\sim ΔPQR$

$\dfrac { AB }{ PQ } =\dfrac { BC }{ QR } =\dfrac { CA }{ RP }$

Substituting this value in equation , we get

$\dfrac { ar\triangle ABC }{ ar\triangle PQR } =\dfrac { AB }{ PQ } \times \dfrac { AD }{ PE }$

we can write

$\dfrac { ar\triangle ABC }{ ar\triangle PQR } ={ \left( \dfrac { AB }{ PQ } \right) }^{ 2 }$

Similarly we can prove

$\dfrac { ar\triangle ABC }{ ar\triangle PQR } ={ \left( \dfrac { AB }{ PQ } \right) }^{ 2 }={ \left( \dfrac { BC }{ QR } \right) }^{ 2 }={ \left( \dfrac { CA }{ RP } \right) }^{ 2 }$

For two acute angled $\triangle ABC$ and $\triangle PQR$ if $\triangle ABC\sim \triangle PQR$ then prove that $\dfrac {area(\triangle ABC)}{area(\triangle PQR)} = \dfrac {AB^{2}}{PQ^{2}} = \dfrac {BC^{2}}{QR^{2}} = \dfrac {AC^{2}}{PR^{2}}$ .

Construction: Draw

$AD$ perpendicular to

$BC$ and

$PS$ perpendicular to

$QR$

$\Delta ABC \sim \Delta PQR$

$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{CA}{RP}$ (1)

Also

$\angle A=\angle P$ ,

$\angle B=\angle Q$ ,

$\angle C=\angle R$ (2)

Now consider

$\Delta ABD$ and

$\Delta PQS$

$\angle B=\angle Q$ (from (2))

$\angle ADB=\angle PSQ=90{^\circ}$ (by construction)

So by AA similarity

$\Delta ABD \sim \Delta PQS$

Therefore

$\dfrac{AB}{PQ}=\dfrac{BD}{QS}=\dfrac{DA}{SP}$ (3)

$area(\Delta ABC)=\dfrac{1}{2}\times AD\times BC$ (4)

$area(\Delta PQR)=\dfrac{1}{2}\times PS\times QR$ (5)

Dividing (4) by (5)

$\dfrac{area(\Delta ABC)}{area(\Delta PQR)}=\dfrac{\dfrac{1}{2}\times AD\times BC}{\dfrac{1}{2}\times PS\times QR}$ (6)

From (1) and (3)

$\dfrac{DA}{SP}=\dfrac{BC}{QR}$ (7)

Substituting (7) in (6)

$\dfrac{area(\Delta ABC)}{area(\Delta PQR)}=\dfrac{\dfrac{1}{2}\times BC\times BC}{\dfrac{1}{2}\times QR\times QR}=\dfrac{(BC)^{2}}{(QR)^{2}}$

From (1)

$\dfrac{area(\Delta ABC)}{area(\Delta PQR)}=\dfrac{(BC)^{2}}{(QR)^{2}}=\dfrac{(AB)^{2}}{(PQ)^{2}}=\dfrac{(AC)^{2}}{(PR)^{2}}$

Hence proved

In the given figure, find the value of x which will make DE || AB ?

$DE || AB$ , then

$\Delta CDE \sim \Delta ABC$
By property of similar triangles:

$\dfrac{CD}{AC}=\dfrac{CE}{BC}$

$\implies\dfrac{x+3}{4x+22}=\dfrac{x}{4x+4}$
Cross-multiplying:

$(x+3)(4x+4)=x(4x+22)$

$\implies 4x^2 + 16x + 12 = 4x^2 + 22x$

$12 = 6x \implies \boxed{x = 2}$

Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Given:- A right angled triangle

$ABC$ , right angle at

$B$

To prove:-

$AC^{2}=AB^{2}+BC^{2}$

Proof:- Draw a perpendicular

$BD$ from

$B$ to

$AC$

$\triangle ABC$ and

$\triangle ABD$

$\angle ADB=\angle ABC=90^{o}$

$\angle DAB=\angle BAC$ ..... (Common angle)

$\therefore \triangle ABC \sim \triangle ABD$ ...... (Using AA similarity criteria)

Now,

$\dfrac{AD}{AB}=\dfrac{AB}{AC}$ ..... (Corresponding sides are proportional)

$\Rightarrow AB^{2}=AD\times AC$ ......

$(i)$

Similarly,

$\triangle ABC \sim \triangle BDC$

$\therefore BC^{2}=CD\times AC$ ......

$(ii)$

Adding equations

$(i)$ and

$(ii)$ , we get

$AB^{2}+BC^{2}=AD\times AC+CD\times AC$

$\Rightarrow AB^{2}+BC^{2}=AC(AD+CD)$

$\Rightarrow AB^{2}+BC^{2}=AC\times AC$

$\Rightarrow AB^{2}+BC^{2}=AC^{2}$

$[hence \, proved]$

In the given figure, $AB\perp BC, FG\perp BC$ and $DE\perp AC$ . Prove that $\triangle ADE$ ~ $\triangle GCF$ .

In the given figure we have to prove that

$\Delta$ ADE and

$\Delta$ FGC are similar.

Now from

$\Delta$ ADE and

$\Delta$ GFC we have,

$\angle AED=\angle GFC$ [right-angles],

$\angle DAE=\angle FGC$ [

$\because$

$AD \parallel GF$ and

$AC$ intersector, similar angles].

Then

$\angle ADE=\angle GCF$ [Remaining angles equal by

$180^\circ$ rule].

$\therefore \Delta ADE$ is similar to

$\Delta GFC.$ [By AA test]

$[hence \, proved]$

In a $\triangle ABC, P$ and $Q$ are points on sides $AB$ and $AC$ respectively, such that $PQ \parallel BC$ . If
$AP = 2.4 cm, AQ = 2 cm,QC = 3 cm$ and $BC = 6 cm$ , find $AB$ and $PQ.$

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Hence

$\dfrac{AQ}{QC}=\dfrac{AP}{PB}$

$\dfrac{2}{3}=\dfrac{2.4}{PB}$

$PB=3.6cm$

$AB=AP+PB=2.4+3.6=6cm$

$PQ\parallel BC$

$\angle AQP=\angle ACB$

$\angle APQ=\angle ABC$

So by

$AAA$

$\triangle AQP\sim \triangle ACB$

Hence

$\dfrac{AQ}{AC}=\dfrac{QP}{CB}$

$\dfrac{2}{5}=\dfrac{QP}{6}$

$QP=2.4cm$

$AB=6cm$ and

$QP=2.4cm$

In a $\triangle ABC, D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE\parallel BC$ . If $AD = 2.4\ cm, AE = 3.2\ cm, DE \parallel BC$ . If $AD = 2.4\ cm, AE = 3.2\ cm, DE = 2\ cm$ and $BC = 5\ cm$ , find $BD$ and $CE$ .

If a line is drawn parallel to one side of a triangle to intersect the

other two sides in distinct points, the other two sides are divided in the same ratio.

$DE\parallel BC$

$\angle ADE=\angle ABC$

$\angle AED=\angle ACB$

So by

$AAA$

$\triangle AED\sim \triangle ACB$

Hence

$\dfrac{AE}{AC}=\dfrac{ED}{CB}$

$\therefore\dfrac{3.2}{AC}=\dfrac{2}{5}$

$\Rightarrow AC=8cm$

$AC=8cm=AE+EC=3.2+EC$

$EC=8-3.2=4.8cm$

Also

$\dfrac{AD}{AB}=\dfrac{ED}{CB}$

$\therefore \dfrac{2.4}{AB}=\dfrac{2}{5}$

$AB=6cm=AD+DB=2.4+DB$

$\Rightarrow DB=3.6cm$

$BD=3.6cm$ and

$CE=4.8cm$

$\triangle PQR$ is right angled at $Q, QX\perp PR, XY\perp RQ$ and $XZ \perp PQ$ are drawn. Prove that $XZ^{2} = PZ \times ZQ$ .

In ΔPZX and ΔPQR

∠P=∠P (∵common)

∠PZX=∠PQR=90° (given)

By AA similarity ΔPZX≈ΔPQR → (1)

In ΔXZQ and ΔPXQ

∠XZQ=∠PXQ (90°each)

∠ZQX=∠PQX (∵common)

By AA similarity, ΔXZQ≈ΔPXQ → (2)

In ΔPQR and ΔPXQ

∠PQR=∠PXQ (∵each 90°)

∠P=∠P (∵common)

By AA similarity, ΔPQR ≈ΔPXQ → (3)

From eq (1),(2) and (3) transitivity property

ΔPZX≈ΔXZQ

Hence, AA similarity postulates,

⇒

$\dfrac{PZ}{ZX}=\dfrac{ZX}{ZQ}\Rightarrow ZX^2=PZ \times ZQ$

Hence, proved

In fig $\triangle ACB \sim \triangle APQ$ . If $BC = 10\ cm,\ PQ = 5\ cm,\ BA = 6.5\ cm$ and $AP= 2.8\ cm$ , find $CA$ and $AQ$ . Also find the area $(\triangle ACB)$ : area $(\triangle APQ)$ .

If two triangles are similar then the sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle.

$\triangle ACB\sim \triangle APQ$

$\dfrac{AC}{AP}=\dfrac{CB}{PQ}=\dfrac{AB}{AQ}$

$\dfrac{AC}{2.8}=\dfrac{10}{5}=\dfrac{6.5}{AQ}$

$\dfrac{AC}{2.8}=2=\dfrac{6.5}{AQ}$

Hence

$AC=2\times 2.8=5.6cm$

Also

$AQ=\dfrac{6.5}{2}=3.25cm$

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\dfrac{area(\triangle ACB)}{area(\triangle APQ)}=\dfrac{AC^{2}}{AP^{2}}$

$\dfrac{area(\triangle ACB)}{area(\triangle APQ)}=(\dfrac{5.6}{2.8})^{2}$

$\dfrac{area(\triangle ACB)}{area(\triangle APQ)}=2^{2}=4$

Hence

$\dfrac{area(\triangle ACB)}{area(\triangle APQ)}=4$

ABCD is a trapezium in which AB $\parallel$ CD. The diagonals AC and BD intersect at O. Prove that:
(i) $\triangle AOB \sim  \triangle COD$ (ii) If OA = 6 cm, OC = 8cm,

Find:
(a) $\dfrac { Area\left( \triangle AOB \right) }{ Area\left(  \triangle COD \right) }$

(b) $\dfrac { Area\left( \triangle AOD \right) }{ Area\left(  \triangle COD \right) }$

$ABCD$ is a trapezium in which

$AB\parallel CD$ .

The diagonals

$AC$ and

$BD$ intersect at

$O$ .

$OA=6 \ cm$ and

$OC=8\,cm$

$\triangle AOB$ and

$\triangle COD$ ,

$\angle AOB=\angle COD$ [ Vertically opposite angles ]

$\angle OAB=\angle OCD$ [ Alternate angles ]

$\therefore$

$\triangle AOB\sim\triangle COD$ [ By AA similarity ]

$(a)$

$\dfrac{ar(\triangle AOB)}{ar(\triangle COD)}=\dfrac{OA^2}{OC^2}$ [ By area theorem ]

$\Rightarrow$

$\dfrac{ar(\triangle AOB)}{ar(\triangle COD)}=\dfrac{(6)^2}{(8)^2}$

$\Rightarrow$

$\dfrac{ar(\triangle AOB)}{ar(\triangle COD)}=\dfrac{36}{64}=\dfrac{9}{16}$

$(b)$ Draw

$DP\perp AC$

$\dfrac{ar(\triangle AOD)}{ar(\triangle COD)}=\dfrac{\dfrac{1}{2}\times AO\times DP}{\dfrac{1}{2}\times CO\times DP}$ [ By area theorem ]

$\Rightarrow$

$\dfrac{ar(\triangle AOD)}{ar(\triangle COD)}=\dfrac{AO}{CO}$

$\Rightarrow$

$\dfrac{ar(\triangle AOD)}{ar(\triangle COD)}=\dfrac{6}{8}=\dfrac{3}{4}$

Triangles $ABC$ and $DEF$ are similar.

(i) If area ( $\triangle$ ABC)= $16$ $cm^{2}$ , area( $\triangle$ DEF)= $25$ $cm^{2}$ and BC = $2.34$ cm, find EF.

(ii) If area ( $\triangle$ ABC)= $9$ $cm^{2}$ , area( $\triangle$ DEF) = $64$ $cm^{2}$ and DE= $5.1$ cm, find AB.

(iii) If AC = $19$ cm and DF= $8$ cm, ratio of the area of two triangles.

(iv) If area ( $\triangle$ ABC)= $36$ $cm^{2}$ , area ( $\triangle$ DEF) = $64$ $cm^{2}$ and DE= $6.2$ cm, find AB.

(v) If AB= $1.2$ cm and DE= $1.4$ find the ratio of the areas of triangle ABC and DEF.

$(i)$ We know that the ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\therefore$ $\dfrac{area(\triangle ABC)}{area(\triangle DEF)}=\left(\dfrac{BC}{EF}\right)^2$

$\Rightarrow$ $\dfrac{16}{25}=\left(\dfrac{2.3}{EF}\right)^2$

$\Rightarrow$ $\dfrac{4}{5}=\dfrac{2.3}{EF}$

$\Rightarrow$ $EF=\dfrac{2.3\times 5}{4}$

$\therefore$ $EF=2.875\,cm$

$(ii)$ We know that the ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\therefore$ $\dfrac{area(\triangle ABC)}{area(\triangle DEF)}=\left(\dfrac{AB}{DE}\right)^2$

$\Rightarrow$ $\dfrac{9}{64}=\left(\dfrac{AB}{DE}\right)^2$

$\Rightarrow$ $\dfrac{3}{8}=\dfrac{AB}{5.1}$

$\Rightarrow$ $AB=\dfrac{5.1\times 3}{8}$

$\therefore$ $AB=1.91\,cm$

$(iii)$ We know that the ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\therefore$ $\dfrac{area(\triangle ABC)}{area(\triangle DEF)}=\left(\dfrac{AC}{DF}\right)^2$

$\therefore$ $\dfrac{area(\triangle ABC)}{area(\triangle DEF)}=\left(\dfrac{19}{8}\right)^2$

$\therefore$ $\dfrac{area(\triangle ABC)}{area(\triangle DEF)}=\left(\dfrac{361}{64}\right)$

$(iv)$ We know that the ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides

$\therefore$ $\dfrac{area(\triangle ABC)}{area(\triangle DEF)}=\left(\dfrac{AB}{DE}\right)^2$

$\Rightarrow$ $\dfrac{36}{64}=\left(\dfrac{AB}{DE}\right)^2$

$\Rightarrow$ $\dfrac{6}{8}=\dfrac{AB}{6.2}$

$\Rightarrow$ $AB=\dfrac{6.2\times 6}{8}$

$\therefore$ $AB=4.65\,cm$

$(v)$ We know that the ratio of area of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\therefore$ $\dfrac{area(\triangle ABC)}{area(\triangle DEF)}=\left(\dfrac{AB}{DE}\right)^2$

$\therefore$ $\dfrac{area(\triangle ABC)}{area(\triangle DEF)}=\left(\dfrac{1.2}{1.4}\right)^2$

$\therefore$ $\dfrac{area(\triangle ABC)}{area(\triangle DEF)}=\dfrac{36}{49}$

Triangles ABC and DEF are similar.

(i) If area ( $\triangle ABC$ )= 16 $\text{cm}^{2}$ , area( $\triangle$ DEF)= 25 $\text{cm}^{2}$ and $BC = 2.3\ \text{cm}$ , find $EF$ .

(ii) If area ( $\triangle ABC$ )= 9 $\text{cm}^{2}$ , area( $\triangle$ DEF) = 64 $\text{cm}^{2}$ and $DE= 5.1\ \text{cm}$ , find $AB$ .

(iii) If $AC = 19\ \text{cm}$ and $DF=8\ \text{cm}$ , find ratio of the area of two triangles.

(iv) If area ( $\triangle ABC$ )= 36 $\text{cm}^{2}$ , area ( $\triangle$ DEF) =64 $\text{cm}^{2}$ and $DE= 6.2\ \text{cm}$ , find $AB$ .

(v) If $AB= 1.2\ \text{cm}$ and $DE=1.4\ \text{cm}$ , find the ratio of the areas of triangle $ABC$ and $DEF$ .

1.)given ABC and DEF are similar

therefore square of the ratio of sides =ratio of areas

implies

$\left(\sqrt {\dfrac{16}{25}}\right)=\dfrac{BC}{EF}$

therefore

$EF=\dfrac{11.5}{4}=2.875$

$\because BC=2.3$

2.) according to above

$\left(\sqrt{\dfrac{64}{9}}\right)=\dfrac{DE}{AB}$

therefore

$AB=\dfrac{15.3}{8}=1.9125$

$\because DE=5.1$

3.)RATIO OF TRIANGLES =

$(\dfrac{19}{8})^2$ =

$5.640625$

4.)

$\left(\sqrt{\dfrac{64}{36}}\right)=\dfrac{DE}{AB}$

$\therefore AB=4.65$

$\because DE=6.2$

5.)ratio of areas of

$ABC$ and

$DEF$ =

$\left(\dfrac{1.2}{1.4}\right)^2$ =

$0.605042017$

The area of two similar triangle are $81$ $cm^2$ and $49\ cm^2$ respectively. Find ratio of their corresponding heights. What is the ratio of their corresponding medians?

Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding altitudes and is also equal to the ratio of squares of their corresponding medians.

$\therefore$

$\dfrac{Area\,of\,triangle\,1}{Area\,of\,triangle\,2}=\left(\dfrac{H_1}{H_2}\right)^2$

$\therefore$

$\dfrac{81}{49}=\left(\dfrac{H_1}{H_2}\right)^2$

$\therefore$

$\dfrac{H_1}{H_2}=\dfrac{9}{7}$

$\therefore$ Ratio of corresponding heights is

$9:7$ and ratio of corresponding medians is also

$9:7$

The area of two similar triangles are 169 $cm^{2}$ and 121 $cm^{2}$ respectively, If the longest side of the largest triangle is 26 cm, find the longest side of the smaller triangle?

Given the area of

$\Delta_1 = 169 \mathrm{cm}^{2}$ and area of

$\Delta_2 = 121 \mathrm{cm}^{2}$

Let longest side of the smaller triangle be

$x$ .

For two similar triangles, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\implies \dfrac{Area\,of \,\Delta_1}{Area\, of \, \Delta_2} = \left ( \dfrac{Longest\,side\,of\, \Delta_1}{Longest\, side\, of \, \Delta_2}\right)^2$

Substituting the values, we get

$\implies {\dfrac{169}{121}}=\left (\dfrac{26}{x}\right )^2$

$\implies \sqrt{\dfrac{169}{121}}=\dfrac{26}{x}$

$\dfrac{13}{11}=\dfrac{26}{x}$

$x=22\, \mathrm {cm}$

The longest side of smaller triangle is

$22\, \mathrm{cm}$

In the given figure, $DE \parallel BC$

(i) If DE = 4 cm, BC= 6 cm and Area ( $\triangle$ ADE)= 16 $cm^{2}$ , find the area of $\triangle$ ABC.

(ii) If DE = 4 cm BC = 8 cm and Area ( $\triangle$ ADE)= 25 $cm^{2}$ , find the area of $\triangle$ ABC.

(iii) If DE : BC = 3 :Calculate the ratio of the areas of $\triangle$ ADE and the area of BCED.

In $\triangle ABC$ , $DE\parallel BC$

$\Rightarrow$ $\angle B=\angle D$ [ Corresponding angles ]

$\Rightarrow$ $\angle C=\angle E$ [ Corresponding angles ]

$\Rightarrow$ $\angle A=\angle A$ [ Common angle]

$\therefore$ $4\triangle ABC\sim\triangle ADE$ [ By AAA criteria ]

$(i)$ $\dfrac{area(\triangle ABC)}{area(\triangle ADE)}=\left(\dfrac{BC}{DE}\right)^2$ [ Ratio of areas of two similar triangles is equal the ratio of squares of their corresponding sides. ]

$\therefore$ $\dfrac{area(\triangle ABC)}{16}=\left(\dfrac{6}{4}\right)^2$

$\therefore$ $area(\triangle ABC) =\dfrac{36\times 16}{16}$

$\therefore$ $area(\triangle ABC)=36\,cm^2$

$(ii)$ $\dfrac{area(\triangle ABC)}{area(\triangle ADE)}=\left(\dfrac{BC}{DE}\right)^2$ [ Ratio of areas of two similar triangles is equal the ratio of squares of their corresponding sides. ]

$\therefore$ $\dfrac{area(\triangle ABC)}{25}=\left(\dfrac{8}{4}\right)^2$

$\therefore$ $area(\triangle ABC) =\dfrac{64\times 25}{16}$

$\therefore$ $area(\triangle ABC)=100\,cm^2$

$(iii)$ $\dfrac{area(\triangle ADE)}{area(\triangle ABC)}=\left(\dfrac{DE}{BC}\right)^2$ [Ratio of areas of two similar triangles is equal the ratio of squares of their corresponding sides. ]

$\dfrac{area(\triangle ADE)}{area(\triangle ABC)}=\left(\dfrac{3}{5}\right)^2$

$\dfrac{area(\triangle ADE)}{area(\triangle ABC)}=\left(\dfrac{9}{25}\right)$

$\Rightarrow$ $\dfrac{25}{9}\times area(\triangle ADE)= area(\triangle ABC)$

$\Rightarrow$ $Area\, of\, trapezium\, BCED=area(\triangle ABC)-area(\triangle ADE)$

$\therefore$ $\dfrac{area(\triangle ADE)}{Area\,of\,BCED}=\dfrac{area(\triangle ADE)}{area(\triangle ABC)-area(\triangle ADE)}$

${Area\, of\, BCED}$ $=\dfrac{area(\triangle ADE)}{(\dfrac{25}{9}-1)area(\triangle ADE)}$

$=\dfrac{1}{\dfrac{16}{9}}$

$=\dfrac{9}{16}$

Two isosceles triangles have equal vertical angles and their area are in the ratio of the $16 : 25$ . Find the ratio of their corresponding heights.

$AB=AC\quad$ (given)

$\therefore \cfrac { AB }{ AC } =1\quad -(1)$

$DE=DF$ (given)

$\cfrac { DE }{ DF } =1\quad -(2)$

comparing

$(1)\& (2)$

$\cfrac { AB }{ AC } =\cfrac { DE }{ DF } \quad \implies\quad \cfrac { AB }{ DE } =\cfrac { AC }{ DF }$

$In \ \triangle ABC \ \& \ \triangle DEF$

$\cfrac { AB }{ DE } =\cfrac { AC }{ DF } \quad \implies\quad \angle A=\angle D$

$\therefore$ By using SAS similar condition

$\triangle ABC\sim \triangle DEF$

$\cfrac { ar(\triangle ABC) }{ ar(\triangle DEF) } =\cfrac { { AD }^{ 2 } }{ { DQ }^{ 2 } }$

$\therefore$ The area of two similar

$\triangle les$ are in the ratio of the square of the corresponding altitude.

$\cfrac { 16 }{ 25 } =\cfrac { { AP }^{ 2 } }{ { DQ }^{ 2 } }$

$\implies\quad \cfrac { AP }{ DQ } =\cfrac { 4 }{ 5 }$

1013143_969356_ans_540634fb96c548d9a4851944818a5388.png

The area of two similar triangles are $36\ cm^{2}$ and $25\ cm^{2}$ . If an altitude of the first triangle is $2.4\ cm$ , find the corresponding altitude of the other triangle.

given two triangles are similar

therefore ratio of their areas =square of ratios of corresponding altitudes

therefore

$\left(\sqrt {\dfrac{25}{36}}\right)=\dfrac{2.4}{x}$

impies

$x=2.88$

The area of two similar triangle are 100 $cm^{2}$ and 49 $cm^{2}$ respectively. If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the other.

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding altitudes.

The area of two similar triangle are 1

$00 cm^{2}$ and

$49 cm^{2}$ respectively.

The altitude of the bigger triangle is

$5 cm$

Let the altitude of the smaller triangle be

$x$ .

Therefore

$\sqrt{\dfrac{100}{49}}=\dfrac{5}{x}$

$\dfrac{10}{7}=\dfrac{5}{x}$

$x=3.5cm$

The altitude of the smaller triangle is

$3.5cm$ .

In each of the figure, a line segment is drawn parallel to one side of the triangle and the lengths of certain line-segments are marked. Find the value of $x$ in each of the following :

Basic proportionality theorm :
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

$(i)$

$\dfrac{1}{c}=\dfrac{d}{x}$

$\therefore$

$x=cd$

$(ii)$

$\dfrac{a}{1}=\dfrac{b}{x}$

$\therefore$

$x=\dfrac{b}{a}$

$(iii)$

$\dfrac{1}{g}=\dfrac{g}{x}$

$\therefore$

$x=g^2$

$(iv)$

$\dfrac{1}{x}=\dfrac{h}{1}$

$\therefore$

$x=\dfrac{1}{h}$

$ABC$ is a triangle and $PQ$ is a straight line meeting $AB$ in $P$ and $AC$ in $Q$ . If $AP = 1$ cm, $PB = 3$ cm, $AQ = 1.5$ cm, $QC = 4.5$ m, prove that area of $\triangle$ $APQ$ is one- sixteenth of the area of $\triangle$ $ABC$ .

$AP=1\,cm,\,PB=3\,cm,\,AQ=1.5\,cm,\,QC=4.5\,cm$ [ Given ]

Then, $AB=AP+PB=1+3=4\,cm$

In $\triangle APQ$ and $\triangle ABC$ ,

$\angle A=\angle A$ [ Common angle ]

$\dfrac{AP}{AB}=\dfrac{AQ}{AC}$ [ Each equal to $\dfrac{1}{4}$ ]

$\therefore$ $\triangle APQ\sim \triangle ABC$ [ By SAS similarity ]

$\therefore$ $\dfrac{area(\triangle APQ)}{area(\triangle ABC)}=\dfrac{AP^2}{AB^2}$ [ By area of similar triangle theorem ]

$\therefore$ $\dfrac{area(\triangle APQ)}{area(\triangle ABC)}=\dfrac{1^2}{4^2}$

$\therefore$ $\dfrac{area(\triangle APQ)}{area(\triangle ABC)}=\dfrac{1}{16}$

$\therefore$ $area(\triangle APQ)=\dfrac{1}{16}\times area(\triangle ABC)$

In $\triangle$ ABC, points P and Q are on CA and CB, respectively such that CA = 16 cm, AC = 10 cm, CB = 30 cm and CQ = 25 cm. Is PQ $\parallel$ AB?

1634961_969537_ans_dfdced953c044ae8a9ceca1c7efff55a.jpeg

In $\triangle$ ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ $\parallel$ BC and PQ divides $\triangle$ ABC into two parts equal in area. Find $\dfrac { BP }{ AB }$ .

Since the line $PQ$ divides $\triangle ABC$ into two equal parts,

$area(\triangle APQ)=area(\triangle BPQC)$

$\Rightarrow$ $area(\triangle APQ)=area(\triangle ABC)-area(\triangle APQ)$

$\Rightarrow$ $2area(\triangle APQ)=area(\triangle ABC)$

$\therefore$ $\dfrac{area(\triangle ABC)}{area(\triangle APQ)}=\dfrac{2}{1}$ ---- ( 1 )

Now, in $\triangle ABC$ and $\triangle APQ$ ,

$\angle BAC=\angle PAQ$ [ Common angles ]

$\angle ABC=\angle APQ$ [ Corresponding angles ]

$\therefore$ $\triangle ABC\sim \triangle APQ$ [ By AA similarity ]

$\therefore$ $\dfrac{area(\triangle ABC)}{area(\triangle APQ)}=\dfrac{AB^2}{AP^2}$

$\therefore$ $\dfrac{2}{1}=\dfrac{AB^2}{AP^2}$ [ From ( 1 ) ]

$\Rightarrow$ $\dfrac{AB}{AP}=\dfrac{\sqrt{2}}{1}$

$\Rightarrow$ $\dfrac{AB-BP}{AB}=\dfrac{1}{\sqrt{2}}$

$\Rightarrow$ $1-\dfrac{BP}{AB}=\dfrac{1}{\sqrt{2}}$

$\Rightarrow$ $\dfrac{BP}{AB}=1-\dfrac{1}{\sqrt{2}}$

$\therefore$ $\dfrac{BP}{AB}=\dfrac{\sqrt{2}-1}{\sqrt{2}}$

If D is a point on the side AB of $\triangle$ ABC such that AD : DB = 3:2 and E is a point on BC such that DE $\parallel$ AC. Find the ratio of areas of $\triangle$ ABC and $\triangle$ BDE.

$\dfrac{AD}{DB}=\dfrac{3}{2}$ [ Given ]

$\therefore$

$\dfrac{AD}{DB}+1=\dfrac{3}{2}+1$

$\therefore$

$\dfrac{AD+DB}{DB}=\dfrac{3+2}{2}$

$\therefore$

$\dfrac{AB}{DB}=\dfrac{5}{2}$ --- ( 1 )

$\triangle BDE$ and

$\triangle ABC$

$\angle B=\angle B$ [ Common ]

$\angle BAC=\angle BDE$ [

$\because DE\parallel AC$ ]

$\therefore$

$\triangle ABC\sim\triangle BDE$ [ By AA similarity criteria ]

$\dfrac{ar(\triangle ABC)}{ar(\triangle BDE)}=\dfrac{AB^2}{DB^2}=\dfrac{5^2}{2^2}=\dfrac{25}{4}$ [ From ( 1 ) ]

$\therefore$

$ar(\triangle ABC):ar(\triangle BDE)=25:4$

In $\triangle$ ABC, P and Q are points on sides AB and AC respectively such that PQ $\parallel$ BC. If AP = 4 cm, PB = 6 cm and PQ = 3 cm, determine BC.

$\triangle ABC$ ,

$PQ\parallel BC$

So, by proportionality theorem

$\dfrac{AP}{PB}=\dfrac{AQ}{QC}=\dfrac{PQ}{BC}$

$\dfrac{4}{6}=\dfrac{AQ}{QC}=\dfrac{3}{BC}$

$\Rightarrow$

$\dfrac{4}{6}=\dfrac{3}{BC}$

$\therefore$

$BC=4.5\,cm$

If $\triangle ABC \sim \triangle DEF$ such that AB = 5 cm, area $(\triangle ABC ) = 20\ { cm }^{ 2 }$ and area $(\triangle DEF ) = 45\ { cm }^{ 2 }$ , determine DE.

Given

$\triangle ABC \sim \triangle DEF$

Hence the ratio of their areas is the ratio of squares of their sides

$\therefore \dfrac {ar(\triangle ABC)} {ar(\triangle DEF)} =\dfrac {AB ^{2}}{DE^{2}}$

$\Rightarrow \dfrac {20}{45}=\dfrac{25}{DE^{2}}$

$\therefore DE =7.5$ cm

In $\triangle ABC ,\ \angle A$ is obtuse, $PB \perp AC$ and $QC \perp AB.$
Prove that:
(I) $AB \times AQ=AC\times AP$
(ii) ${ BC }^{ 2 } = ( AC \times CP + AB \times BQ )$

$\triangle BPA$ and

$\triangle AQC$

$\angle P = \angle Q = 90^o$

$\angle PAB = \angle QAC$ [Vertically opposite angles]

and

$\angle A$ is the common angle

$\therefore$

$\triangle BPA \sim \triangle AQC$ [AAA similarity criterion]

$\Rightarrow$

$\dfrac{AB}{AC}=\dfrac{AP}{AQ}$

$\therefore$

$AB\times AQ=AC\times AP$ --- ( 1 ) [ Hence proved ]

Consider, right angled

$\triangle BCQ$

$\Rightarrow$

$BC^2 = CQ^2 + BQ^2$ [By Pythagoras theorem]

$\Rightarrow$

$BC^2 = CQ^2 + (AB + AQ)^2$ [Since BQ = AB + AQ]

$\Rightarrow$

$BC^2 = [CQ^2 + AQ^2] + AB^2 + 2AB \times AQ$ ----- (2)

In right

$\triangle ACQ$ ,

$CQ^2 + AQ^2 = AC^2$ [By Pythagoras theorem]

Hence equation (2) becomes,

$\Rightarrow$

$BC^2 = AC^2 + AB^2 + AB \times AQ + AB \times AQ$

$\Rightarrow$

$BC^2 = AC^2 + AB^2 + AB \times AQ + AP \times AC$ [From (1)]

$\Rightarrow$

$BC^2 = AC^2 + AP \times AC + AB^2 + AB \times AQ$

$\Rightarrow$

$BC^2 = AC (AC + AP) + AB (AB + AQ)$

$\Rightarrow$

$BC^2 = AC × CP + AB × BQ$ [From the figure]

Hence Proved.

In $\triangle$ PQR, M and N are points on sides PQ and PR respectively such that PM = 15 cm and NR = 8 cm. If PQ = 25 cm and PR = 20 cm state whether MN $\parallel$ QR.

We have,

$PM=15\,cm,NR=8\,cm,PQ=25\,cm$ and

$PR=20\,cm$

Now,

$MQ=PQ-PM=25-15=10\,cm$ and

$PN=PR-NR=20-8=12\,cm$

Hence,

$\dfrac{PM}{MQ}=\dfrac{15}{10}=\dfrac{3}{2}$

Also,

$\dfrac{PN}{NR}=\dfrac{12}{8}=\dfrac{3}{2}$

$\therefore$

$MN\parallel QR$ [ By converse proportionality theorem ]

In the given figure, DE $\parallel$ CB. Determine AC and AE.

Given In

$\triangle ACB, DE\parallel CB$

$\therefore \angle ADC = \angle ACB$ and

$\angle AED = \angle ABC$

$\triangle ADE \sim \triangle ACB$ (by AAA similarity )

$\therefore \dfrac {AD} {AC} =\dfrac {DE} {CB}$

$\implies \dfrac {14}{AC}=\dfrac {12}{15}$

$\therefore AC =17.5$ cm

Similarly,

$\dfrac {AE} {AB} =\dfrac {DE} {CB}$

$\dfrac {AE} {4}=\dfrac {12}{15}$

$\therefore AE=3.2$ cm

ABCD is a trapezium in which AB $\parallel$ DC. P and Q are points on sides AD and BC such that PQ $\parallel$ AB. If PD = 18cm, BQ = 35cm and QC = 15cm, find AD.

Construction: Join

$BD$

$\triangle ABD$ ,

$PO\parallel AB$ [

$\because$

$PQ\parallel AB$ ]

$\Rightarrow$

$\dfrac{DP}{AP}=\dfrac{DO}{OB}$ [ By basic proportionality theorem ] ------ ( 1 )

$\Rightarrow$ In

$\triangle BDC$ ,

$OQ\parallel DC$ [

$\because$

$PQ\parallel DC$ ]

$\dfrac{BQ}{QC}=\dfrac{OB}{OD}$ [ By basic proportionality theorem ]

$\Rightarrow$

$\dfrac{QC}{BQ}=\dfrac{OD}{OB}$ ----- ( 2 )

$\Rightarrow$

$\dfrac{DP}{AP}=\dfrac{QC}{BQ}$ [ From ( 1 ) and ( 2 ) ]

$\Rightarrow$

$\dfrac{18}{AP}=\dfrac{15}{35}$

$\therefore$

$AP=\dfrac{18\times 35}{15}=42$

$\therefore$

$AD=AP+DP=42+18=60\,cm$

Prove that if in two triangles, corresponding angles are equal then their corresponding sides are in the same ratio and hence the two triangles are similar.

$Two\triangle les\quad \triangle ABC\& \triangle DEF\quad such\quad that$

$\angle A=\angle D,\angle B=\angle E\& \angle C=\angle F$

$Draw\quad P\quad and\quad Q\quad on\quad DE\& DF\quad such\quad that\quad DP=AB$

$\& DQ=AC\quad resp.Join\quad PQ$

$In\triangle ABC\& \triangle DPQ AB=DP\quad (by\quad construction)$

$\angle A=\angle D\quad (given)$

$AC=DQ\quad (by\quad construction)$

$\implies\quad \triangle ABC\cong \triangle DPQ\quad (by\quad SAS\quad criterion)$

$\angle B=\angle P\quad \{ by\quad CPCT\}$

$\angle B=\angle E\quad (given)\quad implies\quad \angle P=\angle E$

$For\quad lines\quad PQ\quad and\quad EF\quad with\quad transversal\quad PE,$

$\angle P\& \angle E\quad are\quad corresponding\quad angles\quad and\quad they$

$are\quad equal.Hence,\quad PQ||EF.$

1013433_1000534_ans_271d653546a646a7a8ac72c1f2538ce3.png

In given figure, If $EF||DC||AB$ . prove that $\dfrac { AE }{ ED } =\dfrac { BF }{ FC }$ .

Given $EF||DC||AB$ in the given figure.

PROVE $\dfrac { AE }{ ED } =\dfrac { BF }{ FC }$

CONSTRUCTION Produce DA and CB to meet at p (say).

PROOF In $\triangle PEF$ , we have
$AB||EF$

$\therefore$ $\dfrac { PA }{ AE } =\dfrac { PB }{ BF }$ [By Thale's Theorem]

$\Rightarrow$ $\dfrac { PA }{ AE } +1=\dfrac { PB }{ BF } +1$ [Adding 1 on both sides]

$\Rightarrow$ $\dfrac { PA+AE }{ AE } =\dfrac { PB=BF }{ BF }$

$\Rightarrow$ $\dfrac { PE }{ AE } =\dfrac { PF }{ BF }$ ............(i)

In $\triangle PDC$ , we have

$EF||DC$

$\therefore$ $\dfrac { PE }{ ED } =\dfrac { PF }{ FC }$ [By Basic Proportionality Theorem]........(ii)

On dividing equation (i) by equation (ii),, we get

$\dfrac { \dfrac { PE }{ AE } }{ \dfrac { PE }{ ED } } =\dfrac { \dfrac { PF }{ BF } }{ \dfrac { PF }{ FC } }$

$\Rightarrow$ $\dfrac { ED }{ AE } =\dfrac { FC }{ BF }$

$\Rightarrow$ $\dfrac { AE }{ ED } =\dfrac { BF }{ FC }$

In given figure $EF||AB||DC$ . Prove that $\dfrac { AE }{ ED } =\dfrac { BF }{ FC }$

We have

$EF||AB||DC$ and

$EP||DC$

Thus, in

$\triangle ADC$ , We have

$EP||DC$

Therefore, by basic proportionality theorem, we have

$\dfrac { AE }{ ED } =\dfrac { AP }{ PC }$ .............(i)

Again,

$EF||AB||DC$

$\Rightarrow$

$FP||AB$

Thus, in

$\triangle CAB$ , we have

$FP||BA$

Therefore, by basic proportionality theorem, we have

$\dfrac { BF }{ FC } =\dfrac { AP }{ PC }$ ...............(ii)

From (i) and (ii), we have

$\dfrac { AE }{ ED } =\dfrac { BF }{ FC }$

In a given $\triangle ABC, DE||BC$ and $\dfrac { AD }{ DB } =\dfrac { 3 }{ 5 }$ . If $AC=5.6$ , find AE.

$\triangle ABC$ , we have

$DE||BC$

$\therefore$

$\dfrac { AD }{ DB } =\dfrac { AE }{ EC }$ [By Thale's Theorem]

$\Rightarrow$

$\dfrac { AD }{ DB } =\dfrac { AE }{ AC-AE }$

$\Rightarrow$

$\dfrac { 3 }{ 5 } =\dfrac { AE }{ 5.6-AE }$

$[\because AC=5.6]$

$\Rightarrow$

$3(5.6-AE)=5AE$

$\Rightarrow$

$16.8-3AE=5AE$

$\Rightarrow$

$8AE=16.8$

$\Rightarrow$

$AE=\dfrac { 16.8 }{ 8 } cm=2.1$ cm

In the given figure, $\triangle AMB \sim \triangle CMD$ ; determine $MD$ in terms of $x,\;y$ and $z$ .

$\triangle AMB\sim \triangle CMD$ [ Given ]

$\therefore$

$\dfrac{AM}{CM}=\dfrac{MB}{MD}$ [ By proportionality theorem ]

$\therefore\;\dfrac{y}{z}=\dfrac{x}{MD}$

$\therefore\;MD=\dfrac{xz}{y}$

In $\triangle$ ABC, P and Q are points on sides AB and AC respectively such that PQ $\parallel$ BC. If AP = 3 cm, PB = 5 cm and AC = 8 cm, find AQ.

1568602_969539_ans_c5d8b89a8d4a4b15b7bca0b77c288b7a.jpg

ABCD is a parallelogram, P is a point on side BC such that DP produced meets AB produced at L.. Prove that

(i) $\dfrac { DP }{ PL } =\dfrac { DC }{ BL }$

(ii) $\dfrac { DL }{ DP } =\dfrac { AL }{ DC }$

Given A parallelogram ABCD in which P is a point on side BC such that DP produced meets AB produced at L.

PROOF (i) In

$\triangle ALD$ , we have

$BP||AD$

$\therefore$

$\dfrac { LB }{ BA } =\dfrac { LP }{ PD }$

$\Rightarrow$

$\dfrac { BL }{ AB } =\dfrac { PL }{ DP }$

$\Rightarrow$

$\dfrac { BL }{ DC } =\dfrac { PL }{ DP }$ [because AB=DC]

$\Rightarrow$

$\dfrac { DP }{ PL } =\dfrac { DC }{ BL }$ [Taking reciprocals of both sides]

$[Hence \ proved]$

(ii) From (i), we have

$\dfrac { DP }{ PL } =\dfrac { DC }{ BL }$

$\Rightarrow$

$\dfrac { PL }{ DP } =\dfrac { BL }{ DC }$ [Taking reciprocals of both sides]

$\Rightarrow$

$\dfrac { PL }{ DP } =\dfrac { BL }{ AB }$ [because DC=AB]

$\Rightarrow$

$\dfrac { PL }{ DP } +1=\dfrac { BL }{ AB } +1$ [Adding 1 on both sides]

$\Rightarrow$

$\dfrac { DP+PL }{ DP } =\dfrac { BL+AB }{ AB }$

$\Rightarrow$

$\dfrac { DL }{ DP } =\dfrac { AL }{ AB }$

$\Rightarrow$

$\dfrac { DL }{ DP } =\dfrac { AL }{ DC }$

$[\because AB=DC]$

$[Hence \ proved]$

Let X be any point on the side BC of a triangle ABC. If XM, XN are drawn parallel to BA and CA, meeting CA, BA in M, N respectively. M and N are joined and produced to meet the line passing through BC at T as shown in the figure. Prove that ${ TX }^{ 2 }=TB\times TC$ .

In $\triangle TXM$ , we have

$XM||BN$ { $\because XM||AB$ }

$\therefore$ $\dfrac { TB }{ TX } =\dfrac { TN }{ TM }$ ........(i)

In $\triangle TMC$ , we have

$XN||CM$ { $\because XN||AC$ }

$\therefore$ $\dfrac { TX }{ TC } =\dfrac { TN }{ TM }$ .........(ii)

From equation (i) and (ii), we get

$\dfrac { TB }{ TX } =\dfrac { TX }{ TC }$

$\Rightarrow$ ${ TX }^{ 2 }=TB\times TC$ (hence proved)

In given figure, (i) and (ii), $PQ||BC$ . Find $QC$ in (i) and $AQ$ in (ii).

In figure (i), we have

$PQ||BC$

Therefore, by basic proportionality theorem, we have

$\dfrac { AP }{ PB } =\dfrac { AQ }{ QC }$

$\Rightarrow$

$\dfrac { 1.5 }{ 3 } =\dfrac { 1.3 }{ QC }$

$\Rightarrow$

$\dfrac { 1 }{ 2 } =\dfrac { 1.3 }{ QC }$

$\Rightarrow$

$QC=2.6 cm$

In figure (ii), it is given that

$PQ||BC$ .

Therefore, by basic proportionlity theorem, we have

$\dfrac { AP }{ PB } =\dfrac { AQ }{ QC }$

$\Rightarrow$

$\dfrac { 3 }{ 6 } =\dfrac { AQ }{ 5.3 }$

$\Rightarrow$

$\dfrac { 1 }{ 2 } =\dfrac { AQ }{ 5.3 }$

$\Rightarrow$

$AQ=\dfrac { 5.3 }{ 2 } =2.65 cm$

Any point X inside $\triangle DEF$ is joined to its vertices. From a point $P$ on $DX,\ PQ$ is drawn parallel to $DE$ meeting $XE$ at $Q$ and $QR$ is drawn parallel to $EF$ meeting $XF$ at $R.$ Prove that $PR\parallel DF.$

Given

$\triangle DEF$ and a point

$X$ inside it. Point

$X$ is joined to the vertices

$D,\ E$ and

$F.\ P$ is any point on

$DX.\ PQ\parallel DE$ and

$QR\parallel EF$

To prove:

$PR\parallel DF$

Proof:

Let us join

$PR$

$\triangle XED,$

we have,

$PQ\parallel DE$

$\therefore$

$\dfrac { XP }{ PD } =\dfrac { XQ }{ QE }$ . . . (i)

$\triangle XEF,$

we have,

$QR\parallel EF$

$\therefore$

$\dfrac { XQ }{ QE } =\dfrac { XR }{ RF }$ . . . (ii)

From (i) and (ii),

we have,

$\dfrac { XP }{ PD } =\dfrac { XR }{ RF }$

Thus, in

$\triangle XFD$ , points

$R$ and

$P$ are dividing sides

$EF$ and

$XD$ in the same ratio.

Therefore, by the converse of Basic Proportionality Theorem, we get

$PR\parallel DE.$

In the given figure, If PQ||BC and PR||CD. Prove that

(i) $\dfrac { AR }{ AD } =\dfrac { AQ }{ AB }$

(ii) $\dfrac { QB }{ AQ } =\dfrac { DR }{ AR }$

$\triangle ABC$ , we have

$PQ||BC$

Therefore, by basic proportionality theorem, we have

$\dfrac { AQ }{ AB } =\dfrac { AP }{ AC }$ ........(i)

$\triangle ACD$ , we have

$PR||CD$

Therefore, by basic proportionality theorem, we have

$\dfrac { AP }{ AC } =\dfrac { AR }{ AD }$

From (i) and (ii), we obtain that

$\dfrac { AQ }{ AB } =\dfrac { AR }{ AD }$ or

$\dfrac { AR }{ AD } =\dfrac { AQ }{ AB }$

$[Hence\ proved]$

$\Rightarrow$

$\dfrac { AB }{ AQ } =\dfrac { AD }{ AR }$

$\Rightarrow$

$\dfrac { AQ+QB }{ AQ } =\dfrac { AR+RD }{ AR }$

$\Rightarrow$

$1+\dfrac { QB }{ AQ }$

$=1+\dfrac { RD }{ AR }$

$\Rightarrow \dfrac { QB }{ AQ } =\dfrac { DR }{ AR }$

$[Hence\ proved]$

In given figure, $DE||AC$ and $DC||AP$ . Prove that $\dfrac { BE }{ EC } =\dfrac { BC }{ CP }$

In $\triangle BPA$ , we have

$DC||AP$ [Given]

Therefore, by basic proportionality theorem, we have

$\dfrac { BC }{ CP } =\dfrac { BD }{ DA }$

In $\triangle BCA$ , we have

$DE||AC$

Therefore, by basic proportionality theorem, we have

$\dfrac { BE }{ EC } =\dfrac { BD }{ DA }$

From (i) and (ii), we get

$\dfrac { BC }{ CP } =\dfrac { BE }{ EC }$ or $\dfrac { BE }{ EC } =\dfrac { BC }{ CP }$ $[Hence \ proved]$

In given figure $D$ and $E$ are respectively the points on the sides $AB$ and $AC$ of a $\triangle ABC$ such that $AB=5.6\ cm, AD=1.4\ cm, AC=7.2\ cm$ and $AE=1.8\ cm$ . Then show that $DE||BC$ .

We have

$AB=5.6 cm, AD=1.4 cm, AC=7.2 cm \ and \ AE=1.8 cm.$

$\therefore$

$BD=AB-AD=(5.6-1.4)$ cm

$=$

$4.2$ cm
and ,

$EC=AC-AE=(7.2-1.8)$ cm

$=$

$5.4$ cm

Now,

$\dfrac { AD }{ DB } =\dfrac { 1.4 }{ 4.2 } =\dfrac { 1 }{ 3 }$ and

$\dfrac { AE }{ EC } =\dfrac { 1.8 }{ 5.4 } =\dfrac { 1 }{ 3 }$

$\Rightarrow$

$\dfrac { AD }{ DB } =\dfrac { AE }{ EC }$

Thus, DE divides sides AB and AC of

$\triangle ABC$ in the same ratio.
Therefore, by the converse of Basic Proportionality Theorem, we have

$DE||BC$

In given figure, $DE||BC$ and $CD||EF$ . Prove that ${ AD }^{ 2 }=AB\times AF$

$\triangle ABC$ , we have

$DE||BC$

$\Rightarrow$

$\dfrac { AB }{ AD } =\dfrac { AC }{ AE }$ ........(i)

$\triangle ADC$ , we have

$FE||DC$

$\Rightarrow$

$\dfrac { AD }{ AF } =\dfrac { AC }{ AE }$ ..........(ii)

From (i) and (ii), we get

$\dfrac { AB }{ AD } =\dfrac { AD }{ AF }$

$\Rightarrow$

${ AD }^{ 2 }=AB\times AF$

$[Hence \ proved]$

In the given figure ABC is a triangle in which AB=AC. Points D and E are points on the sides AB and AC respectively such that AD=AE. Show that the points B,C,E and D are concyclic.

In order to prove that the points B,C,E and D are concyclic, it is sufficient to show that

$\angle ABC+\angle CED=180$0$ and

$\angle ACB+\angle BDE=180^0$ .

$\triangle ABC$ , we have

$AB=AC$ and

$AD=AE$

$\Rightarrow$

$AB-AD=AC-AE$

$\Rightarrow$

$DB=EC$

Thus, we have

$AD=AE$ and

$DB=EC$

$\Rightarrow$

$\dfrac { AD }{ DB } =\dfrac { AE }{ EC }$

$\Rightarrow$

$DE||BC$ [By the converse of Thale's Theorem]

$\Rightarrow$

$\angle ABC=\angle ADE$ [Corresponding angles]

$\Rightarrow$

$\angle ABC+\angle BED=\angle ADE+\angle BDE$ [Adding

$\angle BDE$ both sides]

$\Rightarrow$

$\angle ABC+\angle BDE=180^0$

$\Rightarrow$

$\angle ACB+\angle BDE=180^0$ [

$\because AB=AC\therefore\angle ABC=\angle ACB$ ]

Again,

$DE||BC$

$\Rightarrow$

$\angle ACB=\angle AED$

$\Rightarrow$

$\angle ACB+\angle CED=\angle AED+\angle CED$ [Adding

$\angle CE$ on both sides]

$\Rightarrow$

$\angle ACB+\angle CED=180^0$

$\Rightarrow$

$\angle ABC+\angle CED=180^0$ [

$\because \angle ABC=\angle ACB$ ]

Thus, BDEC is a cyclic quadrilateral. Hence, B,C,E and D are concyclic points.

In given figure, $\angle CAB=90^0$ and $AD \bot BC$ . If AC=75 cm. AB=1 m and BD=1.25 m, find AD

We have,

AB=1 m=100 cm, AC=75 cm and BD=125 cm.

$\triangle BAC$ and

$\triangle BDA$ , we have

$\angle BAC=\angle BDA$ [Each equal to

$90^\circ$ ]

and,

$\angle B\ =\angle B$

So, by AA-criterion of similarity, we have

$\triangle BAC\sim \triangle BDA$

$\Rightarrow$

$\dfrac { BA }{ BD } =\dfrac { AC }{ AD }$

$\Rightarrow$

$\dfrac { 100 }{ 125 } =\dfrac { 75 }{ AD }$

$\Rightarrow$

$AD=\dfrac { 125\times 75 }{ 100 } cm=93.75 cm$

In the figure given below, $\angle P = \angle RTS$ .
Prove that $\triangle RPQ \sin \triangle RTS$ .

In triangles RPQ and RTS, we have

$\angle RPQ=\angle RTS$ [Given]

$\angle PRQ=\angle TRS$ [Each equal to R]

Therefore, by AA-criterion of similarity, we have

$\triangle RPQ\sim \triangle RTS$ .

$[Hence \ proved]$

$D$ and $E$ are points on sides $AB$ and $AC$ respectively such that $BD=CE$ . If $\angle B=\angle C$ , show that $DE||BC$ .

$\triangle ABC$ , we have

$\angle B=\angle C$ .

$\Rightarrow AC=AB$ [Sides opposite to equal angles are equal]

$\Rightarrow AE+EC=AD=+DB$

$\Rightarrow AE+CE=AD+BD$

$\Rightarrow AE+CE=AD+CE$ [because BD=CE]

$\Rightarrow AE=AD$

Thus, we have

$AD=AE$ and

$BD=CE$

$\therefore \dfrac { AD }{ BD } =\dfrac { AE }{ CE }$

$\Rightarrow \dfrac { AD }{ DB } =\dfrac { AE }{ EC }$

$\Rightarrow DE||BC$ [By the converse of Basic Proportionality Theorem]

$[hence \, proved]$

1038101_1008479_ans_1492aa6ad216445eacbfe24e715a36b5.png

Two triangles ABC and DBC lie on the same side of the base BC. From a point P on BC, PQ||AB and PR||BD are drawn. They meet AC in Q and DC in R respectively. Prove that QR||AD.

Given Two triangles ABC and DBC lie on the same side of the base BC. Points P,Q and R are points on BC,AC and CD respectively such that PR||BD and PQ||AB.

To prove QR||AD

Proof In

$\triangle ABC$ , we have

$PQ||AB$

$\therefore$

$\dfrac { CP }{ PB } =\dfrac { CQ }{ QA }$ ........(i) [By Basic proportionality Theorem]

$\triangle BCD$ , we have

$PR||BD$

$\therefore$

$\dfrac { CP }{ PB } =\dfrac { CR }{ RD }$ ........(ii) [By Thale's Theorem]

From (i) and (ii), we have

$\dfrac { CQ }{ QA } =\dfrac { CR }{ RD }$

Thus, in

$\triangle ACD$ , Q and R are points on AC and CD respectively such that

$\dfrac { CQ }{ QA } =\dfrac { CR }{ RD }$

$\Rightarrow$

$QR||AD$ [By the converse of Basic Proportionality Theorem]

Let $ABC$ be a triangle and $D$ and $E$ be two points on side $AB$ such that $AD=BE$ . If $DP||BC$ and $EQ||AC$ , then prove that $PQ||AB$ .

$\triangle ABC$ , we have

$DP||BC$ and

$EQ||AC$

$\therefore$

$\dfrac { AD }{ DB } = \dfrac { AP }{ PC }$ and

$\dfrac { BE }{ EA } = \dfrac { BQ }{ QC }$

$\Rightarrow$

$\dfrac { AD }{ DB } = \dfrac { AP }{ PC }$ and

$\dfrac { AD }{ DB } = \dfrac { BQ }{ QC }$

$\Rightarrow$

$\dfrac { AP }{ PC } = \dfrac { BQ }{ QC }$

$\Rightarrow$ In a

$\triangle ABC$ , P and Q divide sides CA and CB respectively in the same ratio.

$\therefore$

$PQ||AB$ [By the converse of Basic Proportionality Theorem]

In given figure, If $\angle A=\angle C$ , then prove that $\triangle AOB\sim \triangle COD$ .

In triangles AOB and COD, we have

$\angle A=\angle C$ [Given]

and,

$\angle 1=\angle 2$ [Vertically opposite angles]

Therefore, by AA-criterion of similarity, we have

$\triangle AOB\sim \triangle COD$

$[Hence \ proved]$

In given figure, If $\dfrac { AD }{ DC } =\dfrac { BE }{ EC }$ and $\angle CDE=\angle CED$ , prove that $\triangle CAB$ is isosceles.

$\triangle ABC$ , we have

$\dfrac { AD }{ DC } =\dfrac { BE }{ EC }$ [Given]

Therefore, by the converse of basic proportionality theorem, we have,

$DE||AB$

$\Rightarrow$

$\angle CDE=\angle CAB$ and

$\angle CED=\angle CBA$ [Corresponding angles]

But,

$\angle CDE=\angle CED$ [Given]

$\therefore$

$\angle CAB=\angle CBA$

$\Rightarrow$

$\angle A=\angle B$

$\Rightarrow$

$BC=AC$ [

$\because$ Sides opposite to equal angles are equal]

$\Rightarrow$

$\triangle CAB$ is isosceles.

In $\triangle ABC$ , if AD is the bisector of $\angle A$ , prove that

$\dfrac { Area(\triangle ABD) }{ Area(\triangle ACD) } =\dfrac { AB }{ AC }$

$\triangle ABC$ , AD is the bisector of

$\angle A$ .

$\therefore$

$\dfrac { AB }{ AC } =\dfrac { BD }{ DC }$ .......(i)

From A, draw AL

$\bot$ BC.

$\therefore$

$\dfrac { Area(\triangle ABD) }{ Area(\triangle ACD) } =\dfrac { (1/2)BD.AL }{ (1/2)DC.AL }$

$\Rightarrow$

$\dfrac { Area(\triangle ABD) }{ Area(\triangle ACD) } =\dfrac { BD }{ DC }$

$\Rightarrow$

$\dfrac { Area(\triangle ABD) }{ Area(\triangle ACD) } =\dfrac { AB }{ AC }$ .........

$[From \ (i)]$

$[Hence \ proved]$

The side BC of a triangle ABC is bisected at D; O is any point in AD. BO and CO produced meet AC and AB in E and F respectively and AD is produced to X so that D is the mid-point of OX. Prove that AO:AX=AF:AB and show that FE||BC.

According to the question,

${\rm{BD}} = {\rm{CD}}$ and ${\rm{OD}} = {\rm{DX}}$

Therefore, BC and OX bisect each other and OBXC is a parallelogram.

This gives ${\rm{BX}}\parallel {\rm{CO}}$ and ${\rm{CX}}\parallel {\rm{BO}}$

or ${\rm{BX}}\parallel {\rm{CF}}$ and ${\rm{CX}}\parallel {\rm{BE}}$

or ${\rm{BX}}\parallel {\rm{OF}}$ and ${\rm{CX}}\parallel {\rm{OE}}$

In $\Delta {\rm{ABX}}$ , as ${\rm{BX}}\parallel {\rm{OF}}$ , then,

$\dfrac{{{\rm{AO}}}}{{{\rm{AX}}}} = \dfrac{{{\rm{AF}}}}{{{\rm{AB}}}}$ .............(1)

In $\Delta {\rm{ACX}}$ , as ${\rm{CX}}\parallel {\rm{OE}}$ , then,

$\dfrac{{{\rm{AO}}}}{{{\rm{AX}}}} = \dfrac{{{\rm{AE}}}}{{{\rm{AC}}}}$ ............. (2)

From equation (1) and (2),

$\dfrac{{{\rm{AF}}}}{{{\rm{AB}}}} = \dfrac{{{\rm{AE}}}}{{{\rm{AC}}}}$

Hence, as E and F divides AB and AC respectively in the same ratio, so, ${\rm{FE}}\parallel {\rm{BC}}$ .

In given figure D is a point on the side BC of $\triangle ABC$ such that $\angle ADC=\angle BAC$ . Prove that

$\dfrac { CA }{ CD } =\dfrac { CB }{ CA }$ or ${ CA }^{ 2 }=CB\times CD$ .

$\triangle ABC$ and

$\triangle DAC$ , we have

$\angle ADC=\angle BAC$ and

$\angle C=\angle C$

Therefore, by AA-criterion of similarity, we have

$\triangle ABC\sim \triangle DAC$

$\Rightarrow$

$\dfrac { AB }{ DA } =\dfrac { BC }{ AC } =\dfrac { AC }{ DC }$

$\Rightarrow$

$\dfrac { CB }{ CA } =\dfrac { CA }{ CD }$

$[Hence \ proved]$

In given figure ABC is a triangle in which AB=AC and D is a point on AC such that ${ BC }^{ 2 }=AC\times CD$ . Prove that BD=BC.

Given

$\triangle ABC$ in which

$AB=AC$ and D is a point on the side AC such that

${ BC }^{ 2 }=AC\times CD$

To prove:

$BD=BC$

construction:

$join \ BD$

we have,

${ BC }^{ 2 }=AC\times CD$

$\Rightarrow$

$\dfrac { BC }{ CD } =\dfrac { AC }{ BC }$ ............(i)

Thus, in

$\triangle ABC$ and

$\triangle BDC$ , we have

$\dfrac { AC }{ BC } =\dfrac { BC }{ CD }$ [From (i)] and,

$\angle C=\angle C$ [Common]

$\therefore$

$\triangle ABC\sim \triangle BDC$ [By SAS criterion of similarity]

$\Rightarrow$

$\dfrac { AB }{ BD } =\dfrac { BC }{ DC }$

$\Rightarrow$

$\dfrac { AC }{ BD } =\dfrac { BC }{ CD }$ [

$\because$ AB=AC]

$\Rightarrow$

$\dfrac { AC }{ BC } =\dfrac { BD }{ CD }$ .............(ii)

From (i) and (ii), we get

$\dfrac { BC }{ CD } =\dfrac { BD }{ CD }$

$\Rightarrow BD=BC$

$[Hence \ proved]$

In the given figure $P$ and $Q$ are points on sides $AB$ and $AC$ respectively of $\triangle ABC.$ If $AP=3 \;cm, \;PB= 6\; cm, \;AQ=5\; cm$ and $QC=10\; cm,$ show that $BC=3 PQ.$

We have,

$AB=AP+PB$

$=(3+6)\;cm$

$=9 \;cm$

$AC=AQ+QC$

$=(5+10)\;cm$

$=15 \;cm$

$\therefore$

$\dfrac { AP }{ AB } =\dfrac { 3 }{ 9 }$

$=\dfrac { 1 }{ 3 }$

And,

$\dfrac { AQ }{ AC } =\dfrac { 5 }{ 15 }$

$=\dfrac { 1 }{ 3 }$

$\Rightarrow$

$\dfrac { AP }{ AB } =\dfrac { AQ }{ AC }$

Thus, in

$\triangle APQ$ and

$\triangle ABC,$ we have

$\dfrac { AP }{ AB } =\dfrac { AQ }{ AC }$ and

$\angle A=\angle A$ [Common]

Therefore, by

$SAS$ -criterion of similarity, we have

$\triangle APQ\sim \triangle ABC$

$\Rightarrow$

$\dfrac { AP }{ AB } =\dfrac { PQ }{ BC } =\dfrac { AQ }{ AC }$

$\Rightarrow$

$\dfrac { PQ }{ BC } =\dfrac { 1 }{ 3 }$

$\Rightarrow BC=3PQ$

Hence proved.

In given figure, considering triangles BEP and CPD, prove that

$BP\times PD=EP\times PC$

Given a

$\triangle ABC$ in which

$BD\bot AC$ and

$CE\bot AB$ and BD and CE intersect at P.

To prove

$BP\times PD=EP\times PC$

Proof In

$\triangle EPB$ and

$\triangle DPC$ , we have

$\angle PEB=\angle PDC$ [Each equal to

$90^\circ$ ]

$\angle EPB=\angle DPC$ [Vertically opposite angles]

Thus, by AA-criterion of similarity, we have

$\triangle EPB\sim \triangle DPC$

$\dfrac { EP }{ DP } =\dfrac { PB }{ PC }$

$\Rightarrow$

$BP\times PD=EP\times PC$

$[Hence \ proved]$

In given figure through the vertex D of a parallelogram ABCD, a line is drawn to intersect the sides BA and BC produced at E and F respectively. Prove that

$\dfrac { DA }{ AE } =\dfrac { FB }{ BE } =\dfrac { FC }{ CD }$

$\triangle 's EAD$ and

$DCF$ , we have

$\angle 1=\angle 2$ [

$\because AB||DC\therefore$ Corresponding angles are equal]

$\angle 3=\angle 4$ [

$\because AD||BC\therefore$ Corresponding angles are equal]

Therefore, by AA-criterion of similarity, we have

$\triangle EAD\sim \triangle DCF$

$\Rightarrow$

$\dfrac { EA }{ DC } =\dfrac { AD }{ CF } =\dfrac { DE }{ FD }$

$\Rightarrow$

$\dfrac { EA }{ DC } =\dfrac { AD }{ CF }$

$\Rightarrow$

$\dfrac { AD }{ AE } =\dfrac { CF }{ CD }$ ...........(i)

Now, in

$\triangle 's EAD$ and

$EBF$ , we have

$\angle 1=\angle 1$ [Common angle]

$\angle 3=\angle 4$

So, by AA-criterion of similarity, we have

$\triangle EAD\sim \triangle EBF$

$\Rightarrow$

$\dfrac { EA }{ EB } =\dfrac { AD }{ BF } =\dfrac { ED }{ EF }$

$\Rightarrow$

$\dfrac { EA }{ EB } =\dfrac { AD }{ BF }$

$\Rightarrow$

$\dfrac { AD }{ AE } =\dfrac { FB }{ BE }$

From (i) and (ii), we have

$\dfrac { AD }{ AE } =\dfrac { FB }{ BE } =\dfrac { CF }{ CD }$

$[Hence \ proved]$

In the figure given below, $E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$ . Show that $\triangle ABE \sim \triangle CFB$ .

$\triangle 's ABE$ and

$CFB$ , we have

$\angle AEB=\angle CBF$ [Alternate angles]

$\angle A=\angle C$ [Opposite angles of a parallelogram]

Thus, by AA-criterion of similarity, we have

$\triangle ABE\sim \triangle CFB$ .

$[Hence \ proved]$

In given figure the diagonal $BD$ of a parallelogram $ABCD$ intersects the segment $AE$ at the point $F$ , where $E$ is any point on the side BC. Prove that $DF\times EF=FB\times FA$

$\triangle AFD$ and

$\triangle BFE$ , we have

$\angle 1=\angle 2$ [Vertically opposite angles]

$\angle 3=\angle 4$ [Alternate angles]

So, by AA-criterion of similarity,we have

$\triangle FBE\sim \triangle FDA$

$\Rightarrow$

$\dfrac { FB }{ FD } =\dfrac { FE }{ FA }$

$\Rightarrow$

$\dfrac { FB }{ DF } =\dfrac { FE }{ FA }$

$\Rightarrow$

$DF\times EF=FB\times FA$

$[Hence \ proved]$

In given figure two triangles BAC and BDC, right angled at A and D respectively, are drawn on the same base BC and on the same side of BC. If AC and DB intersect at P, prove that $AP\times PC=DP\times PB$

$\triangle APB$ and

$\triangle DPC$ , we have

$\angle A=\angle D=90^\circ$ and

$\angle APB=\angle DPC$ [Vertically opposite angles]

Thus, by AA-criterion of similarity, we have

$\triangle APB\sim \triangle DPC$

$\Rightarrow$

$\dfrac { AP }{ DP } =\dfrac { PB }{ PC }$

$\Rightarrow$

$AP\times PC=DP\times PB$

$[Hence \ proved]$

In a $\triangle ABC$ BD and CE are the altitudes. Prove that $\triangle ADB$ and $\triangle AEC$ are similar. Check whether $\triangle CDB\ \ and\ \triangle BEC$ are similar.

$\triangle ABD$ and

$\triangle AEC$ , we have

$\angle ADB=\angle AEC$ [Each equal to

$90^\circ$ ]

$\angle BAD=\angle EAC$ [Common]

So, by AA-criterion of similarity, we have

$\triangle BDA\sim \triangle CEA$ or,

$\triangle ADB\sim \triangle AEC$ .

Clearly,

$\triangle CDB$ is not similar to

$\triangle BEC$ , because they are not equiangular.

In given figure, express x in terms of a, b and c.

$\triangle KPN$ and

$\triangle KLM$ , we have

$\angle KNP=\angle KML=46^\circ$ [Given]

$\angle K=\angle K$ [Common]

$\therefore$

$\triangle KNP\sim \triangle KML$ [By AA-criterion of similarity]

$\Rightarrow$

$\dfrac { KN }{ KM } =\dfrac { NP }{ ML }$ [

$\because$ Corresponding sides of similar triangles are proportional]

$\Rightarrow$

$\dfrac { c }{ b+c } =\dfrac { x }{ a }$

$\Rightarrow x=\dfrac { ac }{ b+c }$

In $\Delta ABC$ , if $EF || AB$ and $ar(\Delta CEF) = ar(\Box EFBA)$ , then find ratio of $CA$ and $EA$ .

Draw a perpendicular from

$C$ onto

$AB$ to meet

$AB$ at

$G$ and

$EF$ at

$H$

Area

$\triangle ABC=1/2\times AB\times CG$

$=ar(EFBA)+ar(CEF)$

$=2\times \triangle CEF=EF\times CH$

$(because\ ar(EFBA)=ar(CEF)$ and

$ar(CEF)=1/2\times CH\times EF)$

$AB*CG=2\times EF\times CH ..... (i)$

$\triangle ABC$ and

$\triangle CEF$ are similar triangles. So, the ratios of their corresponding sides and altitudes are equal.

$EF/AB=CE/CA=CF/CB=CH/CG$

$CH=EF*CG/AB$

$(1)= > AB*CG=2\ EF^{2}*CG/AB$

$AB^{2}=2\ EF^{2}$

$AB\surd{2}EF$

$AB/EF=CA/CE=\surd{2}$

The ratio of the sides in the two triangles is

$\surd{2}$

$CA=\surd{2}CE$

$CA/EA=CA/(CA-CE)=\surd{2}CE/(\surd{2}CE-CE)$

$CA/EA=\surd{2}/(\surd{2}-1)=\surd{2}(\surd{2}+1)$

1432095_1027944_ans_20ea3f45c24142d38b43fc01c8ff728c.png

If the areas of two similar triangles are equal, prove that they are concurrent.

$Let\quad \triangle les\quad be\triangle ABC\quad \& \quad \triangle DEF.$

$Both\quad \triangle les\quad are\quad similar.$

$\implies\quad \triangle ABC\sim \triangle DEF$

$areas\quad are\quad equal.So,ar\triangle ABC=ar\triangle DEF$

$We\quad know\quad that\quad if\quad two\quad \triangle les\quad are\quad similar,$

$Ratio\quad of\quad area\quad is\quad equal\quad to\quad square\quad of\quad ratio$

$of\quad its\quad corresponding\quad sides.$

$\cfrac { ar(\triangle ABC) }{ ar(\triangle DEF) } ={ (\cfrac { BC }{ EF } ) }^{ 2 }={ (\cfrac { AB }{ DE } ) }^{ 2 }={ (\cfrac { AC }{ DF } ) }^{ 2 }$

$(given\quad as\quad \triangle ABC=\triangle DEF)$

$1={ (\cfrac { BC }{ EF } })^{ 2 }={ (\cfrac { AB }{ DE } ) }^{ 2 }={ (\cfrac { AC }{ DF } ) }^{ 2 }$

$1={ (\cfrac { BC }{ EF } ) }^{ 2 }$

$\implies\quad 1=\cfrac { BC }{ EF }$

$\therefore BC=EF$

$1={ (\cfrac { AB }{ DE } ) }^{ 2 }$

$1=\cfrac { AB }{ DE }$

$AB=DE$

$1={ (\cfrac { AB }{ DE } ) }^{ 2 }$

$AC=DF$

$In\triangle ABC\& \triangle DEF$

$EF=BC,\quad AB=DE,\quad AC=DF$

$Hence\quad by\quad SSS\quad congruency.$

$\triangle ABC\cong \triangle DEF$

$hence\quad proved$

1021834_1030646_ans_a35f904a00f44a578818eca8f552181f.png

In given figure, DEFG is a square and $\angle BAC=90^0$ . Prove that

(i) $\triangle AGF\sim \triangle DBG$

(ii) $\triangle AGF\sim \triangle EFC$

(iii) $\triangle DBG\sim \triangle EFC$

(iv) ${ DE }^{ 2 }=BD\times EC$

(i) In $\triangle AGF$ and $\triangle DBG$ , we have

$\angle GAF=\angle BDG$ [EAch equal to $90^0$ ] and,

$\angle AGF=\angle DBG$ [Corresponding angles]

$\therefore$ $\triangle AGF\sim \triangle DBG$ [By AA-criterion of similarity] $[Hence \ proved]$

(ii) In $\triangle AGF$ and $\triangle EFC$ , we have

$\angle FAG=\angle CEF$ [Each equal to $90^0$ ] and,

$\angle AFG=\angle ECF$ [Corresponding angles]

$\therefore$ $\triangle AGF\sim \triangle EFC$ [By AA-criterion of similarity] $[Hence \ proved]$

(iii) Since $\triangle AGF\sim \triangle DBG$ and $\triangle AGF\sim \triangle EFC$

$\therefore$ $\triangle DBG\sim \triangle EFC$ $[Hence \ proved]$

(iv) we have,

$\triangle DBG\sim \triangle EFC$ [Using (iii)]

$\therefore$ $\dfrac { BD }{ EF } =\dfrac { DG }{ EC }$

$\Rightarrow$ $\dfrac { BD }{ DE } =\dfrac { DE }{ EC }$ [ $\because$ DEFG is a square $\therefore$ EF=DE, DG=DE]

$\Rightarrow$ ${ DE }^{ 2 }=BD\times EC$ $[Hence \ proved]$

In the given figure, $AB||CD||EF$ given $AB=7.5\ cm$ $DC=ycm$ $EF=4.5\ cm$ , $BC=x\ cm$ . Find the value of $x$ and $y$

Consider

$\triangle ACB$ and

$\triangle CEF$

Both triangles are similar

$\therefore \dfrac{AB}{EF}=\dfrac{x}{3}$

$\Rightarrow \dfrac{7.5}{4.5}=\dfrac{x}{3}$

$x=\dfrac{7.5}{4.5}\times 3$

$\therefore x=5$ cm

Consider

$\triangle BCD$ and

$\triangle BFE$

$\dfrac{x}{y}=\dfrac{x+3}{4.5}$

$\dfrac{5}{y}=\dfrac{5+3}{4.5}$

$\Rightarrow y=\dfrac{5\times 4.5}{8}$

$\therefore y=\dfrac{45}{16}$ cm

Given $\triangle ABC \sim \triangle PQR$ , if $\dfrac{AB}{PQ}=\dfrac{1}{3}$ , then find $\dfrac{ar (\triangle ABC)}{ar( \triangle PQR)}$ .

$\triangle ABC\sim \triangle PQR$

$\dfrac{AB}{PQ}=\dfrac{1}{3}$

$\dfrac{ar\triangle ABC}{ar\triangle PQR}=\left ( \dfrac{AB}{PQ} \right )^2=\left ( \dfrac{BC}{QR} \right )^2=\left ( \dfrac{AC}{PR} \right )^2$

$\dfrac{ar\triangle ABC}{ar\triangle PQR}=\left (\dfrac{1}{3} \right )^2=\dfrac{1}{9}$

Prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

$\triangle ABC$ ,

$P$ and

$Q$ are mid points of side

$AB$ and

$AC$ , respectively.

$AP = PB$ and

$AQ = QC$

$\dfrac{AP}{PB} = \dfrac{AQ}{QC} = 1$

Using converse theorem of Proportionality (If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.) we get,

$PQ\parallel BC$ Hence Proved.

957182_1019871_ans_b286427e8558464fa53454c099ce9d2c.JPG

In given figure if a perpendicular is drawn from the right angle vertex of a right triangle to the hypotenuse, then prove that the triangle on each side of the perpendicular is similar to each other and to the original triangle. Also, prove that the square of the perpendicular is equal to the product of the lengths of the two parts of the hypotenuse.

Given a right triangle

$ABC$ right angled at

$B$ and

$BD\bot AC$

To prove

(i)

$\triangle ADB\sim \triangle BDC$

(ii)

$\triangle ADB\sim \triangle ABC$

(iii)

$\triangle BDC\sim \triangle ABC$

(iv)

${ BD }^{ 2 }=AD\times DC$

Solution:

(i) In

$\triangle ADB$ and

$\triangle BDC$ , we have:

$\angle ABD+\angle DBC=90^0$

Also

$\angle C+\angle DBC+\angle BDC=180^0$

$\Rightarrow$

$\angle C+\angle DBC+90^0=180^0$

$\Rightarrow$

$\angle C+\angle DBC=90^0$

But,

$\angle ABD+\angle DBC=90^0$

$\therefore$

$\angle ABD+\angle DBC=\angle C+\angle DBC$

$\Rightarrow$

$\angle ABD=\angle C$ .........(1)

Thus, in

$\triangle ADB$ and

$\triangle BDC$ , we have:

$\angle ABD=\angle C$ [From (1)]

and,

$\angle ADB=\angle BDC$ [Each equal to

$90^0$ ]

So, by AA-similarity criterion,

$\triangle ADB\sim \triangle BDC$

$[Hence \ proved]$

(ii) In

$\triangle ADB$ and

$\triangle ABC$ , we have:

$\angle ADB=\angle ABC$ [Each equal to

$90^0$ ]

and,

$\angle A=\angle A$ [Common]

So, by AA-similarity criterion,

$\triangle ADB\sim \triangle ABC$

$[Hence \ proved]$

(iii) In

$\triangle BDC$ and

$\triangle ABC$ , we have:

$\angle BDC=\angle ABC$ [Each equal to

$90^0$ ]

$\angle C=\angle C$ [Common]

So, by AA-similarity criterion,

$\triangle BDC\sim \triangle ABC$

$[Hence \ proved]$

(iv) From (i), we have

$\triangle ADB\sim \triangle BDC$

$\therefore$

$\dfrac { AD }{ BD } =\dfrac { BD }{ DC }$

$\Rightarrow$

${ BD }^{ 2 }=AD\times DC$

$[Hence \ proved]$

In given figure, PB and Qa are perpendiculars to segment AB. If PO=5 cm, QO=7 cm and Area $\triangle POB=150 {cm}^{2}$ find the area of $\triangle QOA$ .

$\triangle OAQ$ and

$\triangle OBP$ , we have

$\angle A=\angle B$ [Each equal to

$90^\circ$ ]

$\angle AOQ=\angle BOP$

So, by AA-criterion of similarity, we have

$\triangle AOQ\sim \triangle BOP$

$\Rightarrow$

$\dfrac { Area(\triangle AOQ) }{ Area(BOP) } =\dfrac { { OQ }^{ 2 } }{ { OP }^{ 2 } }$

$\Rightarrow$

$\dfrac { Area(\triangle AOQ) }{ 150 } =\dfrac { { 7 }^{ 2 } }{ { 5 }^{ 2 } }$

$\Rightarrow$

$Area(\triangle AOQ)=\dfrac { 49 }{ 25 } \times 150{cm}^{2}=294{cm}^{2}$

In given figure, $\dfrac { OA }{ OC } =\dfrac { OD }{ OB }$ . then prove that $\angle A=\angle C$ and $\angle B=\angle D$ .

$\triangle 's AOD$ and

$COB$ , we have

$\dfrac { OA }{ OC } =\dfrac { OD }{ OB }$

$\Rightarrow$

$\dfrac { OA }{ OD } =\dfrac { OC }{ OB }$

Also,

$\angle 1=\angle 2$ [Vertically opposite angles]

So, by SAS-criterion of similarity, we have

$\triangle AOD\sim \triangle COB$

$\Rightarrow$

$\angle A=\angle C$ and

$\angle B=\angle D$

$[Hence \ proved]$

In given figure, $\triangle FEC\cong \triangle GBD$ and $\angle 1=\angle 2$ . then prove that $\triangle ADE\sim \triangle ABC$ .

We have,

$\triangle FEC\cong \triangle GBD$

$\Rightarrow$

$EC=BD$ ..........(i)

It is given that

$\angle 1=\angle 2$

$\Rightarrow$

$AD=AE$ [Sides opposite to equal angles are equal].......(ii)

From (i) and (ii), we have

$\dfrac { AE }{ EC } =\dfrac { AD }{ BD }$

$\Rightarrow$

$DE||BC$ [By the converse of basic proportionality theorem]

$\Rightarrow$

$\angle 1=\angle 3$ and

$\angle 2=\angle 4$

Thus, in

$\triangle 's ADE$ and

$ABC$ , we have

$\angle A=\angle A$

$\angle 1=\angle 3$

$\angle 2=\angle 4$

So, by AAA-criterion of similarity, we have

$\triangle ADE\sim \triangle ABC$

$[Hence \ proved]$

State and prove pythagorus theorum.

Pythagoras theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides

Given :

$\triangle ABC$ right angles at

$B$

To prove :

$AC^2=AB^2+AC^2$

Construction : Draw

$DB\bot AC$

Proof: Since

$BD\bot AC$

Therefore by AAS silimalrity criterion,

$\triangle ADB\sim \triangle ABC\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad | \triangle BDC\sim \triangle ABC$

Since sides of similar triangles are in same ratio

$|$ since similar triangle sides are in same ratio then

$\dfrac{AD}{AB}=\dfrac{AB}{AC}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad | \dfrac{CD}{BC}=\dfrac{BC}{AC}$

$AD.AC=AB^2------(1)$

$|$

$AC.CD=BC^2-------(2)$

Add

$(1)$ and

$(2)$ we get,

$AD.AC+CD.AC=AB^2+BC^2$

$AC(AD+DC)=AB^2+BC^2$

$AC\times AC=AB^2+BC^2$

$AC^2=AB^2+BC^2$

Hence proved

1426867_1051078_ans_23ed8027bfcd44f4ba0deaa8206eeaf5.png

In the given figure, $CM$ and $RN$ are respectively the medians of $\triangle ABC$ and $\triangle PQR$ . If $\triangle ABC \sim \triangle PQR$ , prove that:
(i) $\triangle AMC \sim \triangle PQR$
(ii) $CM/ RN = AB/ PQ$
(iii) $\triangle CMB \sim \triangle RNQ$ .

Given :

$\triangle ABC \sim \triangle PQR$ 7

$CM$ and

$RN$ are medians of

$\triangle ABC$ &

$\triangle PQR$ respectively.

To prove :

$(i) \triangle AMC \sim \triangle PNR, (ii) \dfrac{CM}{RN} =\dfrac{AB}{PQ}, (iii) \triangle CMB \sim \triangle RNQ$

Proof :

$(i)$ since

$\triangle ABC \sim \triangle PQR.$

So,

$\angle A = \angle P$

$\angle B= \angle Q, \angle C= \angle R$

$\dfrac{AB}{PQ}= \dfrac{BC}{QR} =\dfrac{CA}{RP}, \dfrac{AB}{PQ}= \dfrac{x+x}{y+y} = \dfrac{2x}{2y} =\dfrac{x}{y}$

Let

$Am=MB= x$ (Given)

$PN= NQ= y$ (Given)

$\triangle AMC$ &

$\triangle PNR$

$\dfrac{x}{y}=\dfrac{AM}{PN}, \dfrac{x}{y} =\dfrac{AB}{PQ}=\dfrac{CA}{RP}$

So,

$\dfrac{AM}{PN}= \dfrac{CA}{RP}$ &

$\angle A= \angle P$

Therefore,

$\triangle AMC \sim \triangle PNR$ (by side angle side similarity)

$(ii)$ As

$\triangle AMC \sim \triangle PNR$

So,

$\dfrac{CM}{RN}= \dfrac{AM}{PN}$ Proved (by similar parts of similar triangles are similar)

$\dfrac{CM}{RN}= \dfrac{AM}{PN}=\dfrac{x}{y} =\dfrac{AB}{PQ}$ Proved

$(iii)$ In

$\triangle CMB$ &

$\triangle RNQ$

$\dfrac{BC}{QR}=\dfrac{AB}{PQ}= \dfrac{x}{y} = \dfrac{BM}{QN}= \dfrac{x}{y}$

So,

$\dfrac{BC}{QR} =\dfrac{BM}{QN}, \angle B= \angle Q$

Therefore ,

$\triangle CMB \sim \triangle RNQ$

(by

$SAS$ similarly )

1426768_1052192_ans_0aca58bd99ac45918f9c7007fde91475.png

State and Prove the relation between areas of two similar triangles.

Given :

$\triangle ABC \sim \triangle PQR$

To prove,

$\dfrac{Area(\triangle ABC)}{Area(\triangle PQR)}=\dfrac{BC^2}{QR^2}=\dfrac{AB^2}{PQ^2}=\dfrac{AC^2}{PR^2}$

Construction : Draw segment

$AD\bot$ side

$BC \,\,\&\,\,PS\bot$ side

$QR$

Proof:

$\triangle ABC \sim \triangle PQR$ (given)

$\therefore \dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AC}{PR}..........(i)$

$\angle B=\angle Q\,\,\, \&\,\,\, \angle C=\angle R...........(ii)$

Now in

$\angle ADB\,\,\&\,\, \triangle PSQ,$

$\angle ABD=\angle PQS$ (from ii)

$\angle ADB=\angle PSQ$ (

$=90^0$ )

$\therefore \triangle ADB\sim \triangle PSQ$ (A-A similarity)

$\therefore \dfrac{AB}{PQ}=\dfrac{BD}{QS}=\dfrac{AD}{PS}$

$\therefore$ From (i) and (iii)

$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PS}$ ......(iv)

Area of

$\triangle =\dfrac{1}{2}\times b\times h=\dfrac{\dfrac{1}{2}\times AD \times BC}{\dfrac{1}{2}\times PS\times QR}=\dfrac{Area(\triangle ABC)}{Area(\triangle PQR)}$

$=\dfrac{AD\times BC}{PS\times QR}=\dfrac{AD}{PS}\times \dfrac{BC}{QR}=\dfrac{BC}{QR}\times \dfrac{BC}{QR}[\because\dfrac{AD}{PS}=\dfrac{BC}{QR}]$

$=\dfrac{BC^2}{QR^2}$

$\therefore \dfrac{A\,of\, (\triangle ABC)}{A\,of\, (\triangle PQR)}=\dfrac{BC^2}{QR^2}=\dfrac{AB^2}{PQ^2}=\dfrac{AC^2}{PR^2}$

$\left[\because \dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AC}{PR}(from\,\,i)\right]$

1427651_1046644_ans_0242de9b2822448480e0fa8307c3e0d0.png

$AX$ and $DY$ are altitiudes of two triangles $\triangle ABC$ and $\triangle DEF$ . Prove that $AX:DY=AB:DE$ .

$\Delta ABC\, \, \& \, \, \Delta DEF\, \, are\, \, similar$

they are equangular

Draw

$\begin{array}{l} A\times \bot BC\, \, \& \, \, DY\bot EF \\ Now,\, \, \Delta A\times B\, \, \& \, \, \Delta DyE\, \, are\, \, equi-angles \\ by\, \, AAA\, \, swimilarity \\ AA\times B\, \& \, \, ADYE\, \, are\, \, similar \end{array}$

Their corresponding sides are proportional

$\begin{array}{l} \frac { { Ax } }{ { dy } } =\frac { { xB } }{ { yE } } =\frac { { AB } }{ { DE } } \\ \frac { { Ax } }{ { Dy } } =\frac { { AB } }{ { DE } } \end{array}$

How to prove areas of similar triangles theorem

Draw

$AM \bot BC\,\,\& \,\,PN \bot QR$

and

$\Delta ABC \sim \Delta PQR$

$\begin{array}{l} \frac { { ar\left( { \Delta ABC } \right) } }{ { ar\left( { \Delta PQR } \right) } } =\frac { { \frac { 1 }{ 2 } \times BC\times AM } }{ { \frac { 1 }{ 2 } \times QR\times PN } } & \\ { { Sin } }ce,\, \, \Delta ABC\sim \Delta PQR & \\ \therefore \frac { { AB } }{ { PQ } } =\frac { { BC } }{ { QR } } =\frac { { AC } }{ { PR } } & \\ In\, \, \Delta ABM\, \, \& \, \, \Delta PQN & \\ \angle B=\angle Q & \left[ { \because \Delta ABC\sim \Delta PQR } \right] \\ \angle M=\angle N=90^{ \circ } & \\ \therefore \Delta ABM\sim \Delta PQN & \\ \therefore \frac { { AB } }{ { PQ } } =\frac { { AM } }{ { PN } } & \\ \therefore \frac { { ar\left( { \Delta ABC } \right) } }{ { ar\left( { \Delta PQR } \right) } } =\frac { { AB } }{ { PQ } } \times \frac { { AB } }{ { PQ } } ={ \left( { \frac { { AB } }{ { PQ } } } \right) ^{ 2 } }={ \left( { \frac { { BC } }{ { QR } } } \right) ^{ 2 } }={ \left( { \frac { { AC } }{ { PR } } } \right) ^{ 2 } } & \end{array}$

1218930_1052933_ans_a0b43a65655444c5922d75d8aa888c51.PNG

In the Figure, if $LM \parallel CB$ and $LN \parallel CD$ , prove that $\dfrac{AM}{AB}=\dfrac{AN}{AD}$ .

In $\Delta ABC$ , as $LM\parallel BC$ , so by basic proportionality theorem,

$\dfrac{{AL}}{{AC}} = \dfrac{{AM}}{{AB}}$ ........... $(1)$

In $\Delta ADC$ , as $LN\parallel CD$ , so by basic proportionality theorem,

$\dfrac{{AL}}{{AC}} = \dfrac{{AN}}{{AD}}$ ........... $(2)$

From equation $(1)$ and $(2),$

$\dfrac{{AM}}{{AB}} = \dfrac{{AN}}{{AD}}$ .

990243_1056995_ans_a8b172b7516346df9326c3763b510bfe.PNG

In the given figure, $\dfrac{{QR}}{{QS}} = \dfrac{{QT}}{{PR}}\,\,and\,\,\angle 1 = \angle 2.$ Show that $\Delta PQS$ $\sim$ $\Delta TQR$

$\angle 1 = \angle 2 \Rightarrow PQ = PR$ (Property of Isosceles Triangle)

$Also,{\dfrac {QR} {QS}} = { \dfrac {QT} {PR}}$

$\Rightarrow {\dfrac {QR} {QS}} = {\dfrac {QT} {PQ}}$

$In \Delta PQS \ and \ \Delta TQR$

$\angle PQS = \angle TQR$ (common)

$\Delta PQS \sim \Delta TQR$ ........using SAS similarity criteria

$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\Delta PQR$ . For each of the following cases state whether $EF\parallel \,QR$
(1) $PE = 3.9\,cm,\,EQ = 3\,cm,\,PF = 3.6\,cm\,\,and\,\,FR = 2.4$
(2) $PE = 4\,cm,\,QE = 4.5\,cm,\,PF = 8\,cm\,\,\,and\,\,\,RF = 9\,cm$
(3) $PQ = 1.28\,cm\,{\mathop{\rm PR}\nolimits} = 2.56\,cm,\,PE = 0.18\,cm\,\,\,and\,PF = 0.36\,cm$

By basic proportionality theorm ,

$EF||QR$ then

$\dfrac{PE}{EQ}=\dfrac {PF}{FR}$

(1)

$PE = 3.9\,cm,\,EQ = 3\,cm,\,PF = 3.6\,cm\,\,and\,\,FR = 2.4$

$\dfrac {3.9}{3}=\dfrac {3.6}{2.4}\\\dfrac {1.3}{1}\neq\dfrac {3}{2}$

So EF is not parallel to QR

(2)

$PE = 4\,cm,\,QE = 4.5\,cm,\,PF = 8\,cm\,\,\,and\,\,\,RF = 9\,cm$

$\dfrac {4}{4.5}=\dfrac {8}{9}\\\dfrac {8}{9}=\dfrac {8}{9}$

So EF is parallel to QR

(3)

$PQ = 1.28\,cm\,{\mathop{\rm PR}\nolimits} = 2.56\,cm,\,PE = 0.18\,cm\,\,\,and\,PF = 0.36\,cm$

$\dfrac {1.28}{2.56}=\dfrac {0.18}{0.36}\\\dfrac {1}{2}=\dfrac {1}{2}$

So EF is parallel to QR

In A, B and C are points on OP, OQ and OR respectively such that AB $||$ PQ and AC $|$ PR. Show that BC $||$ QR.

$\triangle OPQ$ , we have

$AB\parallel PQ$

Therefore, by using basic proportionality theorem , we have

$\dfrac{OA}{AP}=\dfrac{OB}{BQ}.................(i)$

$\triangle OPR$ , we have

$AC\parallel PR$

Therefore, by using basic proportionality theorem , we have

$\dfrac{OC}{CR}=\dfrac{OA}{AP}.................(ii)$

Comparing (i)&(ii), we get

$\dfrac{OB}{BQ}=\dfrac{OC}{CR}$

Therefore, by using converse of basic proportionality theorem, we get

$BC\parallel QR$

1026567_1057375_ans_7344bca9939d454491f7cf721378c353.png

Write the statement of Basic proportionality theorem.

Basic Proportionality Theorem states that if a line is drawn parallel to any one side of the triangle that intersects the other two sides in two distinct points, then the line divides those two sides in the same ratio.

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR.
Sow that $\Delta \,ABC\, \sim \,\Delta \,PQR$ .

(Image 1)
Given:

$\triangle ABC$ &

$\triangle PQR$
AD is the median of

$\triangle ABC$
PM is the median of

$\triangle PQR$

$\cfrac { AB }{ PQ } =\cfrac { AC }{ PR } =\cfrac { AD }{ PM } \rightarrow 1$
To prove:

$\triangle ABC\sim \triangle PQR$
Proof:Let us extend AD to point D such that that

$AD=DE$ and PM upto point L such that

$PM=ML$
Join B to E, C to E, & Q to L and R to L
(Image 2)
We know that medians is the bisector of opposite side
Hence

$BD=DC$ &

$AD=DE$ *By construction)
Hence in quadrilateral ABEC, diagonals AE and BC bisect each other at point D

$\therefore ABEC$ is a parallelogram

$\therefore AC=BE$ &

$AB=EC$ (opposite sides of a parallelogram are equal)

$\rightarrow 2$
Similarly we can prove that

$PQLR$ is a parallelogram.

$PR=QL,PQ=LR$ (opposite sides of a parallelogram are equal)

$\rightarrow 3$
Given that

$\cfrac { AB }{ PQ } =\cfrac { AC }{ PR } =\cfrac { AD }{ PM }$ (frim

$1$ )

$\Rightarrow \cfrac { AB }{ PQ } =\cfrac { BE }{ QL } =\cfrac { AD }{ PM }$ (from

$2$ and

$3$ )

$\Rightarrow \cfrac { AB }{ PQ } =\cfrac { BE }{ QL } =\cfrac { 2AD }{ 2PM }$

$\Rightarrow \cfrac { AB }{ PQ } =\cfrac { BE }{ QL } =\cfrac { AE }{ PL }$ (As

$AD=DE,AE=AD+DE=AD+AD=2AD$ &

$PM=ML,PL=PM+ML=PM+PM=2PM$ )

$\therefore \triangle ABE\sim \triangle PQL$ (By SSS similarity criteria)
We know that corresponding angles of similar triangles are equal

$\therefore \angle BAE=\angle QPL\rightarrow 4$
Similarly we can prove that

$\triangle AEC\sim \triangle PLR$
We know that corresponding angles of similar triangles are equal

$\angle CAE=\angle RPL\rightarrow 5$
Adding

$4$ and

$5$ , we get

$\angle BAE+\angle CAE=\angle QPL+\angle RPL$

$\Rightarrow \angle CAB=\angle RPQ\rightarrow 6$
In

$\triangle ABC$ and

$\triangle PQR$

$\cfrac { AB }{ PQ } =\cfrac { AC }{ PR }$ (from

$1$ )

$\angle CAB=\angle RPQ$ (from

$6$ )

$\therefore \triangle ABC\sim \triangle PQR$ (By SAS similarity criteria)
Hence, proved

1139802_1063617_ans_3da7544d22554674b290ed92a64e7366.png

State converse of Pythagoras theorem.

Statement: Converse of phythogoren theorem states that in a triangle , if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle.

CM and RN are respectively the medians of $\Delta \,ABC$ and $\Delta \,PQR$ . If $\Delta \,ABC \sim \Delta \,PQR$ , prove that:

(A) $\Delta \,AMC \sim \,\Delta \,PNR$

(B) $\dfrac{{CM}}{{RN}} = \dfrac{{AB}}{{PQ}}$

$(A)$ Given,

$\triangle ABC$ and

$\triangle PQR$
CM is the median of

$\triangle ABC$
and RN is the median of

$\triangle PQR$

Also,

$\triangle ABC \sim \triangle PQR$

To prove:

$\triangle AMC \sim \triangle PNR$

Proof:
CM is the median of

$\triangle ABC$
So,

$AM=MB=\cfrac { 1 }{ 2 } AB\quad ..... (1)$
Similarly, RN is the median of

$\triangle PQR$
So,

$PN=QN=\cfrac { 1 }{ 2 } PQ\quad ..... (2)$

Given,

$\triangle ABC\sim \triangle PQR$

$\therefore \ \cfrac { AB }{ PQ } =\cfrac { BC }{ QR } =\cfrac { CA }{ RP }$ (Corresponding sides of similar triangle are proportional)

$\Rightarrow \cfrac { AB }{ PQ } =\cfrac { CA }{ RP } \Rightarrow \cfrac { 2AM }{ 2PN } =\cfrac { CA }{ RP }$ [From

$(1)$ and

$(2)$ ]

$\Rightarrow \cfrac { AM }{ PN } =\cfrac { CA }{ RP } \quad .....(3)$

Also, since

$\triangle ABC\sim \triangle PQR$

$\angle A=\angle P$ (corresponding angles of similar triangles are equal)

$\ \ ......(4)$

$\triangle AMC$ &

$\triangle PNR$ ,

$\angle A=\angle P$ [From

$(4)$ ]

Also,

$\cfrac { AM }{ PN } =\cfrac { CA }{ RP }$ [From

$(3)$ ]

Hence, by SAS similarity,

$\triangle AMC\sim \triangle PNR$

$[Hence\ proved]$

$(B)$

$\triangle AMC\sim \triangle PNR$

So,

$\cfrac { CM }{ RN } =\cfrac { AC }{ PR } =\cfrac { AM }{ PN }$ (corresponding sides of similar triangle are proportional)

Therefore,

$\cfrac { CM }{ RN } =\cfrac { AM }{ PN }$

$\cfrac { CM }{ RN } =\cfrac { 2AM }{ 2PN }$

$\Rightarrow \cfrac { CM }{ RN } =\cfrac { AB }{ PQ }$

$[Hence\ proved]$

1140029_1066826_ans_b98d218426a1464988a21b0f4c58ae6d.png

In the adjoining figure $AB \parallel CD$ , prove that $\Delta AOB \sim \Delta DOC$

$AB\parallel CD$ ( given )

$\triangle AOB$ and

$\triangle DOC$ ,

$\angle AOB=\angle DOC$ (Vertically opposite angles)

$\angle OAB=\angle ODC$ (Alternate angles)

$\angle ABO=\angle DCO$ (Alternate angles)

From

$AAA$ (angle-angle-angle) similarity theorem,

$\triangle AOB\sim \triangle DOC$

In the figure if $PQ \parallel RS$ , prove that $\triangle POQ \sim \triangle SOR$ .

Solution:-

Given:-

$PQ \parallel RS$

To prove:-

$\triangle{POQ} \sim \triangle{SOR}$

Proof:-

$\triangle{POQ}$ and

$\triangle{SOR}$

$\angle{POQ} = \angle{SOR} \; \left[ \because \text{Vertically opposite angles} \right]$

$\angle{P} = \angle{S} \; \left[ \because \text{As } PQ \parallel RS, \text{ Alternate angles} \right]$

$\angle{Q} = \angle{R} \; \left[ \because \text{As } PQ \parallel RS, \text{ Alternate angles} \right]$

Using AAA similarity function criteria,

$\triangle{POQ} \sim \triangle{SOR}$

Hence proved.

1011069_1063781_ans_ca292f2e3b3b44d79caa3110ebbd579c.png

i) $\triangle ABC\sim \triangle AMP$ ii) $\cfrac { CA }{ PA } =\cfrac { BC }{ MP }$

Given figure : (Image )

i) To prove

$\triangle ABC$ is ismilar to triangle AMP

We can prove this using angle-angle similarity postulate

In triangles AMP and ABC

$\angle AMP=\angle ABC=90°$

$\angle MAP=\angle BAC=$ same angle (common angle)

Hence, two angles of the required triangles are same

$\therefore$ By angle-angle (AA) similarity:

We can say that

$\triangle ABC\sim \triangle AMP$

ii) To prove

$\rightarrow \cfrac { CA }{ PA } =\cfrac { BC }{ MP }$

$\because$ In part i) we already proved

that

$\angle AMP=\angle ABC=90°$

and

$\angle MAP=\angle BAC=$ common angle

and sum of

$3$ angles of a triangle

$=180°$

$\therefore \angle APM=\angle ACB$

From side-angle-side similarity theorem,

we can that the sides including the angles which are equal must be proportional to each other

Sides including

$\angle APM=AP$ and

$MP$

Sides including

$\angle ACB=AC$ &

$BC$

Using S-A-S similarity theorem

$\cfrac { AP }{ MP } =\cfrac { AC }{ BC }$

$\Rightarrow \cfrac { BC }{ MP } =\cfrac { AC }{ AP }$

$\cfrac { CA }{ PA } =\cfrac { BC }{ MP }$

Hence proved.

1037228_1078608_ans_0659e3960ead4901bc88972aedda8d49.JPG

In the given figure, $\Delta$ ABC and $\Delta$ AMP are right angled at B and M respectively.
Given AC $=10$ cm, AP $=15$ cm and PM $=12$ cm.
Prove that : $\Delta$ ABC $\sim\Delta$ AMP.

$\triangle ABC,\triangle AMP$

$\angle AMP=\angle ABC=90^{\circ}$

$\angle MAP=\angle BAC=\angle A$

As two angles are equal automatically third angle will be equal since sum of angles in a triangle is constant

$AAA$ similarity

$\triangle ABC\sim\triangle AMP$

In the given figure, $\Delta$ ABC and $\Delta$ AMP are right angled at B and M respectively.
Given AC $=10$ cm, AP $=15$ cm and PM $=12$ cm. Find : AB and BC.

$\triangle ABC,\triangle AMP$

$\angle AMP=\angle ABC=90^{\circ}$

$\angle MAP=\angle BAC=\angle A$

As two angles are equal automatically third angle will be equal since sum of angles in a triangle is constant

$AAA$ similarity

$\triangle ABC\sim\triangle AMP$

Given

$AC=10\ cm,AP=15\ cm,PM=12\ cm$

$\dfrac{AP}{PM}=\dfrac{AC}{CB}\implies \dfrac{10}{BC}=\dfrac{15}{12}$

$\implies BC=8\ cm$

$\triangle ABC,\angle B=90^{\circ}$

$\implies AC^2=AB^2+BC^2\implies BC=\sqrt{AC^2-BC^2}=\sqrt{10^2-8^2}=6\ cm$

In $\Delta ABC$ , $DE\ II\ BC$ , $AD=2.5\ cm$ , $DB=5\ cm$ , $AE=3\ cm$ and $EC=x\ cm$
Find the value of $x$ ?

Solution:

$\dfrac{AD}{AD+BD}=\dfrac{AE}{AE+EC}$

$\dfrac{2.5}{7.5}=\dfrac{3}{3+x}$

$3+x=9$

$x=6$

1111711_1102112_ans_aefffc0df3d140e6881551708dc53519.jpg

In fig. $S$ and $T$ are the points on the sides $PQ$ and $PR$ respectively of $\Delta PQR$ , such that $PT = 4$ cm, $TR = 4$ cm & $ST \parallel QR$ . Find the ratio of areas of $\Delta PST$ & $\Delta PQR$ .

Given:

$ST \parallel QR$

PT= 4 cm

TR = 4cm

$\triangle PST$ and

$\triangle PQR$ ,

$\angle SPT = \angle QPR$ (Common)

$\angle PST = \angle PQR$ (Corresponding angles)

$\triangle PST \sim \triangle PQR$ (By AA similarity criterion)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\dfrac{area\triangle PST}{area\triangle PQR}= \dfrac{PT^2}{PR^2}$

$\dfrac{area\triangle PST}{area\triangle PQR}= \dfrac{4^2}{(PT+TR)^2}$

$\dfrac{area\triangle PST}{area\triangle PQR}= \dfrac{16}{(4+4)^2}= \dfrac{16}{8^2}=\dfrac{16}{64}= \dfrac{1}{4}$

Thus, the ratio of the areas of

$\triangle PST$ and

$\triangle PQR$ is

$1:4$ .

If $\triangle ABC$ is right-angled at $B$ . then prove that ${ AB }^{ 2 }+{ BC }^{ 2 }=AC^{ 2 }$

We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, then triangle on both sides of the perpendicular are similar to the whole triangle and to each other.

We have

$\triangle ADB\sim\triangle ABC$

$\therefore \dfrac{AD}{AB}=\dfrac{AB}{AC}$ (In similar triangles corresponding sides are equal)

$AB^2=AD\times AC.........(1)$

Also

$\triangle BDC\sim\triangle ABC$

$\therefore \dfrac{CD}{BC}=\dfrac{BC}{AC}$ (In similar triangles corresponding sides are equal)

$BC^2=CD\times AC.........(2)$

Adding (1) and (2) we get

$AB^2+BC^2=AD\times AC+CD\times AC$

$AB^2+BC^2=AC(AD+A=CD)$

From the figure

$AD+CD=AC$

$AB^2+BC^2=AC.AC$

Therefore,

$AB^2+BC^2=AC^2$

1026783_1114706_ans_b480b512934549b58293680210123520.png

In figure, the line segment XY is parallel to side AC of $\Delta ABC$ and it divides the triangle into two parts of equal areas. Find the ratio $\dfrac{AX}{AB}$ .

REF.Image

2 Area (xy) = Area (ABC)

$\Delta ABC$ &

$\Delta XBY$

$\angle ABC=\angle XBY$ (common)

$\angle ACB=\angle XYB$ (since XY||AC, angles are equal)

$\Delta ABC\sim \Delta XBY$ (AA similarity)

$(\dfrac{AB}{XB})^{2}=\dfrac{Area(\Delta ABC)}{Area(\Delta BXY)}=2$

$(AB)^{2}=2(XB)^{2}$

$AB=\sqrt{2}\times B$

$AX=AB-XB=AB-\dfrac{AB}{\sqrt{2}}=(\dfrac{\sqrt{2}-1}{\sqrt{2}})AB$

$\dfrac{AX}{AB}=\dfrac{\sqrt{2}-1}{\sqrt{2}}$

1068405_1141314_ans_e2f50a5b75b84f8298242331e78acd46.png

Prove that the ratio of the area of two similar triangles is equal to the ratio of squares on the corresponding sides.

$Given:\triangle ABC\sim\triangle PQR$

To prove :

$\dfrac{ar\left(ABC\right)}{ar\left(PQR\right)}={\left(\dfrac{AB}{PQ}\right)}^{2}={\left(\dfrac{BC}{QR}\right)}^{2}={\left(\dfrac{AC}{PR}\right)}^{2}$

Proof:-

$arABC=\dfrac { 1 }{ 2 } \times base\times height=\dfrac { 1 }{ 2 } \times BC\times AM---(i)$

$ar\left( PQR \right) =\dfrac { 1 }{ 2 } \times base\times hight=\dfrac { 1 }{ 2 } \times QP\times PN----(ii)$

$\dfrac{ar\left(ABC\right)}{ar\left(PQR\right)}=\dfrac {BC \times AM}{QR\times PN}----(A)$

$\triangle ABM$ and

$\triangle PQN, <B=<Q, <M=<N$

$\triangle ABM\sim \triangle PQN$

$\therefore \dfrac{AB}{PQ}=\dfrac{AM}{PN}---(B)$

From

$\left(A\right)$

$\dfrac{ar\left(ABC\right)}{ar\left(PQR\right)}= \dfrac{BC\times AM}{QR\times PN}= \dfrac{BC}{QR}\times \dfrac{AB}{PQ}---(C)$

Now, given

$\triangle ABC \sim \triangle PQR \Rightarrow \dfrac{AB}{PQ}=\dfrac{BC}{QP}= \dfrac{AC}{PR}$

Putting in

$\left(C\right) \quad \dfrac{ar\left(ABC\right)}{ar\left(PQR\right)}= {\left(\dfrac{AB}{PQ}\right)}^{2}$

Similarly

$\Rightarrow\dfrac{ar\left(ABC\right)}{ar\left(PQR\right)}= {\left(\dfrac{AB}{PQ}\right)}^{2}{\left(\dfrac{BC}{QR}\right)}^{2}={\left(\dfrac{AC}{PR}\right)}^{2}$

Hence proved

1378457_1146691_ans_783dcd07848a422a9d8b0da03f7f2b0e.png

Cut out two identical right angled triangles. Name the vertices of the A, B, C on both sides .Draw the altitude on the hypotenuse of one of the foot of the perpendicular as D. cut the triangle on its altitude an triangles. state the correspondences by which the three triangle an one another.

We have,

Given that:-

$\Delta ABC$ right angled at point $B$ .and perpendicular from $B$ intersecting $AC$ at point $D$ .

To prove :-

$\Delta ADB\sim \Delta ABC$

$\Delta BDC\sim \Delta ABC$

$\Delta ADB\sim \Delta BDC$

Proof:-

$In\,\Delta BDC\,\,and\,\,\Delta ABC$

$\angle C=\angle C\,\,\,\left( Common \right)$

$\angle BDC=\angle ABC\,\,\,\left( Each\,{{90}^{o}} \right)$

$\text{Then,}$

$\Delta BDC\sim \Delta ABC\,\,\,(By\,AA\,similarity)$

$In\,\Delta ADB\,\,and\,\,\Delta ABC$

$\angle A=\angle A\,\,\,\left( Common \right)$

$\angle ADB=\angle ABC\,\,\,\left( Each\,{{90}^{o}} \right)$

$\text{Then,}$

$\Delta ADB\sim \Delta ABC\,\,\,(By\,AA\,similarity)$

Hence proved.
1169346_1156054_ans_47b16c8351e94776bb44b9a0664379ae.png

Corresponding sides of two similar triangles are $3cm$ and $4cm$ . If the area of the larger triangle is $48cm^{2}$ , find the area of the smaller triangle.

Let

$A_1$ and

$A_2$ be the area of triangle and

$S_1$ and

$S_2$ be the their corresponding sides.

By the property of area of two similar triangle,

Ratio of area of both triangles = (Ratio of their corresponding sides)

$^2$

$\cfrac { { A }_{ 1 } }{ { A }_{ 2 } } ={ \left( \cfrac { { S }_{ 1 } }{ { S }_{ 2 } } \right) }^{ 2 }\quad$

$\cfrac { { A }_{ 1 } }{ 48 } ={ \left( \cfrac { 3 }{ 4 } \right) }^{ 2 }\Rightarrow { A }_{ 1 }=48\times \cfrac { 9 }{ 16 } =27{ cm }^{ 2 }$

In figure $\angle {1}=\angle{2}$ and $\triangle {NSQ}\cong \triangle{MTR}$ , then prove that $\triangle PTS \sim \triangle {PRQ}$

Given

$\triangle{NSQ}\cong\triangle{MTR}$

$\therefore \angle{NQS}=\angle{MRT}$ [CPCT]

$i.e. \angle{PQR}=\angle{PRQ}.........(1)$

$\triangle{PST}$

$\angle{P}+\angle{1}+\angle{2}=180^{o}$ [Angle sum property]

Since

$\angle{1}=\angle{2}$

$\angle{P}+\angle{1}+\angle{1}=180^{o}$

$\angle{P}+2\angle{1}=180^{o}..........(2)$

$\triangle{PQR}$

$\angle{P}+\angle{PQR}+\angle{PRQ}=180^{o}$ [Angle sum property]

From (1)

$\angle{PQR}=\angle{PRQ}$

$\angle{P}+\angle{PQR}+\angle{PQR}=180^{o}$

$\angle{P}+2\angle{PQR}=180^{o}..........(3)$

Right side of (2) and (3) are same

So, from (2) and (3)

$\angle{P}+2\angle{1}=\angle{P}+2\angle{PQR}$

$2\angle{1}=2\angle{PQR}$

$\triangle{PTS}\ \& \ \triangle{PRQ}$

$\angle{P}=\angle{P}$ [Common angle]

From (4)

$\angle{PST}=\angle{PQR}$

$\therefore \triangle{PTS}\sim\triangle{PRQ}$ [AA Similarity]

Hence proved.

Prove that if a line is drawn parallel to one side of a triangle intersecting the other two side,then it divides the two sides in the same ratio.

Given

$DE\parallel BC\\ In\quad \triangle ADE\quad and\triangle ABC\\ \angle DAE=\angle BAC\quad (common\quad angle)\\ \angle ADE=\angle ABC\quad \left[ \because DE\parallel BC\\ \therefore corresponding\quad angles\quad are\quad equal \right] \\ \angle AED=\angle ACB\\ \therefore \triangle ADE\sim \triangle ABC\\ Hence\quad \dfrac { AD }{ AB } =\dfrac { AE }{ AC } \left( In\quad similar\quad triangle,\quad corresponding\quad sides\quad are\quad in\quad same\quad ratio \right)$

$\because \dfrac { AD }{ AB } =\dfrac { AE }{ AC } \\ \Rightarrow \dfrac { AB }{ AD } =\dfrac { AC }{ AE } \\ \Rightarrow \dfrac { AB }{ AD } -1=\dfrac { AC }{ AE } -1\\ \Rightarrow \dfrac { AB-AD }{ AD } =\dfrac { AC-AE }{ AE } \\ \Rightarrow \dfrac{ DB }{ AD } =\dfrac { EC }{ AE } \\ \Rightarrow \dfrac { AD }{ DB } =\dfrac { AE }{ EC }$

It is proved that if a line is drawn parallel to one side of a triangle intersecting the other two side, then it divides the two sides in the same ratio.

1176956_1194317_ans_f85028eb11694720acfc23c09381d4ec.png

In the given figure $DE\parallel AC$ and $DF\parallel AE$ . Prove that.
$\cfrac{BF}{FE}=\cfrac{BE}{EC}$

Given:

$D$ is any point on side

$AB$ of

$\triangle ABC$ and

$E$ and

$F$ are two points on side

$BC$ , line segment

$DF, DE$ and

$AE$ are drawn

To prove that:

$\cfrac{BF}{FE}=\cfrac{BE}{EC}$

Proof: In

$\triangle BCA$

$DE\parallel AC$ (given)
By basic proportionality theorem, if two triangles are similar then their corresponding sides are proportional.

$\cfrac{BE}{EC}=\cfrac{BD}{DA}$ ...(i)

Again in

$\triangle BEA$ ,

$DF\parallel AE$ (given)

$\cfrac{BF}{FE}=\cfrac{BD}{DA}$ ....(ii) (by basic proportionality theorem)

From equation (i) and (ii), we can write

$\cfrac{BF}{FE}=\cfrac{BE}{EC}$

1071138_1183319_ans_ffd64c83ec1f42039aa3a70675c07c31.png

If $\angle{A}$ and $\angle {P}$ are acute angles such that $\sin{A}=\sin{P}$ then prove that $\angle {A}=\angle {P}$

Given

$\sin A=\sin P$

$\therefore \dfrac{BC}{AC}=\dfrac{RQ}{PR}$

Let

$\dfrac{BC}{RQ}=\dfrac{AC}{PR}=k$

$BC=kRQ$ and

$AC=kPR$

Now,

$\dfrac{AB}{PQ}=\dfrac{\sqrt{AC^2-BC^2}}{\sqrt{PR^2-RQ^2}}$

$\dfrac{AB}{PQ}=\dfrac{\sqrt{k^2PR^2-k^2RQ^2}}{\sqrt{PR^2-RQ^2}}$

$\dfrac{AB}{PQ}=\dfrac{k\sqrt{PR^2-RQ^2}}{\sqrt{PR^2-RQ^2}}$

$\dfrac{AB}{PQ}=k$

$\therefore \dfrac{AB}{PQ}=\dfrac{BC}{RQ}=\dfrac{AC}{PR}$

$\therefore \triangle{ABC}\sim \triangle{PQR}$ [By SSS similarity]

$\therefore \angle{A}=\angle{P}$

$\triangle {ABC}$ and $\triangle {DBC}$ are two isosceles triangles on the same base $BC$ (in figure). Show that $\angle{ABD}=\angle{ACD}$ .

$\triangle{ABC}$ is isosceles

So,

$AB=AC.........(1)$

$\triangle{DBC}$ is isosceles

So,

$DB=DC.........(2)$

Let us join AD.

$\triangle{ABD}$ and

$\triangle{ACD}$

$AB=AC$ [From (1)]

$BD=CD$ [From (2)]

$AD=AD$ [Common]

$\therefore \triangle{ABD}\cong\triangle{ACD}$ [SSS congruency]

$\therefore \angle{ABD}=\angle{ACD}$ [CPCT}

Hence proved.

1080589_1186467_ans_c2075d768f494bfb875f74ecda164ee1.png

In the figure,line $PQ|| side BC,AP=2.4cm,PB=7.2cm,QC=5.4$ cm then find $AQ$ ?

$\because PQ\parallel BC\\ \Rightarrow \angle APQ=\angle ABC\\ and\quad \angle AQP=\angle ACB\quad (coresponding\quad angle\quad are\quad same)\\ \therefore \triangle APQ\sim \triangle ABC\quad (AA\quad criteria)\\ hence,\quad \dfrac { AP }{ AB } =\dfrac { AQ }{ AC } \\ \Rightarrow \dfrac { AB }{ AP } =\dfrac { AC }{ AQ } \\ \Rightarrow \dfrac { AB }{ AP } -1=\dfrac { AC }{ AQ }- 1\\ \Rightarrow \dfrac { PB }{ AP } =\dfrac { QC }{ AQ } \\ \Rightarrow \dfrac { 7.2 }{ 2.4 } =\dfrac { 5.4 }{ AQ } \\ \Rightarrow AQ=1.8cm\\$

If the areas of two similar triangles are in ratio $25 : 64 ,$ write the ratio of their corresponding sides.

Let the triangle be

$\triangle$ ABC and

$\triangle$ DEF.

We know that

$\dfrac{ar(ABC)}{ar(DEF)}$ =

$[\dfrac{AB}{DE}]^2$

$\dfrac{25}{64}$ =

$[\dfrac{AB}{DE}]^2$

$\dfrac{AB}{DE}$ =

$\dfrac{5}{8}$

Diagonals of a trapezium $ABCD$ with $AB\parallel DC$ intersect each other at the point $O$ . If $AB=2CD$ , find the ratio of the areas of triangles $AOB$ and $COD$

$\triangle{AOB}$ and

$\triangle{COD}$

$\angle{AOB}=\angle{COD}$ [Vertically opposite angles]

$\angle{OAB}=\angle{OCD}$ [Since AB||CD with AC as traversal alternate angle are equal]

Hence,

$\triangle{AOB}\sim\triangle{COD}$ [AA similarity]

We know that if two triangle re similar, then

Ration of areas is equal to square of ratio of its corresponding sides

Hence,

$\dfrac{Area\ \triangle{AOB}}{Area\ \triangle{COD}}=\left(\dfrac{AB}{CD}\right)^2$

$\dfrac{Area\ \triangle{AOB}}{Area\ \triangle{COD}}=\left(\dfrac{2CD}{CD}\right)^2$

$\dfrac{Area\ \triangle{AOB}}{Area\ \triangle{COD}}=\left(\dfrac{2}{1}\right)^2$

$\dfrac{Area\ \triangle{AOB}}{Area\ \triangle{COD}}=\dfrac{4}{1}$

$\therefore Area\ \triangle{AOB}:Area\ \triangle{COD}=4:1$

Hence ratio of areas is 4:1.

1080854_1186888_ans_001bf1f5ebaf46e98ce74ac3f9a7205e.png

In the figure given below, $\angle PQR = \angle PRS$ . Prove that triangles $PQR$ and $PRS$ are similar. If $PR = 8\ cm, PS = 4\ cm$ , calculate $PQ$ .

From the figure,

$\angle P = \angle P$ (common angle for both triangles)

$\angle PQR = \angle PRS$ [from the question]

So,

$\triangle PQR \sim \triangle PRS$

Then,

$PQ/PR = PR/PS = QR/SR$

Consider

$PQ/PR = PR/PS$

$PQ/8 = 8/4$

$PQ = (8 \times 8)/4$

$PQ = 64/4$

$PQ = 16 cm$ .

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Let the two triangles be:

$\Delta \mathrm{ABC} \ and \ \Delta \mathrm{PQR}$

Area of

$\Delta A B C=\dfrac{1}{2} \times B C \times A M \ldots \ldots \ldots \ldots \ldots . .(1)$

Area of

$\Delta P Q R=\dfrac{1}{2} \times Q R \times P N \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .(2)$

Dividing (1) by (2)

$\dfrac{\operatorname{ar}(A B C)}{\operatorname{ar}(P Q R)}=\dfrac{\dfrac{1}{2} \times B C \times A M}{\dfrac{1}{2} \times Q R \times P N}$

$\dfrac{a r(A B C)}{a r(P Q R)}=\dfrac{B C \times A M}{Q R \times P N} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .(1)$

$\Delta A B M$ and

$\Delta P Q N$

$\angle B=\angle Q$ (Angles of similar triangles)

$\left.\angle M=\angle N \text { (Both } 90^{\circ}\right)$

Therefore,

$\Delta A B M \sim \Delta P Q N$

So,

$\dfrac{A B}{A M}=\dfrac{P Q}{P N} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .(2)$

From 1 and 2

$\dfrac{a r(A B C)}{a r(P Q R)}=\dfrac{B C}{Q R} \times \dfrac{A M}{P N}$

$\Rightarrow \dfrac{a r(A B C)}{a r(P Q R)}=\dfrac{B C}{Q R} \times \dfrac{A B}{P Q} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .(3)$

$\dfrac{A B}{P Q}=\dfrac{B C}{Q R}=\dfrac{A C}{P R} \ldots \ldots \ldots \ldots .(\Delta A B C \sim \Delta P Q R)$

Putting in ( 3 )

$\dfrac{a r(A B C)}{a r(P Q R)}=\dfrac{A B}{P Q} \times \dfrac{A B}{P Q}=\left(\dfrac{A B}{P Q}\right)^{2}$

$\Rightarrow \dfrac{a r(A B C)}{a r(P Q R)}=\left(\dfrac{A B}{P Q}\right)^{2}=\left(\dfrac{B C}{Q R}\right)^{2}=\left(\dfrac{A C}{P R}\right)^{2}$

1679381_1198134_ans_2a6dccda82b84950be72ad13dc42a6c7.png

Sides $A B$ and $B C$ and median $AD$ of a triangle $A B C$ are respectively proportional to sides $P Q$ and $Q R$ and median $P M$ of $\Delta P Q R .$ Show that $\Delta A B C$ $\sim \Delta P Q R$ ?

Given

$\dfrac{AB}{PQ}=\dfrac{AD}{PM}=\dfrac{BC}{QR}$ .

$\dfrac{BC}{QR}=\dfrac{BC/2}{QR/2}=\dfrac{BD}{QM}$ .

$\therefore \dfrac{AB}{PQ}=\dfrac{AD}{PM}=\dfrac{BD}{QM}$

$\Rightarrow \triangle ABD \sim \triangle PQM$ ( sss similarly).

Hence

$\angle B=\angle Q$ .

From

$\triangle ABC$ and

$\triangle PQR$ , we have

$\dfrac{AB}{PQ}=\dfrac{BC}{QR}$ and

$\angle B=\angle Q$ .

$\Rightarrow \triangle ABC\sim \triangle PQR$

Hence Proved.

1353950_1204309_ans_d3ae39281f40417a9bb6bdfcae303de5.png

Given $\Delta A B C \sim \Delta P Q R , \text { if } \dfrac { A B } { P Q } = \frac { 1 } { 3 } \text { find } \dfrac { a r \Delta A B C } { a r \Delta P Q R }$

Given

$\dfrac {AB}{PQ}=\dfrac 13$

we know that area of similar triangles are equal to the proportion of squares of proportional sides this

$\displaystyle \frac{ar\triangle ABC}{ar \triangle PQR} = \frac{AB^{2}}{PQ^{2}} = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$

Area of two similar triangle are $144\text{ cm}^2$ and $81\text{ cm}^2$ . If one side of the first triangle is $6\text{ cm}$ , then find the corresponding side of the second triangle.

Area of similar triangles are

$144 \text{ cm}^{ 2 }$ and

$81 \text{ cm}^{2}$

Since one side of the triangle is

$6 \text{ cm}$ ,

Hence the other corresponding side be

$x$ and can be find out by the following theorem;

$\dfrac { ar\left( { \triangle }_{ 1 } \right) }{ ar\left( { \triangle }_{ 2 } \right) } ={ \left( \dfrac { { s }_{ 1 } }{ { s }_{ 2 } } \right) }^{ 2 }$

where,

${ s }_{ 1 },{ s }_{ 2 }$ are the corresponding sides of similar triangles

$\Rightarrow \dfrac { 144 }{ 81 } ={ \left( \dfrac { 6 }{ x } \right) }^{ 2 }\\ \Rightarrow x=4.5 \text{ cm}$

$\Delta ABC \sim \Delta DEF$ , area of $\Delta ABC=64\ cm^{2}$ and area of $\Delta DEF=121\ cm^{2}$ . If $EF=15.4\ cm$ , find $BC$ .

area of ABC / area of DEF = BC / EF

64 / 121 = BC / 15.4

BC = (64/121 )* 15.4

BC = 8.14 cm

In the adjoining figure, DEFG is a square and $\angle BAC = {90^0}$ .
Prove that:
${\rm{D}}{{\rm{E}}^{\rm{2}}}{\rm{ = BD \times EC}}$

The ratio of the corresponding altitudes of two similar triangles is $\dfrac{3}{5}$ . Is it correct to say that ratio of their areas is $\dfrac{6}{5}$ ?Why ?

False: If two triangles are similar, then the ratio of areas of two triangles will be equal to the square of the ratio of their corresponding sides or altitudes or angle bisectors.

$\Delta ABC\sim \Delta PQR$ , then

Refer image,

$\dfrac{ar(\Delta ABC)}{ar(\Delta PQR)}=\left(\dfrac{AD}{PM}\right)^2$ (By area theorem)

$\Rightarrow \dfrac{ar(\Delta ABC)}{ar(\Delta PQR)}=\left(\dfrac{3}{5}\right)^2$

$=\dfrac{9}{25}\neq \dfrac{6}{5}$

So, the given statement is false.

1789921_1253982_ans_d6ef5b0dfed844aaaa11a0aa09728570.png

It is possible to draw i) an obtuse-angled equilateral triangle ii)a right-angled equilateral triangle iii)a triangle with two right angles.

i) We can't draw an obtuse angled equilateral triangle, because for a triangle to be equilateral all the sides should be equal and all the corresponding angles also should be equal.

We know that sum of angles in a triangle is

$180°$

$\therefore$ Each angle should be

$60°<90°$

$\therefore$ We can't draw an obtuse angled equilateral triangle.

ii) We cannot draw a right angled equilateral triangle. Reason is as stated in i)

iii) We cannot draw a triangle with two right angles. If two angles are

$90,90$ sum is

$180$ and the third angle should be

$0°$ (Then the triangle won't be formed )

$\therefore$ we cannot draw a triangle with two right angles.

In the given figure, $ST\parallel RQ,PS=3$ cm and $SR=4$ cm. Find the ratio of the areas of $\triangle{PST}$ to the area of $\triangle{PRQ}$

In figure ST || QR

and PS= 3 cm & RS= 4cm

$\Delta PST$ &

$PQR$

$QPR = \angle TPS$ [Common]

$\angle PRQ=\angle PST$ [corresponding]

$\Delta PQR \sim \Delta PST$ [AA similarity]

Now

In similar triangles ratios of the area of triangles is equal to the ratio of the square of corresponding sides.

$\frac {Ar. \, of \Delta PST}{Ar.\ of \Delta PRQ}$

$(\dfrac{3}{3+4})^2$

$=\dfrac {9}{49}$

Hence

$\left [\frac {Ar. \, of \Delta PST}{Ar. \, of \Delta PRQ}= \frac {9}{49} \right ]$

$A$ and $B$ are respectively the points on the sides $PQ$ and $PR$ of a triangle $PQR$ such that $PQ = 12.5\ cm, PA = 5\ cm, BR = 6\ cm$ and $PB = 4\ cm$ . Is $AB \parallel QR$ ? Give reasons for your answer.

From the dimensions given in the question,
Consider the

$\triangle PQR$
So,

$PQ/PA = 12.5/5$

$= 2.5/1$

$PR/PB = (PB + BR)/PB$

$= (4 + 6)/4$

$= 10/4$

$= 2.5$
By comparing both the results,

$2.5 = 2.5$
Therefore,

$PQ/PA = PR/PB$
So,

$AB$ is parallel to

$QR$ .

1700067_1262018_ans_91d8c2d7c596495f98b22160094557c6.PNG

In the figure, it is given $TS=TR, \angle 1=2\angle =$ and $\angle 4=2\angle 2\angle 3$ . Prove that $\triangle RBT\cong \triangle SAT$ .

Given,

$TS=TR$ ...(i)

$\angle 1=2\angle 2$ ...(ii)

$\angle 4=2\sqrt{3}$ ...(iii)

$\angle 1=\angle 4$ Vertically Opposite Angles

$\therefore 2\angle 2=2\angle 3\Rightarrow \angle 2=\angle 3$ ...(iv)

$\angle \therefore \angle TRS=\angle TSR$ ...(A) Isosceles triangle

$\angle TRS=\angle TRB+\angle 2$ ...(B)

$\angle TSR=\angle TSA+\angle 3$ ...(C)

substitute (B) and (C) in (A), we get,

$\angle TRB+\angle 2=\angle TSA+\angle B$

$\Rightarrow \angle TRB=\angle TSA$

$RT=ST$ ...From (i)

$\angle TRB=\angle TSA$ ...From (iv)

$\angle RTB=\angle STA$

$\therefore \triangle RBT\cong \triangle SAT$ From ASA

$\triangle ABC$ is right angled at A and $AD\bot BC$ . If $BC=13$ cm and $AC=5$ cm, find the ratio of the areas of $\triangle ABC$ and $\triangle ADC$ .

1869856_1293650_ans_eaeee1d05bb4445086fbf673efb5efd4.jpg

In figure, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that
$\dfrac{CA}{PA}=\dfrac{BC}{MP}$

$\triangle {ABC}$ and

$\triangle {AMP}$

$\angle CAB=\angle MAP$ (common angle)

$\angle {ABC}=\angle {AMP}$ (both

${90}^{o}$ )

using AA similarity

$\triangle {ABC}\sim \triangle {AMP}$

We know that if two triangles are similar, the ratio of their corresponding sides are proportional

$\cfrac{CA}{PA}=\cfrac{BC}{MP}=\cfrac{AB}{AM}$

$\therefore$

$\cfrac{CA}{PA}=\cfrac{BC}{MP}$

1343099_1253068_ans_69244d1be0a44054956608a1f44e5e7e.PNG

In $\triangle ABC, DE\parallel BC$ , find the value of $x$ .

Since the

$\Delta ADE$ and

$\Delta ABC$ are similar.
1215758_1402241_ans_2f1985d1f7d747f9807c46080d15a00f.jpg

In the given figure $AP=12$ , $RP=4$ , $PS=6$ . Find $BS$

From figure,

$\Delta PAB\sim \Delta PRS$

$\implies \dfrac{AP}{RP}=\dfrac {BP}{PS}\\BP=\dfrac{12}{4}\times 6=18\\BP=PS+BS\\BS=18-6=12cm$

In the diagrams of the following figure, same the triangles which is represent to the $\triangle ABC$ keeping the letters. In the right order. State the congruence conditions wise.

From

$\triangle{ABC}$ and

$\triangle{BQS}$

$\angle{A}=\angle{B}$ (given)

$AB=BQ$ (given)

$\angle{B}=\angle{Q}$ (given)

$\therefore \triangle{ABC}\cong\triangle{BQS}$ by ASA congruence property.

In the figure, altitudes $AD$ and $CE$ of $\triangle{ABC}$ intersect each other at the point $P$ .Show that $\triangle{AEP}\sim\triangle{CDP}$

Given:

$\triangle{ABC}$ and altitude

$AD$ and

$CE$ of triangle intersects each other at the point

$P$ .

To prove:

$\triangle{AEP}\sim\triangle{CDP}$

Proof:In

$\triangle{AEP}$ and

$\triangle{CDP}$

$\angle{AEP}=\angle{CDP}$ since

$AD$ and

$CE$ are altitudes.

$\angle{APE}=\angle{CPD}$ (vertically opposite angles)

Hence,

$\triangle{AEP}\sim\triangle{CDP}$ by AAA similarity

Hence proved.

STATE BPT THEOREM AND PROVE IT.

Basic proportionality theorem :

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

Given:

$\triangle$ ABC in which

$DE \parallel BC$ and DE intersects AB and AC at D and E respectively.

To prove :

$\dfrac{AD}{DB}=\dfrac{AE}{EC}$

Construction :

Join BE and CD.

Draw

$EL \perp AB$ and

$DM \perp AC$ .

Proof :

We have,

$ar(\triangle ADE)=\dfrac{1}{2} \times AD \times EL$

$ar(\triangle DBE) =\dfrac{1}{2} \times DB \times EL$

Therefore,

$\dfrac{ar(\triangle ADE)}{ar(\triangle DBE)}=\dfrac{AD}{DB}$ ...(1)

Again,

$ar(\triangle ADE)=ar(\triangle AED)=\dfrac{1}{2} \times AE \times DM$

$ar(\triangle ECD)=\dfrac{1}{2} \times EC \times DM$

Therefore,

$\dfrac{ar(\triangle ADE)}{ar(\triangle ECD)}=\dfrac{AE}{EC}$ ....(2)

Now,

$\triangle DBE$ and

$\triangle ECD$ being on the same base DE and between the same parallels DE and BC, we have,

$ar(\triangle DBE)=ar(\triangle ECD)$ ....(3)

From 1, 2 & 3, we have,

$\dfrac{AD}{DB}=\dfrac{AE}{EC}$

1277636_1371126_ans_462cd1cea8534b22bf3ba8bd9ae98299.png

In the adjacent figure $XY\parallel BC$ find the length of $AY$ and $CY$

$XY\parallel BC$

$\Rightarrow \angle{AXY}=\angle{ABC}$ and

$\angle{AYX}=\angle{ACB}$ (corresponding angles)

$\Rightarrow \triangle{AXY}\sim\triangle{ABC}$ by AA similarity

$\Rightarrow \dfrac{AX}{AB}=\dfrac{AY}{AC}=\dfrac{XY}{BC}$

$\Rightarrow \dfrac{1}{AX+XB}=\dfrac{AC-YC}{AC}$

$\Rightarrow \dfrac{1}{4}=\dfrac{6-YC}{6}$

$\Rightarrow 24-4YC=6$

$\Rightarrow -4YC=6-24=-18$

$\therefore YC=\dfrac{-18}{-4}=4.5$ cm

$AY=AC-YC=6-4.5=1.5$ cm

$\therefore AY=1.5$ cm and

$CY=4.5$ cm

Side $AB$ and $BC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $QR$ and median $PM$ of $PQR$ . Show that $ABC\sim PQR$ .

Given :

$\triangle$ ABC where AD is the median,

$\triangle$ PQR where PM is the median.

And,

$\dfrac{AB}{PQ} = \dfrac{BC}{QR}=\dfrac{AD}{PM}$

To prove :

$\triangle ABC \sim \triangle PQR$

Proof :

Since AD is the median,

$BD=CD=\dfrac{1}{2}BC$

Similarly, PM is the median,

$QM=RM=\dfrac{1}{2}QR$

Given that

$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PM}$

$\dfrac{AB}{PQ}=\dfrac{2BD}{2QM}=\dfrac{AD}{PM}$

$\dfrac{AB}{PQ}=\dfrac{BD}{QM}=\dfrac{AD}{PM}$

Since all 3 sides are proportional

$\triangle ABD \sim \triangle PQM$ [By SSS similarity]

Hence,

$\angle B =\angle Q$

$\triangle ABC$ and

$\triangle PQR$

$\angle B = \angle Q$

$\dfrac{AB}{PQ}=\dfrac{BC}{QR}$

Hence, by SAS similarity,

$\triangle ABC \sim \triangle PQR$

1281166_1366681_ans_58b432ef51614054ba365df8f8097b0f.jpg

$D$ and $E$ are respectively the points on the sides $AB$ and $AC$ of a $\triangle{ABC}$ such that $AB=2.6\ cm,AD=1.3\ cm$ and $AE=1.5\ cm$ ,Show that $DE\parallel BC$

Here

$\dfrac{AB}{AD}=\dfrac{2.6}{1.3}=2$

$\dfrac{AC}{AE}=\dfrac{3}{1.5}=2$

$\therefore \dfrac{AB}{AD}=\dfrac{AC}{AE}$

By the converse of BPT,

$DE\parallel BC$

Hence proved.

1333573_1373283_ans_1698c014191047fcba8bd60702825efe.png

In $\triangle{PQR},\,ST\parallel QR$ and $\dfrac{PS}{SQ}=\dfrac{3}{5}$ if $PR=6$ cm find $PT$

$\triangle{PQR},$ we have

$ST\parallel QR$

By Basic proportionality theorem,

$\dfrac{PS}{SQ}=\dfrac{PT}{TR}$

$\Rightarrow \dfrac{PS}{SQ}=\dfrac{PT}{PR-PT}$

$\Rightarrow \dfrac{3}{5}=\dfrac{PT}{6-PT}$

$\Rightarrow 18-3PT=5PT$

$\Rightarrow 8PT=18$

$\Rightarrow PT=\dfrac{18}{8}=2.25$ cm

1333597_1371235_ans_8220c861cc6545c5bbb81007d54c9e29.PNG

$ABCD$ is a rhombus . $AC$ is a diagonal. Is $\angle{BAC}\cong\angle{DAC}$ ? Give reason.

Given :

$ABCD$ is a Rhombus

$\Rightarrow$ All sides are equal in length

$AB=BC=CD=DA\Rightarrow$ Opposite angles are equal while adjacent angles are Supplementary

$\angle{DAB}=\angle{BCD},\,\,\angle{ABC}=\angle{ADC}$

$\Rightarrow$ diagonals bisect each other perpendicularly

$\triangle{ABC}$ and

$\angle{ADC},\,\, AB=CD$ (given)

$BC =AD$ (given)

$AC=AC$ ..... (common side)

$\triangle{ABC}\cong\triangle{ADC}$

$\angle{BAC}=\angle{DAC}$ (As

$AC$ is a diagonal of Rhombus and diagonals bisect the angles.)

If the ratio of the perimeter of two similar triangles is $4 : 25$ , then find the ratio of the areas of the similar triangles .

$\because$ ratio of perimeter of two triangles

$=4:25$

$\therefore$ Ratio of corresponding sides of the two triangles

$=4:25$

The ratio of area of two triangles

$=$ Ratio of square of its corresponding sides

$=\dfrac{{4}^{2}}{{25}^{2}}=\dfrac{16}{625}$

In the following figure you are given two triangles answer weather two triangles are similar? Give reason also.

$\triangle ABC$ and

$\triangle AEF$ ,

$\angle BAC = \angle EAF$ (Common)

And,

$\angle ABC= \angle AEF$ (Given)

Hence,

$\triangle ABC \sim \triangle AEF$ (By AA similarity criterion)

In figure, AD $=$ DC and AB $=$ BC. State the three parts of matching pairs you have used to answer.(Is $\Delta$ ABD $\cong\Delta$ CBD).

1615418_1678058_ans_c25cc32b9b4246e7aca9d079aa46f6af.jpg

In the given figure, $AD=DC$ and $AB=BC.$ Is $\Delta ABD \cong\Delta CBD?$

Given:

$AD = CD$

$AB = CB$

$BD = BD$ (Common side)

By SSS congruence rule

$\Delta ABD \cong \Delta CBD$

In the following triangle, the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic form.

From figure ,

$AB=FE= 5\ cm$

$BD=EC=7\ cm$

$AD=FC=10.5\ cm$

So, by

$SSS$ test

$\Delta ABD\cong \Delta FEC$

Hence,

$\angle A\cong \angle F$

$\angle B\cong \angle E$

$\angle D\cong \angle C$

If $\Delta$ PQR $\cong\Delta$ EFD, which angles of $\Delta$ PQR equals $\angle E$ ?

As,

$\Delta PQR\cong\Delta EFD$

So, corresponding angles are equal

Here, corresponding to

$E$ is

$P$ from

$PQR$

Hence,

$\angle P$ is equals to

$\angle E$

ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If $\angle BAC=40^o$ and $\angle BDC=100^o$ ; then find $\angle ADB$ .

1615884_1678065_ans_70a2514331a94fa7ba4f40648db852a6.jpg

In Fig. Prove that:
$CD + DA + AB > BC$

$\Delta ACD,$ we have

$CD + DA > CA$

$\Rightarrow CD + DA + AB > CA + AB$

$\Rightarrow CD + DA + AB > BC$

$[\because AB + AC > BC]$

In the following triangle, the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic form.

1615368_1678056_ans_fa4f171682a048edb61bdb67a7a7147b.jpg

In figure, $AB||DC$ , AB $=$ DC and BC $=$ AD. You have used some fact to say theat $\triangle ADC\cong\triangle CBA$ , not given in the question, what is that?

It is given that,

$AB$ =

$DC \text{ and } BC$ =

$AD.$

Also,

$AC$ is common side of both triangles

$ABC$ and

$ADC$

So, all three sides of both triangles are equal.

And

$\angle BAC=\angle DCA$ , because alternate angles as

$AC$ is transversal.

So, by

$SAS$ test

$\triangle ADC\cong\triangle CBA$

Hence,

$\angle BAC=\angle DCA$ and

$AC$ is common side is information that was not given in question but was used.

In figure, AB $=$ DC and BC $=$ AD. What congruence condition have you used?

1615484_1678060_ans_2d2c712089604b7ba0c4c0aa690228c3.jpg

In $DE||AC$ and $DC||AP$ . Prove that $\dfrac{BE}{EC}=\dfrac{BC}{CP}$

Given

$DE || AC \, \& \, DC || AP$

To prove :-

$\dfrac{BE}{EC} = \dfrac{BE}{CP}$

Proof :-

$\Delta BPA$ , we have

$DC || AP$

So, by Basic proportionality theorem (B . P. T.), we get

$\dfrac{BC}{CP} = \dfrac{BD}{DA}$ ... (i)

$\Delta BCA$ , we have

$DE || AC$ (given)

$\dfrac{BE}{EC} = \dfrac{BD}{DA}$ ___(ii) (by B.P. T.)

Comparing (i) & (ii) , we get

$\dfrac{BC}{CP} = \dfrac{BD}{DA} = \dfrac{BE}{EC}$

$\Rightarrow \dfrac{BE}{EC} = \dfrac{BC}{CP}$

1501232_1700961_ans_b647e7b4553c471dad0e5d7559a2bb49.png

Fill in the blank.
Given $\Delta ABC \sim \Delta PQR$ , if $\dfrac{AB}{PQ} = \dfrac{1}{3} ,$ then $\dfrac{ar (\Delta ABC)}{ar (\Delta PQR)} =$ ______ .

We know that

$\Delta ABC \sim \Delta PQR$

$\therefore \dfrac{ar (\Delta ABC)}{ar (\Delta PQR)} = \dfrac{(AB)^2}{(PQ)^2} = \dfrac{(AC)^2}{(PR)^2} = \dfrac{(BC)^2}{(QR)^2}$

$\therefore \dfrac{ar (\Delta ABC)}{ar (\Delta PQR)} = \left(\dfrac{AB}{PQ} \right)^2 = \left(\dfrac{1}{3} \right)^2 = \dfrac{1}{9}$

$D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\Delta ABC$ such that $DE || BC$ . if $\dfrac{AD}{AB} = \dfrac{8}{15}$ and $EC = 3.5 cm$ , find $AE$ .

Given :

$\dfrac{AD}{AB} = \dfrac{8}{15}$ and

$EC = 3.5 \ cm$

Find

$AE$

By BPT:

$\dfrac{AD}{AB} = \dfrac{AE}{AC}$

$\dfrac{8}{15} = \dfrac{AE}{(AE+EC)} = \dfrac{AE}{(AE+3.5)}$

$8(AE + 3.5) = 15AE$

$7 AE = 28$

$AE = 4 \ cm$

Solve the following :
In $\Delta PQR , \, NM || RQ$ . If $PM = 15, MQ = 10, NR = 8$ , then find $PN$ .

Given

$\, NM || RQ$ and

$PM = 15, MQ = 10, NR = 8$

since

$\, NM || RQ$ then by basic proportionality theorem

$\dfrac {PN}{NR}=\dfrac {PM}{MQ}$

$\dfrac {PN}{8}=\dfrac {15}{10}$

$PN=\dfrac {15}{10}\times 8$

$PN=\dfrac 32 \times 8$

$\therefore\ PN=12$

$D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\Delta ABC$ such that $DE || BC$ . Find the value of $x$ , when $AD = x cm, DB = (x - 2) cm, AE = (x + 2) cm$ and $EC = (x - 1) cm$ .

Given :

$AD = x \ cm, DB = (x - 2) \ cm, AE = (x + 2) \ cm$ and

$EC = (x - 1) \ cm$

To find :

$x$

From figure,

$D$ and

$E$ are the points on the sides

$AB$ and

$AC$ respectively and

$DE || BC$

Then

$\dfrac{AD}{DB} = \dfrac{AE}{EC}$

$\dfrac{x}{(x-2)} = \dfrac{(x+2)}{(x-1) }$

$x(x-1) = (x+2)(x-2)$

Solving above equation, we get

$x = 4 \ cm$

$D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\Delta ABC$ such that $DE || BC$ . If $\dfrac{AD}{DB} = \dfrac{4}{7}$ and $AC = 6.6 cm$ , find $AE$ .

Given :

$\dfrac{AD}{DB} = \dfrac{4}{7}$ and

$AC = 6.6 \ cm$

Find :

$AE$ .

By BPT :

$\dfrac{AD}{DB} = \dfrac{AE}{EC }$

Add 1 on both sides

$\dfrac{AD}{DB} + 1 = \dfrac{AE}{EC} + 1$

$\dfrac{(AD + DB)}{DB} = \dfrac{(AE + EC)}{EC}$

$\dfrac{(4+7)}{7} = \dfrac{AC}{EC} = \dfrac{6.6}{EC}$

$EC = \dfrac{(6.6 \times 7)}{11}$

$EC = 4.2 \ cm$

And,

$AE = AC - EC$

$AE = 6.6 - 4.2$

$AE = 2.4 \ cm$

$D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\Delta ABC$ such that $DE || BC$ . If $AD = 3.6 cm, AB = 10 cm$ and $AE = 4.5 cm$ , find $EC$ and $AC$ .

From given triangle, points

$D$ and

$E$ are on the sides

$AB$ and

$AC$ respectively such that

$DE || BC$ .

$AD = 3.6 \ cm, \ AB = 10 \ cm$ and

$AE = 4.5 \ cm$ .

By Basic Proportonality Theorem:

$\dfrac{AD}{DB} = \dfrac{AE}{EC}$

Here

$DB = AB - AD = 10 - 3.6 = 6.4 \ cm$

$\Rightarrow EC = \dfrac{4.5}{3.6}\times 6.4$

$EC = 8 \ cm$

And,

$AC = AE + EC$

$AC = 4.5 + 8 = 12.5 \ cm$

$BL$ nad $CM$ are median of a $\triangle ABC$ , right angled at $A$ , Prove that $4(BL^{2}+CM^{2})=5BC^{2}$

$\triangle ABC$ is right angled at

$A.$ Where,

$BL$ and

$CM$ are the medians.

Since,

$BL$ is the median,

$\Rightarrow$

$AL=CL=\dfrac{1}{2}AC$ ------ ( 1 )

Similarly,

$CM$ is the median,

$\Rightarrow$

$AM=MB=\dfrac{1}{2}AB$ ----- ( 2 )

In right-angled

$\triangle BAC,$

$\Rightarrow$

$(BC)^2=(AB)^2+(AC)^2$ [ By Pythagoras theorem ]

$\triangle BAL,$

$\Rightarrow$

$(BL)^2=(AB)^2+(AL)^2$ [ By Pythagoras theorem ] ------- ( 3 )

$\Rightarrow$

$(BL)^2=(AB)^2+\left(\dfrac{AC}{2}\right)^2$

$\Rightarrow$

$(BL)^2=(AB)^2+\dfrac{(AC)^2}{4}$

$\Rightarrow$

$(BL)^2=\dfrac{4(AB)^2+(AC)^2}{4}$

$\Rightarrow$

$4(BL)^2=4(AB)^2+(AC)^2$ ------ ( 4 )

$\triangle MAC,$

$\Rightarrow$

$(CM)^2=(AM)^2+(AC)^2$ [ By Pythagoras theorem ]

$\Rightarrow$

$(CM)^2=\left(\dfrac{AB}{2}\right)^2+(AC)^2$

$\Rightarrow$

$(CM)^2=\dfrac{(AB)^2}{4}+(AC)^2$

$\Rightarrow$

$(CM)^2=\dfrac{(AB)^2+4(AC)^2}{4}$

$\Rightarrow$

$4CM^2=AB^2+4AC^2$ ------- ( 5 )

Adding ( 4 ) and ( 5 ),

$\Rightarrow$

$4BL^2+4CM^2=4AB^2+AC^2+AB^2+4AC^2$

$\Rightarrow$

$4(BL^2+CM^2)=5AB^2+5AC^2$

$\Rightarrow$

$4(BL^2+CM^2)=5(AB^2+AC^2)$

From ( 3 ),

$\Rightarrow$

$4(BL^2+CM^2)=5BC^2$

1454837_1688016_ans_67a9bec7757541d0a1baeecac49ed8f6.png

$D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\Delta ABC$ such that $DE || BC$ . Find the value of $x$ , when $AD = 4 \text{ cm}, DB = (x - 4) \text{ cm}, AE = 8 \text{ cm}$ and $EC = (3x - 19) \text{ cm}$ .

Given :

$AD = 4 \text{ cm}, DB = (x - 4) \text{ cm}, AE = 8 \text{ cm}$ and

$EC = (3x - 19) \text{ cm}$ .

To Find :

$x$

We know that, by BPT(Basic proportionality theorem)

$\dfrac{AD}{DB} = \dfrac{AE}{EC}$

$\dfrac{4}{(x-4)} = \dfrac{8}{(3x-19)}$

$4(3x-19) = 8(x-4)$

$12x-76=8x-32$

Solving, we get

$x = 11 \text{ cm}$

$D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\Delta ABC$ such that $DE || BC$ . If $AB = 13.3 cm, AC = 11.9 cm$ and $EC = 5.1 cm$ , find $AD$ .

Given :

$AB = 13.3 \ cm, AC = 11.9 \ cm$ and

$EC = 5.1 \ cm$

Find

$AD$ .

By BPT:

$\dfrac{AD}{DB} =\dfrac{ AE}{EC}$

Add 1 on both sides

$\dfrac{AD}{DB} + 1 = \dfrac{AE}{EC} + 1$

$\dfrac{(AD + DB)}{DB} = \dfrac{(AE + EC)}{EC}$

$\dfrac{AB}{DB} = \dfrac{AC}{EC}$

$DB = \dfrac{(AB\times EC)}{ AC}$

$= \dfrac{(13.3 \times 5.1)}{11.90}$

$DB= 5.7 \ cm$

$\Rightarrow BD = 5.7 \ cm$

And,

$AD = AB - DB AD = 13.3 - 5.7$

$AD = 7.6 \ cm$

In each of the given pairs of triangles, find which pair of triangles are similar. State the similarity criterion and write the similarity relation in symbolic form:

Two triangles are similarity of their corresponding angles are equal and corresponding sides are proportional.

(i) In

$\Delta ABC$ and

$\delta PQR$

$\angle A = \angle Q = 50^o$

$\angle B = \angle P = 60^o$ and

$\angle C = \angle R = 70^o$

$\Delta ABC \sim \Delta QPR$ (By

$AAA$ )

(ii) In

$\Delta ABC$ and

$\Delta DEF$

$\Delta ABC$ and

$\Delta DEF$

$AB = 3 cm, \ BC = 4.5$

$DF = 6 cm, \ DE = 9 cm$

$\Delta ABC$ is not similar to

$\Delta DEF$

(iii) In

$\Delta ABC$ and

$\Delta PQR$

$\Delta ABC$ and

$\Delta PQR$

$AC = 8 cm \ BC = 6 cm$

Included

$\angle C = 80^o$

$PQ = 4.5 cm, \ QR = 6 cm$

and included

$\angle Q = 80^o$

$\dfrac{AC}{QR} = \dfrac{8}{6} = \dfrac{4}{3}$

and

$\dfrac{BC}{PQ} = \dfrac{6}{4.5} = \dfrac{4}{3}$

$\Rightarrow \dfrac{AC}{QR} = \dfrac{BC}{PQ}$

and

$\angle C = \angle Q = 80^o$

$\Delta ABC \sim \Delta PQR$ (By SAS)

(iv) In

$\Delta DEF$ and

$\Delta PQR$

$DE = 2.5, \ DF = 3$ and

$EF = 2$

$PQ = 4, \ PR = 6$ and

$QR = 5$

$\dfrac{DE}{QR} = \dfrac{2.5}{5} = \dfrac{1}{2}$

$\dfrac{DF}{PR} = \dfrac{3}{6} = \dfrac{1}{2}$

and

$\dfrac{EF}{PQ} = \dfrac{2}{4} = \dfrac{1}{2}$

$\Delta DEF \sim \Delta PQR$ (By SSS)

(v) In

$\Delta ABC$ and

$\Delta MNR$

$\angle A = 80^o,\ \angle C = 70^o$

So,

$\angle B = 180^o - (80^o + 70^o) = 30^o$

$\angle M = 80^o, ∠N = 30^o$ , and

$\angle R = 180^o - (80^o + 30^o) = 70^o$

Now, in

$\Delta ABC$

$\angle A = \angle M – 80^o, \angle B = \angle N = 30^o$

and

$\angle C = \angle R = 70^o$

$\Delta ABC \sim \Delta MNR$ (By AAA or AA)

$\Delta ABC \sim \Delta DEF$ and their areas are respectively $64 cm^2$ and $121 cm^2$ . If $EF = 15.4 cm$ , find $BC$ .

Area of

$\Delta ABC = 64 cm^2$ and

area of

$\Delta DEF = 121 cm^2$

$EF = 15.4 cm$

$\dfrac{\text{area of }\ \Delta ABC}{\text{area of} \ \Delta DEF} = \dfrac{BC^2}{EF^2}$

$\dfrac{64}{121} = \dfrac{BC^2}{(15.4)^2}$

$\left(\dfrac{8}{11}\right)^2 = \left(\dfrac{BC}{15.4}\right)^2$

$\dfrac{8}{11} = \dfrac{BC}{15.4}$

$BC = \dfrac{8\times 15.4}{11} = 11.2$

$BC = 11.2cm$

$\Delta ABC \sim \Delta PQR$ and $ar (\Delta ABC) = 4ar (\Delta PQR)$ . If $BC = 12 cm$ , find $QR$ .

$\Delta ABC \sim \Delta PQR$

$ar (\Delta ABC) = 4ar (\Delta PQR)$

$\dfrac{area\ \Delta ABC}{area \ \Delta PQR} =\dfrac{4}{1}$

$\dfrac{area (\Delta ABC)}{area (\Delta PQR)} = \dfrac{BC^2}{QR^2}$

$\dfrac{4}{1} = \dfrac{BC^2}{QR^2}$

$\Rightarrow \left(\dfrac{2}{1}\right)^2 = \left(\dfrac{12}{QR}\right)^2$

$\dfrac{2}{1} = \dfrac{12}{QR}$

$\Rightarrow QR = \dfrac{12\times 1}{2} = 6$

$QR = 6cm$

$D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\triangle ABC$ . In each of the following cases, determine whether $DE || BC$ or not. $AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm$ and $EC = 8 cm$

From figure,

$D$ and

$E$ are the points on the sides

$AB$ and

$AC$ of

$\triangle ABC$

$AD = 5.7 cm, \ DB = 9.5 cm, \ AE = 4.8 cm$ and

$EC = 8 cm$

$\dfrac{AD}{DB} = \dfrac{5.7}{9.5} = \dfrac{3}{5}$

and

$\dfrac{AE}{EC} = \dfrac{4.8}{8} = \dfrac{3}{5}$

$\Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC}$

$\therefore$ by converse of basic proportionality theorem

$\Rightarrow DE || BC$

$D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\Delta ABC$ . In each of the following cases, determine whether $DE || BC$ or not. $AB = 10.8 cm, AD = 6.3 cm, AC = 9.6 cm$ and $EC = 4 cm$ .

Given :

$AB = 10.8 cm, AD = 6.3 cm, AC = 9.6 cm$ and

$EC = 4 cm$ .

$DB = AB - AD \\= 10.8 - 6.3 \\= 4.5 cm$

$AE =AC - EC \\= 9.6 - 4 \\= 5.6 cm$

$\dfrac{AD}{DB} = \dfrac{6.3}{4.5} = \dfrac{7}{5}$

$\dfrac{AE}{EC} = \dfrac{5.6}{4} = \dfrac{7}{5}$

$\Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC}$

BY BPT

$\Rightarrow DE || BC$

Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel to the parallel sides.

Here,

$AB || DC$

$M$ and

$N$ are the mid points of sides

$AD$ and

$BC$ respectively.

$MN$ is joined.

To prove :

$MN || AB$ or

$DC$ .

Produce

$AD$ and

$BC$ to meet at

$P$ .

Now, In

$\triangle PAB$

$DC ||AB$

$\dfrac{PD}{DA} = \dfrac{PC}{CB}$

$\dfrac{PD}{2PM} = \dfrac{PC}{2CN}$

$M$ and

$N$ are midpoints of

$AD$ and

$BC$ respectively.

$\dfrac{PD}{PM} = \dfrac{PC}{CN}$

$MN||DC$ But

$DC ||AB$

Therefore,

$MN || DC ||AB$

1538396_1713864_ans_c9cd51b6a31646769e0de5b0670ba184.png

In a $\Delta ABC,\, M$ and $N$ are points on the sides $AB$ and $AC$ respectively such that $BM = CN$ . If $\angle B = \angle C$ then show that $MN || BC$ .

$\Delta ABC$ ,

$M$ and

$N$ are points on the sides

$AB$ and

$AC$ respectively.

$\angle B = \angle C$ ,we know that, sides opposite to equal angles are equal

$AB = AC$

$BM = CN$ ....... (1)

$AB - BM = AC - CN$

$\Rightarrow AM = AN$ ......... (2)

From

$\Delta ABC$

Dividing (2) by (1), we get

$\dfrac{AM}{MB} = \dfrac{AN}{NC}$

By converse of basic proportionality theorem,

$MN ||BN$

1538473_1713895_ans_3153ba3579fd40bf89543b9ad61b4d68.png

In the given figure, if $\angle ADE = \angle B$ , show that $\Delta ADE \sim \Delta ABC$ . If $AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm$ and $BC = 4.2 cm$ , find $DE$ .

From given figure,

$\angle ADE = \angle B$

To prove:

$\Delta ADE \sim \Delta ABC$ and

find

$DE$

Given:

$AD = 3.8 cm, \ AE = 3.6 cm, \ BE = 2.1 cm$ and

$BC = 4.2 cm$

Now, In

$\Delta ADE$ and

$\Delta ABC$

$\angle ADE = \angle B$ (given)

$\angle A = \angle A$ (common)

$\Delta ADE \sim \Delta ABC$ (By AA)

Again,

$\dfrac{AD}{AB} = \dfrac{DE}{BC}$

$\dfrac{AD}{AE + EB} = \dfrac{x}{4.2}$

$\dfrac{3.8}{3.6 + 2.1} = \dfrac{x}{4.2}$

$\dfrac{3.8}{5.4} = \dfrac{x}{4.2}$

$x = 2.8$

$DE = 2.8cm$

1538594_1713978_ans_0cbe9e27c67d4c969abc5217ed9e8cca.png

$D$ and $E$ are the points on the sides $AB$ and $AC$ respectively of a $\Delta ABC$ . If $AB = 11.7 cm, AC = 11.2 cm, BD = 6.5 cm$ and $AE = 4.2 cm$ then determine whether $DE || BC$ or not.

Given :

$AB = 11.7 \ cm, AC = 11.2 \ cm, BD = 6.5 \ cm$ and

$AE = 4.2 \ cm$ .

$AD = AB - BD = 11.7 - 6.5 = 5.2 \ cm$

and

$EC = AC - AE = 11.2 - 4.2 = 7 \ cm$

$\dfrac{AD}{DB} = \dfrac{5.2}{6.5} = \dfrac{4}{5}$

$\dfrac{ AE}{EC} = \dfrac{4.2}{7} = \dfrac{3}{5}$

$\Rightarrow \dfrac{AD}{DB} \neq \dfrac{AE}{EC}$

By the converse of basic proportionality theorem,

$\Rightarrow DE$ is not parallel to

$BC$

$D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\Delta ABC$ . In each of the following cases, determine whether $DE || BC$ or not. $AD = 7.2 \text{ cm}, AE = 6.4 \text{ cm}, AB = 12 \text{ cm}$ and $AC = 10 \text{ cm}$ .

Given :

$AD = 7.2 \text{ cm}, AE = 6.4\text{ cm}, AB = 12\text{ cm}$ and

$AC = 10\text{ cm}$ .

$DB = AB - AD = 12 - 7.2 = 4.8 \text{ cm}$

$EC = AC - AE = 10 - 6,4 =3.6 \text{ cm}$

$\dfrac{AD}{DB} = \dfrac{7.2}{4.8} = \dfrac{3}{2}$ and

$\dfrac{AE}{EC} = \dfrac{6.4}{3.6} = \dfrac{16}{9}$

$\Rightarrow \dfrac{AD}{DB}\neq \dfrac{AE}{EC}$

By the converse of the basic proportionality theorem

$\Rightarrow DE$ is not parallel to

$BC$

The areas of two similar triangles are $81 cm^2$ and $49 cm^2$ respectively. If the altitude of the first triangle is $6.3 cm$ , find the corresponding altitude of the other.

Areas of two similar triangles are

$81 cm^2$ and

$49 cm^2$

Altitude of the first triangle

$= 6.3 cm$

Let altitude of second triangle

$= x cm$

Area of

$\Delta ABC = 81 cm^2$ and area of

$\Delta DEF = 49cm^2$

Altitude

$AL = 6 - 3 cm$

Let altitude

$DM = x cm$

$\dfrac{\text{area of}\ \Delta ABC}{\text{area of}\ \Delta DEF} = \dfrac{AL^2}{DM^2}$

$\dfrac{81}{49} = \dfrac{(6.3)^2}{x^2}$

$\left(\dfrac{9}{7}\right)^2= \left(\dfrac{6.3}{x}\right)^2$

$\dfrac{9}{7} = \dfrac{6.3}{x}$

$x = 4.9$

Altitude of second triangle is

$4.9 cm$

The areas of two similar triangles are $100 cm^2$ and $64 cm^2$ respectively. If a median of the smaller triangle is $5.6 cm$ , find the corresponding median of the other.

Areas of two similar triangles are

$100 cm^2$ and

$64 cm^2$

Median

$DM$ of

$\Delta DEF = 5.6 cm$

Let median

$AL$ of

$\Delta ABC = x$

$\dfrac{\text{area of} \, \Delta ABC}{\text{area of}\ \Delta DEF} = \dfrac{AL^2}{DM^2}$

$\dfrac{100}{64} = \dfrac{x^2}{(5.6)^2}$

$\left(\dfrac{10}{8}\right)^2 = \left(\dfrac{x}{5.6}\right)^2$

$\dfrac{10}{8} = \dfrac{x}{5.6}$

$x = 7$

Corresponding median of the other triangle is

$7 cm$ .

State the SAS-similarity criterion.

In two triangles, if two sides of the one are proportional to the corresponding sides of the other and their included angles are equal, the two triangles are similar.

State the SSS-criterion for similarity of triangles.

In two triangles, if three sides of the one triangle are proportional to the corresponding sides of the other, the triangles are similar

State the AA-similarity criterion.

In two triangles, if two angles of the one triangle are equal to the corresponding angles of the other triangle, then the triangles are similar.

In the given figure, $O$ is a point inside a $\Delta PQR$ such that $\angle POR = 90^o,\ OP = 6 cm$ and $OR = 8 cm$ . If $PQ = 24 cm$ and $QR = 26 cm$ , prove that PQR is right-angled.

$\Delta PQR, \ O$ is a point in it such that

$OP = 6 cm, \ OR = 8 cm$ and

$\angle POR = 90^o$

$PQ = 24 cm, \ QR = 26 cm$

Now,

$\Delta POR,\ \angle O = 90^o$

$PR^2 = PO^2 + OR^2$

$= (6)^2 + (8)^2$

$= 36 + 64$

$= 100$

$PR = 10$

Greatest side

$QR$ is

$26 cm$

$QR^2 = (26)^2 = 676$

and

$PQ^2 + PR^2 = (24)^2 + (10)^2$

$= 576 + 100$

$= 676$

Which implies,

$676 = 676$

$QR^2 = PQ^2 + PR^2$

$\Delta PQR$ is a right angled triangle and right angle at

$P$ .

State the AAA-similarity criterion.

In any two triangles, if three angles of one triangle are equal to the three angles of the other triangle, then the triangles are similar.

In the given figure, $ABC$ is a triangle and $PQ$ is a straight line meeting $AB$ in $P$ and $AC$ in $Q$ . If $AP = 1 cm,\ PB = 3cm, \ AQ = 1.5 cm, \ QC = 4.5 cm$ , prove that area of $\Delta APQ$ is $1/16$ of the area of $\Delta ABC$ .

$\Delta ABC, \ PQ$ is a line which meets

$AB$ in

$P$ and

$AC$ in

$Q$ .

Given:

$AP = 1 cm, \ PB = 3 cm, \ AQ = 1.5 cm\ QC = 4.5 cm$

Now,

$\dfrac{AP}{PB} = \dfrac{1}{3}$ and

$\dfrac{AQ}{QC} = \dfrac{1.5}{4.5} = \dfrac{1}{3}$

$\Rightarrow \dfrac{AP}{PB} = \dfrac{AQ}{QC}$

From figure:

$AB = AP + PB = 1+3 = 4 cm$

$AC = AQ + QC = 1.5 + 4.5 = 6 cm$

$\Delta APQ$ and

$\Delta ABC$ ,

$\dfrac{AP}{AB} = \dfrac{AQ}{AC}$

angle

$A$ (common)

$\Delta APQ$ and

$\Delta ABC$ are similar triangles.

Now,

$\dfrac{\text{area of}\ (\Delta APQ)}{\text{area of} \ (\Delta ABC)} = \dfrac{AP^2}{AB^2} = \dfrac{(1)^2}{(4)^2} = \dfrac{1}{16}$

Which implies,

area of

$\Delta APQ = \dfrac{1}{16}$ of the area of

$\Delta ABC$

Hence Proved.

In the given figure, $DE || BC$ . If $DE = 3 cm, \ BC = 6 cm$ and $ar (\Delta ADE) = 15 cm^2$ , find the area of $\Delta ABC$ .

$DE || BC$

$DE = 3 cm, \ BC = 6 cm$

area

$(\Delta ADE) = 15 cm^2$

Now,

$\Delta ABC$

$DE ||BC$ .

Therefore triangles,

$\Delta ADE$ and

$\Delta ABC$ are similar.

$\dfrac{\text{area of}\ (\Delta ADE)}{\text{area of}\ (\Delta ABC) } = \dfrac{DE^2}{BC^2}$

$\dfrac{15}{\text{Area of}\, \Delta ABC} = \dfrac{(3)^2}{(6)^2} = \dfrac{9}{36}$

area of

$\Delta ABC = \dfrac{15\times 36}{9} = 60$

Area of

$\Delta ABC$ is

$60 cm^2$ .

State the converse of Thales theorem.

If a line divides any two sides of a triangle in the same ratio. Then, the line must be parallel to the third side.

If the equal sides $AB$ and $AC$ of an isosceles triangle be produced to $E$ and $F$ so that
$BE\cdot CF=AB^{2}$ , show that the line $EF$ will always pass through a fixed point.

1564904_1740519_ans_85b42adf39fc4cc19705bac799d3cd92.png

In the given figure, $DE || BC$ such that $AD = x cm,\ DB = (3x + 4) cm,\ AE = (x + 3) cm$ and $EC = (3x + 19) cm$ . Find the value of $x$ .

From figure:

$DE || BC$

$AD = x\ cm, \ DB = (3x + 4) cm$

$AE = (x + 3) cm$ and

$EC = (3x + 19) cm$

$\Delta ABC$

$\dfrac{AD}{DB} = \dfrac{AE}{EC}$

$\dfrac{x}{(3x+4)} = \dfrac{(x+3)}{(3x+19)}$

$3x^2 + 19x - 3x^2- 9x - 4x = 12$

$x = 2$

In the above figure, $CA \parallel BD$ , the lines $AB$ and $CD$ meet at $O.$
(i) Prove that $\triangle ACO \sim \triangle BDO.$
(ii) If $BD = 2.4\text{ cm}, OD = 4\text{ cm}, OB = 3.2\text{ cm}$ and $AC = 3.6\text{ cm}$ , calculate $OA$ and $OC.$

(i) We have to prove that,

$\triangle ACO \sim \triangle BDO$ .
So, from the figure
Consider

$\triangle ACO$ and

$\triangle BDO,$

$\angle AOC = \angle BOD$ [from vertically opposite angles]

$\angle A = \angle B$ [Alternate angles]

So, by

$AA$ criterion of similarity,

$\triangle ACO \sim \triangle BDO$

(ii) As we know that,

$\triangle ACO \sim \triangle BOD$

So, by using basic proportionality theorem we can say that,

$\dfrac{{AO}}{{BO}} = \dfrac{{CO}}{{DO}} = \dfrac{{AC}}{{BD}}$

Consider,

$\begin{aligned}{}\frac{{AO}}{{BO}}& = \frac{{AC}}{{BD}}\\\frac{{AO}}{{3.2}}& = \frac{{3.6}}{{2.4}}\\AO &= \frac{{3.2 \times 3.6}}{{2.4}}\\ &= 4.8 \text{ cm}\end{aligned}$

And,

$\begin{aligned}{}\frac{{CO}}{{DO}} &= \frac{{AC}}{{BD}}\\\frac{{CO}}{4}& = \frac{{3.6}}{{2.4}}\\CO& = \frac{{4 \times 3.6}}{{2.4}}\\ &= 6\text{ cm}\end{aligned}$

State which pairs of triangles in the figure given below are similar. Write the similarity rule used and also write the pairs of similar triangles in symbolic form (all lengths of sides are in cm):

From the

$\triangle ABC$ and

$\triangle PQR$

$AB/PQ = 3.2/4$

$= 32/40$
Divide both numerator and denominator by

$8$ we get,

$= 4/5$

$AC/PR = 3.6/4.5$

$= 36/45$
Divide both numerator and denominator by $$9$ we get,

$= 4/5$

$BC/QR = 3/5.4$

$= 30/54$
Divide both numerator and denominator by

$6$ we get,

$= 5/9$
By comparing all the results, the side are not equal.
Therefore, the triangles are not equal.

Two triangles $ABC$ and $PQR$ are such that $AB = 3 cm, \ AC = 6 cm, \ \angle A = 70^o, \ PR = 9 cm, \ \angle P = 70^o$ and $PQ = 4.5 cm$ . Show that $\Delta ABC \sim \Delta PQR$ and state the similarity criterion.

In two triangles

$\Delta ABC$ and

$\Delta PQR$ ,

$AB = 3 cm, \ AC = 6 cm, \ \angle A = 70^o$

$PR = 9 cm, \ \angle P = 70^o$ and

$PQ= 4.5 cm$

Now,

$\angle A = \angle P = 70^o$ (Same)

$\dfrac{ AC}{PR} = \dfrac{6}{9} = \dfrac{2}{3}$

and

$\dfrac{AB}{PQ} = \dfrac{3}{4.5} = \dfrac{2}{3}$

$\Rightarrow \dfrac{AC}{PR} = \dfrac{AB}{PQ}$

Both

$\Delta ABC$ and

$\Delta PQR$ are similar by SAS

In the given figure, $DB \perp DC, DE\perp AB$ and $AC \perp BC$ . Prove that $BE/ DE = AC/BC$ .

From the figure,

$DB \perp BC, DE \perp AB$ and

$AC \perp BC$
We have to prove that,

$BE/DE = AC/BC$
Consider the

$\triangle ABC$ and

$\triangle DEB$ ,

$\angle C = 90^{\circ}$

$\angle A + \angle ABC = 90^{\circ}$ [from the figure equation (i)]
Now in

$\triangle DEB$

$\angle DBE + \angle ABC = 90^{\circ}$ [from the figure equation (ii)]
From equation (i), we get

$\angle A = \angle DBE$
Then, in

$\triangle ABC$ and

$\triangle DBE$

$\angle C = \angle E$ [both angles are equal to

$90^{\circ}$ ]
So,

$\triangle ABC \sim \triangle DBE$
Therefore,

$AC/BE = BC/DE$
By cross multiplication, we get

$AC/BC = BE/DE$ .

In the figure given below, $PQRS$ is a parallelogram; $PQ = 16\ cm, QR = 10\ cm. L$ is a point on $PR$ such that $RL : LP = 2 : 3 : QL$ produced meets $RS$ at $M$ and $PS$ produced at $N$ .
Name a triangle similar to triangle $RLM$ . Evaluate $RM$ .

From the figure,
Consider

$\triangle RLM$ and

$\triangle QLP$
Then,

$\angle RLM = \angle QLP$ [vertically opposite angles are equal]

$\angle LRM = \angle LPQ$ [alternate angles are equal]
Therefore,

$\triangle RLM \sim \triangle QLP$
Then,

$RM/PQ = RL/LP = 2/3$
So,

$RM/16 = 2/3$

$RM = (16 \times 2)/3$

$RM = 32/3$

$RM = 10\dfrac {2}{3}$ .

$\Delta \text{QPR} \sim \Delta \text{TSM}$ ? Why?

Intriangle

$PQR$ ,

$\angle R =180^{\circ}-(\angle P+\angle Q)$

$=180^{\circ}-\left(55^{\circ}+25^{\circ}\right)=100^{\circ}$

In triangle

$TSM$ ,

$\angle T=180^\circ-(\angle S+\angle M)$

$=180^\circ-(100^\circ+25^\circ)=55^\circ$

Now, in triangles

$PQR$ and

$TSM$ ,

$\angle P=\angle T\, ,\, \angle Q=\angle S\ ,\ \angle R=\angle M\qquad$ [ By AAA criterion ]

$\therefore \Delta QPR \sim \Delta STM$

Hence,

$\Delta Q P R\sim\Delta STM$ .

In Figure, $DE \parallel OQ$ and $DF \parallel O R,$ Show that $EF \parallel QR$

$\Delta P O Q,$ we have

$DE \parallel OQ \quad$ (Given)

By Basic proportionality Theorem, we have

$\dfrac{PE}{E Q}=\dfrac{P D}{D O} \quad \quad \dots (i)$

Similarly,

$\Delta POR$ , we have

$DF \parallel OR \quad$ (Given)

$\therefore \dfrac{PD}{D O}=\dfrac{P F}{F R} \quad \quad \dots (ii)$

Now, from

$(i)$ and

$(ii)$ , we have

$\dfrac{PE}{E Q}=\dfrac{P F}{F R} \quad \\ \Rightarrow E F \parallel Q R$

[Applying the converse of Basic proportionality Theorem in

$\Delta P Q R]$

S and T are points on sides $\mathrm{PR}$ and $\mathrm{QR}$ if $\Delta P Q R$
such that $\angle P=\angle R T S$ . Show that $\Delta \mathrm{RPQ} \sim \Delta \mathrm{RTS}$

$\Delta\mathrm{RPQ}$ and

$\Delta \mathrm{RTS}$ , we have

$\angle\mathrm{RPQ}=\angle \mathrm{RTS} \quad$ (Given)

$\angle P RQ=\angle T R S=\angle R \quad$ (Common)

$\therefore \angle \mathrm{RPQ} \sim \Delta \mathrm{RTS}\qquad$ (By AA criterion of similarity)

$A$ and $B$ are respectively the points on the sides $P Q$ and $P R$ of a $\Delta PQR$ such that $\mathrm{PQ}=12.5 \mathrm{cm}, \mathrm{PA}=5 \mathrm{cm}, \mathrm{BR}=6 \mathrm{cm}$ and $\mathrm{PB}=4 \mathrm{cm} .$ Is $AB\|\mathrm{QR} ?$ Give reason.

Hrere,

$\dfrac{PA}{A Q}=\dfrac{5}{12.5-5}=\dfrac{5}{7.5}=\dfrac{2}{3}$

Similarly,

$\dfrac{P B}{B R}=\dfrac{4}{6}=\dfrac{2}{3}$

$\text {Since, } \ \dfrac{P A}{A Q}=\dfrac{P B}{B R}=\dfrac{2}{3}$

Then by using the converse of basic proportionality theorem,

$\ \ \mathrm{AB}\| \mathrm{QR}$

$E$ and $F$ are points on the sides $P Q$ and $PR$ respectively of a $\Delta P Q R$ . Show that $EF || QR$ . If $P Q=1.28 \mathrm{cm}, P R=2.56 \mathrm{cm}, P E=0.18 \mathrm{cm}$ and $\mathrm{PF}=0.36 \mathrm{cm}$

We have,

$PQ=1.28\ cm, PR=2.56 \ cm$

$P E=0.18\ cm, PF=0.36 \ cm$

Now,

$EQ=PQ-PE=1.28-0.18=1.10 \ cm$

And,

$FR=PR-PF=2.56-0.36=2.20 \ cm$

$\therefore \dfrac{P E}{E Q}=\dfrac{0.18}{1.10}=\dfrac{18}{110}=\dfrac{9}{55}\\ \text{And, } \dfrac{P F}{F R}=\dfrac{0.36}{2.20}=\dfrac{36}{220}=\dfrac{9}{55}\\$

$\therefore \dfrac{P E}{E Q}=\dfrac{P F}{F R}$

Hence,

$EF\parallel QR \quad$ [By the converse of basic proportionality Theorem]

If it is given that $\Delta \mathrm{ABC} \sim \Delta P Q R$ with $\dfrac{BC}{QR}=\dfrac{1}{3},$ then find $\dfrac{ar(\Delta P Q R)}{a r(\Delta A B C)}$

Given,

$\Delta ABC \sim \Delta P Q R$ and

$\dfrac{BC}{Q R}=\dfrac{1}{3}$

We know that,

The ratio of the area of similar triangles is equal to the ratio of the square of its

corresponding side.

$\therefore \dfrac{\operatorname{ar}(\Delta P Q R)}{\operatorname{ar}(\Delta A B C)}=\dfrac{(Q R)^{2}}{(B C)^{2}}\\=\left(\dfrac{QR}{B C}\right)^{2}\\=\left(\dfrac{3}{1}\right)^{2}=\dfrac{9}{1}=9$

In Figure, if $\mathrm{LM} \parallel \mathrm{CB}$ and $\mathrm{LN} \parallel\mathrm{CD},$ prove that $\dfrac{A M}{A B}=\dfrac{A N}{A D}$

$\Delta\mathrm{ABC}$ , we have

$LM\parallel CB$ (Given)

$\therefore$ By Basic proportionality Theorem, we have

$\dfrac{AM}{AB}=\dfrac{AL}{AC} \quad \dots (i)$

Again, in

$\Delta\mathrm{ACD}$ , we have

$LN\parallel CD$ (Given)

$\therefore$ By Basic proportionality Theorem, we have

$\dfrac{AN}{A D}=\dfrac{A L}{A C} \quad \dots (ii)$

Now, from

$(i)$ and

$(ii)$ , we have

$\dfrac{A M}{A B}=\dfrac{A N}{A D}$

$\Delta DEF \sim \Delta ABC,$ if $DE: A B=2: 3$ and $ar(\Delta D E F)$ is equal to 44 square units. Find the area of $(\Delta ABC)$

Given,

$\Delta \mathrm{DEF} \sim \Delta \mathrm{ABC}$ and

$DE:AB=2:3$

We know that,

Ratio of area of similar triangles in equal to the ratio of square of its corresponding side

$\therefore \dfrac{ar(\Delta D E F)}{ar(\Delta A B C}=\dfrac{(D E)^{2}}{(A B)^{2}}\\\Rightarrow \dfrac{44}{ar(\Delta A B C)}=\left(\dfrac{2}{3}\right)^{2} \\\Rightarrow ar(\Delta A B C)=\dfrac{44 \times 9}{4}\\\therefore ar(\Delta ABC)=99\ {cm}^{2}$

Using converse of Basic proportionality Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Given: In

$\triangle ABC,\,D$ and

$E$ are the mid-points of sides

$AB$ and

$A C$ respectively.

$\therefore A D=D B \quad \text { and } \quad A E=E C$

$\Rightarrow \dfrac{AD}{D B}=1 \quad \text { and } \quad \dfrac{A E}{E C}=1$

$\Rightarrow \dfrac{A D}{D B}=\dfrac{A E}{E C}$

Therefore, by the converse of Basic proportionality Theorem,

$DE \parallel B C$

State which pairs of triangles in the following figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.

(i) In

$\Delta ABC$ and

$\Delta PQR$ , we have

$\dfrac{AB}{Q R}=\dfrac{2}{4}=\dfrac{1}{2}, \quad \dfrac{A C}{P Q}=\dfrac{3}{6}=\dfrac{1}{2}$

Hence,

$\dfrac{AB}{Q R}=\dfrac{A C}{P Q}=\dfrac{B C}{P R}$

$\therefore\Delta A B C \sim \Delta Q R P$ by SSS criterion of similarity.

(ii)

$\ln\Delta {LMP}$ and

$\Delta {FED}$ , we have

$\dfrac{LP}{D F}=\dfrac{3}{6}=\dfrac{1}{2}, \quad \dfrac{M P}{D E}=\dfrac{2}{4}=\dfrac{1}{2},\quad \dfrac{L M}{E F}=\dfrac{27}{5}$

Hence,

$\dfrac{LP}{D F}=\dfrac{M P}{D E} \neq \dfrac{L M}{E F}$

$\therefore\Delta L M P$ is not similar to

$\Delta F E D$

(iii) In

$\Delta{NML}$ and

$\Delta {PQR}$ , we have

$\angle M=\angle Q=70^{\circ}$

Now,

$\quad\dfrac{M N}{P Q}=\dfrac{2.5}{6}=\dfrac{5}{12} \quad$ and

$\quad \dfrac{M L}{QR}=\dfrac{5}{10}=\dfrac{1}{2}$

Hence

$\quad\dfrac{M N}{P Q} \neq \dfrac{M L}{Q R}$

$\therefore\Delta N M L$ is not similar to

$\Delta P Q R$

$\mathrm{D}$ is a point on the side $\mathrm{BC}$ of a triangle $\mathrm{ABC}$ such that $\angle \mathrm{ADC}=\angle \mathrm{BAC}$ . Show that $C A^{2}=C B . C D$

$\Delta {ABC}$ and

$\Delta{DAC}$ , we have

$\angle B A C=\angle A D C\quad$ (Given)

and

$\angle C=\angle C \quad$ (Common)

$\therefore\Delta {ABC} \sim \Delta{DAC}$ (By AA criterion of similarity)

$\Rightarrow \dfrac{A B}{D A}=\dfrac{B C}{AC}=\dfrac{A C}{D C}$

$\Rightarrow \dfrac{C B}{C A}=\dfrac{C A}{CD}$

$\Rightarrow C A^{2}=C B \times C D$

In the given figure $, \Delta {ABC}$ and $\Delta{DBC}$ are on the same base $BC$ . If ${AD}$ intersects $B C$ at $O .$ prove that $\dfrac{a r(\Delta A B C)}{a r(\Delta D B C)}=\dfrac{A O}{D O}$

Given:

$\Delta{ABC}$ and

$\Delta {DBC}$ are on the same base

${BC}$

and

$\mathrm{AD}$ intersects

$\mathrm{BC}$ at

$\mathrm{O}$ .

To Prove:

$\dfrac{{ar}(\Delta A B C)}{{ar}(\Delta D B C)}=\dfrac{A O}{D P}$

Construction:

Draw

$AL \perp {BC}$ and

${DM} \perp {BC}$

Proof:

$\Delta{ALO}$ and

$\Delta{DMO}$ , we have

$\angle{ALO}=\angle {DMO}=90^{\circ}$ and

$\angle{AOL}=\angle {DOM} \quad$ (Vertically opposite angles)

$\therefore\Delta{ALO} \sim \Delta {DMO}\quad$ (By AA-Similarity)

$\Rightarrow \dfrac{A L}{D M}=\dfrac{A O}{DO} \quad \quad \dots(i)$

$\therefore\dfrac{\operatorname{ar}(\Delta AB C)}{\operatorname{ar}(\Delta D B C)}=\dfrac{\dfrac{1}{2} B C \times A L}{\dfrac{1}{2}B C \times D M}=\dfrac{A L}{D M}=\dfrac{A O}{D O}\qquad$ [Using

$(i)$ ]

Hence,

$\dfrac{ar(\Delta A B C)}{ar(\Delta D B C)}=\dfrac{A O}{D O}$

In the Fig., we have
$BX = \frac{1}{2} AB$
$BY= \frac{1}{2} BC$ and $AB = BC$ .
Show that $BX = BY$ .

We have

$AB = BC$ [Given]
Now, by Euclid's axiom 7, we have things which are halves of the same things are equal to one another.

$\therefore \ \ \frac{1}{2} AB = \frac{1}{2} BC$
Hence.

$BX = BY$ . [

$\because BX = \frac{1}{2} AB$ and

$BY = \frac{1}{2} BC$ (Given)]

In Figure $, P$ is the mid-point of $B C$ and $Q$ is the mid-point of $A P .$ If $B Q$ when produced meets $A C$ at $R,$ prove that $R A=\dfrac{1}{3} C A$

Given: In

$\Delta A B C, P$ is the mid-point of

$B C, Q$ is the mid-point of

$A P$ such that

$B Q$ produced meets

$A C$ at

$R$

To prove:

${RA}=\dfrac{1}{3} {CA}$

Construction:

Draw

$PS\parallel BR$ , meeting

$AC$ at

$S$ .

Proof:

$\Delta B C R, P$ is the mid-point of

$BC$ and

$PS\parallel BR$ .

$\therefore S$ is the mid-point of

${CR}$

$\Rightarrow {CS}={SR} \quad \dots (i)$

$\Delta A P S, Q$ is the mid-point of

$A P$ and

$Q R \parallel P S$

$\therefore R$ is the mid-point of

$AS$ .

$\Rightarrow A R=R S\quad \quad \dots (ii)$

From

$(i)$ and

$(ii)$ , we get

$A R=R S=S C$

$\Rightarrow A C=A R+R S+S C=3 A R \quad \\\Rightarrow A R=\dfrac{1}{3} A C=\dfrac{1}{3} C A$

Hence, proved.

Using Basic proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Given:

$\Delta A B C$ in which

$D$ is the mid-point of

$A B$ and

$DE$ is drawn parallel to

$BC$ , which meets

$AC$ at

$E$ .

To prove:

$AE=EC$

Proof:

$\Delta ABC$ ,

$DE\parallel BC$

$\therefore$ By Basic proportionality Theorem, we have

$\dfrac{AD}{D B}=\dfrac{A E}{E C} \quad \quad \dots (i)$

Now, since

$D$ is the mid-point of

$A B$

$\Rightarrow A D=B D\quad \quad \dots (ii)$

From

$(i)$ and

$(ii)$ , we have

$\dfrac{B D}{B D}=\dfrac{A E}{E C}$

$\Rightarrow 1=\dfrac{AE}{E C}$

$\Rightarrow A E=E C$

Hence,

$E$ is the mid-point of

$AC$ .

Sides $A B$ and $A C$ and median $A D$ of a triangle $A B C$ are respectively proportional to sides $\mathrm{PQ}$ and $PR$ and median $PM$ of another triangle $PQR$ . Show that $\triangle A B C \sim \triangle P Q R$

Given:

$\Delta A B C$ and

$\Delta P Q R, A D$ and

$P M$ are their medians respectively such that

$\dfrac{A B}{P Q}=\dfrac{A D}{P M}=\dfrac{A C}{P R}\qquad \ldots(i)$

To prove:

$\triangle A B C \sim \Delta P Q R$

Construction:

Produce

$AD$ to

$E$ such that

$A D=D E$ and produce

$PM$ to

$N$ such that,

${PM}={MN}$

Join

$BE, CE, QN, RN.$

Proof:

Since,

$BD=DC$ and

$AD=DE$

$\Rightarrow$ Quadrilateral

$ABEC$ is a

$\parallel$ gm.

Similarly,

$QM=MR$ and

$PM=MN$

$\Rightarrow$ Quadrilateral

$PQNR$ is a

$\parallel$ gm.

$\Rightarrow B E=A C$ and

$Q N=P R$

$\Rightarrow \dfrac{B E}{Q N}=\dfrac{A C}{P R}$

$\Rightarrow \dfrac{B E}{QN}=\dfrac{A B}{P Q} \quad$ [ From

$(i)$ ]

i.e.,

$\dfrac{A B}{P Q}=\dfrac{B E}{Q N} \quad \quad \dots (ii)$

From

$(i)$ ,

$\dfrac{A B}{P Q}=\dfrac{A D}{P M}=\dfrac{2 A D}{2 PM}=\dfrac{AE}{P N}$

$\Rightarrow \dfrac{A B}{P Q}=\dfrac{A E}{P N} \quad \quad \dots (iii)$

From

$(ii)$ and

$(iii)$ ,

$\dfrac{AB}{P Q}=\dfrac{B E}{Q N}=\dfrac{A E}{P N}$

$\Rightarrow \Delta A B E \sim \Delta P Q N$ (SSS similarity criterion)

$\Rightarrow \angle 1=\angle 2\qquad\dots(iv)$

Similarly, we can prove

$\Delta A C E \sim \Delta P R N\\ \Rightarrow \angle 3=\angle 4 \quad \dots({v})$

Adding (iv) and (v), we get

$\angle 1+\angle 3=\angle 2+\angle 4$

$\Rightarrow \angle A=\angle P$

And

$\dfrac{AB}{P Q}=\dfrac{A C}{P R}$ (Given)

$\therefore\Delta A B C \sim \Delta P Q R$ (By SAS criterion of similarity)

1707663_1786753_ans_0818b4535e2a42019b36e1c779f1742d.png

In the adjoining figure, DE ll BC, $$ AD = 2.4 cm, AB =3.2 cm, CE = 4.8 cm $$ , flnd BD.

Given: AD = 2.4cm, AB = 3.2cm and EC = 4.8cm and DEIIBC
Basic Proportionality theorem m which states that if aline is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides divided in the same ratio.So,by basic proportionality theorem.

$\frac{AD}{DB} = \frac {AE}{EC}$

$\frac {2.4}{DB} = \frac {3.2}{4.8}$

$\implies \frac {2.4}{DB} = \frac {2}{3}$

$\implies DB = \frac {2.4×3}{2}$

$\implies DB = 3.6cm$ or BD =3.6cm

In figure, $DE||BC$ . Find AD

Given:

$DB = 7.2$ cm,

$AE = 1.8$ cm and

$EC = 5.4$ cm and

$DE || BC$

$\cfrac {AD}{DB} = \cfrac {AE}{AC}$

[by basic proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same.

$\cfrac {AD}{7.2} = \cfrac {1.8}{5.4}$

$AD = \cfrac {7.2 × 1.8}{5.4}$

$AD = \cfrac {24}{10}$

$\implies AD = 2.4cm$

If $\triangle ABC \sim \triangle DEF, AB= 10 \ cm$ , area $(\triangle ABC)= 20 \ sq. cm$ , area $(\triangle DEF)= 45 \ sq. cm$ . Determine $DE$

We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\therefore \dfrac{ar(\triangle ABC)}{ar(\triangle DEF)}= \dfrac{(AB)^2}{(DE)^2}$

$\implies \dfrac{20}{45}= \dfrac{(10)^2}{(DE)^2}$ [given]

$\implies \dfrac{20}{45}= \dfrac{100}{(DE)^2}$

$\implies (DE)^2= \dfrac{100 \times 45}{20}$

$\implies (DE)^2= 5 \times 45$

$DE= 15 \ cm$

In ABC, P and Q are two points on AB and AC. respectively such that PQ || BC and $\dfrac {AP}{PB} = \dfrac{2}{3}$ then find $\dfrac{AQ}{ QC}$ .

Given: PQ || BC and

$\dfrac{AP}{PB}=\frac{2}{3}$
By Basic Proportionality theorem . which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

$\because \dfrac{AP}{PB} = \dfrac{AQ}{QC}$

$\implies \dfrac {AP}{PB} = \dfrac{AQ}{QC} = \dfrac{2}{3}$
1725744_1814090_ans_9b3029c33d1b4e7fbbe005f9e710b45c.png

In a $\triangle ABC, DE || BC$ , where D is a point on AB and E is a point on AC, then
$\dfrac{EC}{DB}$ =...........

we have

$\dfrac{AB}{DB} = \dfrac{AC}{EC}$

$\dfrac{EC}{DB} = \dfrac{AC}{AB}$
1725864_1815679_ans_2b5cae25ca1f40609d99e2d175dbbbcb.jpg

In figure, $DE||BC$ . Find $AD$

Given:

$DB = 7.2 cm, AE = 1.8 cm$

$EC = 5.4 cm$ and

$DE \parallel BC$ .

By basic proportionality theorem which states that "if a line is drawn parallel to one side of a triangle to other sides in distinct points, then the other two sides are divided in the same ratio", we write:

$\dfrac {AD}{DB} = \dfrac {AE}{AC}$

$\dfrac {AD}{7.2} = \dfrac {1.8}{5.4}$

$AD = \dfrac {7.2 × 1.8}{5.4}$

$AD = \dfrac {24}{10}$

$\implies AD = 2.4\ cm$

State whether the following pair of triangle is similar or not. Write the similarity criterion used and write it in symbolic form.

$\Delta ABC$ and

$\Delta PQR$

$\angle A = \angle Q = 85^o$

$\angle B = \angle P = 60^o$
and

$\angle C = \angle R = 35^o$
so,

$\Delta PQR \sim \Delta LMN$ [by AAA similarity]

In figures DE || BC. Find EC..

Given: DE | | BC . Hence $\dfrac {AD}{DB} = \dfrac {AE}{EC}$

[by basic proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other twos sides are divided in the same ratio

$\implies {1.5}{3} = \dfrac {1}{EC}$

[Given: AD= 1.5cm, DB = 3cm & AE = 1cm ]

$EC = \dfrac {3}{1.5}$

$\implies EC = 2cm$

In the adjoining figure, $AD = 2 cm, DB = 3 cm,AB = 5 cm$ and $DE || BC$ , then find $EC$ .

Given:

$AD = 2cm, DB =3cm, AB = 5cm$
and

$DE~ || BC$

Basic Proportionality theorem which states that. if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio. So, by basic proportionality theorem.

$\dfrac {AD}{DB} =\dfrac {AE}{EC}$

$\dfrac {2}{3} = \dfrac{5}{EC}$

$EC = \dfrac {5\times 3}{2}$

$EC = 7.5cm$

In the above figure, $DE \parallel BC.$ Find $x.$

Given:

$AD = x , DB =16, AB = 34$ and

$EC = 17$

$DE \parallel BC$ Basic Proportionality theorem. which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio. So, by the basic proportionality theorem.

$\begin{aligned}{}\frac{{AD}}{{DB}} &= \frac{{AE}}{{EC}}\\\frac{x}{{16}} &= \frac{{34}}{{17}}\\\frac{x}{{16}} &= 2\\x& = 32\end{aligned}$

In the given $\triangle QPR$ , $LM$ is parallel to $QR$ and $PM: MR = 3: 4$ . Calculate the value of the ratio $PL: PQ$ and then $LM : QR$ .

In $\Delta PLM$ and $\Delta PQR,$

As $LM || QR,$ corresponding angles are equal.

$\angle{PLM} = \angle{PQR}$

$\angle{PML} = \angle{PRQ}$

Hence, $\Delta PLM \sim \Delta PQR$ by $AA$ criterion for similarity.

So, we have

$\dfrac{PM}{ PR} = \dfrac{LM}{ QR}$ (the ratio of the corresponding sides of two similar triangles is in proportion)

$\dfrac{3}{7} = \dfrac{LM}{QR}$ $[\because PM:PR=3:(3+4)=3:7]$

$\therefore LM : QR=3:7$

$\because LM || QR$ then, by basic proportionality theorem we have

$\dfrac{PL}{LQ} = \dfrac{PM}{MR} = \dfrac{3}{4}$

$\implies \dfrac{LQ}{PL} = \dfrac{4}{3}$

$1 + \dfrac{LQ}{PL} = 1 + \dfrac{4}{3}$ (by componendo)

$\implies \dfrac{PL + LQ}{ PL} = \dfrac{3 + 4}{3}$

$\implies \dfrac{PQ}{PL} = \dfrac{7}{3}$

Hence, $\dfrac{PL}{PQ} = \dfrac{3}{7}$

$\therefore PL:PQ=3:7$

In the following figure, $XY$ is parallel to $BC$ , $AX = 9\ cm$ . $XB = 4.5 cm$ and $BC = 18\ cm$ .
What is the value of $\displaystyle\,\dfrac{YC}{AC}$ ?

We have,

$AX/AB = AY/AC$

$9/13.5 = AY/ AC$

$YC/AC = 4.5/13.5 = 1/3$

In the following figure, $AB=AC;BC=CD$ and $DE$ is parallel to $BC$ . Calculate:
$\angle CDE$

$\angle FAB=128^o$ [given]

$\angle BAC+\angle FAB=180^o$ [

$FAC$ is a st. line ]

$\Rightarrow \angle BAC=180^o-128^o$

$\Rightarrow \angle BAC=52^o$
In

$\Delta ABC,$

$\angle A=52^o$

$\angle B=\angle C$ [Given

$AB=AC$ and angles opposite to equal sides are equal]

$\angle A+\angle B+\angle C=180^o$

$\Rightarrow \angle A+\angle B+\angle B=180^o$

$\Rightarrow 52^o+2\angle B=180^o$

$\Rightarrow 2\angle B=128^o$

$\Rightarrow \angle B=64^o=\angle C$ .........(i)

$\angle B=\angle ADE$ [Given DE||BC]
Now,

$\angle ADE+\angle CDE+\angle B=180^o$ [

$ADB$ is a st. line]

$\Rightarrow 64^o+\angle CDE+64^o=190^o$

$\Rightarrow \angle CDE=180^o-128^o$

$\Rightarrow \angle CDE=52^o$

In the following figure, point D divides AB in the ratio $3: 5$ . Find: $BC = 4.8\,cm$ , find the length of DE.

Since, ΔADE ~ ΔABC by AA criterion for similarity

So, we have

AD/AB=DE/BC

3/8=DE/4.8

DE=1.8cm

In the following figure, AB, CD and EF are perpendicular to the straight line BDF.If AB = x and; CD = z unit and EF = y unit, prove that: 1/x + 1/y = 1/z.

In $\Delta$ FDC and $\Delta$ FBA,

$\angle{FDC} = \angle{FBA}$ [As DC || AB]

$\angle{DFC} = \angle{BFA}$ [common angle]

Hence, $\Delta$ FDC ~ $\Delta$ FBA by AA criterion for similarity.

So, we have

$DC/AB = DF/BF$

$z/x = DF/BF$ …. (1)

In $\Delta$ BDC and $\Delta$ BFE,

$\angle{BDC} =\angle{BFE}$ [As DC || FE]

$\angle{DBC} = \angle{FBE}$ [Common angle]

Hence, $\Delta$ BDC ~ $\Delta$ BFE by AA criterion for similarity.

So, we have

$BD/BF = z/y$ ….. (2)

Now, adding (1) and (2), we get

$BD/BF + DF/BF = z/y + z/x$

$1 = z/y + z/x$

$Thus,$

$1/z = 1/x + 1/y$

– Hence Proved

A line PQ is drawn parallel to the side BC of Δ ABC which cuts side AB at P and side AC at Q. If $AB = 9.0\,cm$ , $CA = 6.0\,cm$ and $AQ = 4.2\, cm$ , find the length of AP.

In $\Delta$ APQ and $\Delta$ ABC,

$\angle{ACQ} = \angle{ABC}$ [As PQ || AB, corresponding angles are equal.]

$$\angle{PAQ = \angle{BAC}$$ [Common angle]

Hence, $\Delta$ APQ ~ $\Delta$ ABC by AA criterion for similarity

So, we have

$AP/AB = AQ/AC$

$AP/9 = 4.2/6$

Thus,

$AP = 6.3\,cm$

In the given figure, PQ AB; $CQ = 4.8\,cm$ $QB = 3.6\,cm$ and $AB = 6.3\,cm$ . Find:
If AP = x, then the value of AC in terms of x.

As $\Delta$ CPQ ~ $\Delta$ CAB by AA criterion for similarity

We have,

$\dfrac{CP}{AC} = \dfrac{CQ}{CB}$

$= \dfrac{4.8}{8.4}$

So, if AC is $7$ parts and CP is $4$ parts, then PA is $3$ parts.

Hence, $AC = \dfrac{7}{3} \times PA$

$=\dfrac{7}{3}x$

In the given figure, PQ ‖ AB; $CQ = 4.8\,cm$ $QB = 3.6\,cm$ and $AB = 6.3\,cm$ . Find:
$\dfrac{CP}{PA}$

In ΔCPQ and ΔCAB,

∠PCQ=∠ACB [As PQ || AB, corresponding angles are equal.]

∠C=∠C [Common angle]

Hence, ΔCPQ ~ ΔCAB by AA criterion for similarity

So, we have

CP/CA=CQ/CB

CP/CA=4.8/8.4=4/7

Thus, CP/PA=4/3

In the figure given below, AB EF CD. If $AB = 22.5\,cm$ , $EP = 7.5\,cm$ , $PC = 15\,cm$ and $DC = 27\,cm$ . Calculate AC.

And, as EF || AB

$\Delta$ CEF ~ $\Delta$ CAB by AA criterion for similarity.

$EC/AC = EF/AB$

$22.5/AC = 13.5/22.5$

Thus, $AC = 37.5\,cm$

In the following figure, $\angle{AXY} =\angle{AYX}.$ If $\dfrac{BX}{AX} =\dfrac{ CY}{AY},$ show that $\triangle ABC$ is isosceles.

Given: $\angle{AXY} = \angle{AYX}$

So, $AX = AY$ [Sides opposite to equal angles are equal.]

Also, from Basic Proportionality Theorem, we have

$\Rightarrow\dfrac{BX}{AX} = \dfrac{CY}{AY}$

$\Rightarrow BX=CY$

$\Rightarrow AX + BX = AY + CY$

So, $AB = AC$

Therefore, $\triangle ABC$ is an isosceles triangle.

In the following figure, $OAB$ is a triangle and $AB\parallel DC.$
If the area of triangle $CAD=140\ cm^{2}$ and the area of triangle $ODC=172\ cm^{2},$ find
The area of triangle $ODB$

Given:

$\triangle CAD=140\ cm^{2}$

$\triangle ODC=172\ cm^{2}$

$AB\parallel CD$
As Triangle

$DBC$ and

$\triangle CAD$ have same base

$CD$ and between the same parallel lines Hence,
Area of

$\triangle DBC$ =Area of

$\triangle CAD=140\ cm^{2}$
Area of

$\triangle OAC$ =Area of

$\triangle CAD+$ Area of

$\triangle CDC=140\ cm^{2}+172\ cm^{2}=312\ cm^{2}$
Area of

$\triangle OBD$ =Area of

$\triangle DBC+$ Area of

$\triangle ODC=140\ cm^{2}+172\ cm^{2}=312\ cm^{2}$

In the following figure, $OAB$ is a triangle and $AB\parallel DC.$
If the area of triangle $CAD=140\ cm^{2}$ and the area of triangle $ODC=172\ cm^{2},$ find
The area of triangle $OAC$

Given:

$\triangle CAD=140\ cm^{2}$

$\triangle ODC=172\ cm^{2}$

$AB\parallel CD$
As Triangle

$DBC$ and

$\triangle CAD$ have same base

$CD$ and between the same parallel lines Hence,
Area of

$\triangle DBC$ =Area of

$\triangle CAD=140\ cm^{2}$
Area of

$\triangle OAC$ =Area of

$\triangle CAD+$ Area of

$\triangle CDC=140\ cm^{2}+172\ cm^{2}=312\ cm^{2}$

In the following figure, $OAB$ is a triangle and $AB\parallel DC.$
If the area of triangle $CAD=140\ cm^{2}$ and the area of triangle $ODC=172\ cm^{2},$ find
The area of triangle $DBC$

Given:

$\triangle CAD=140\ cm^{2}$

$\triangle ODC=172\ cm^{2}$

$AB\parallel CD$
As Triangle

$DBC$ and

$\triangle CAD$ have same base

$CD$ and between the same parallel lines Hence,
Area of

$\triangle DBC$ = Area of

$\triangle CAD=140\ cm^{2}$

In the following figure, $AB=AC;BC=CD$ and $DE$ is parallel to $BC$ . Calculate:
$\angle DCE$

$\angle FAB=128^o$ [given]

$\angle BAC+\angle FAB=180^o$ [

$FAC$ is a st. line ]

$\Rightarrow \angle BAC=180^o-128^o$

$\Rightarrow \angle BAC=52^o$
In

$\Delta ABC,$

$\angle A=52^o$

$\angle B=\angle C$ [Given

$AB=AC$ and angles opposite to equal sides are equal]

$\angle A+\angle B+\angle C=180^o$

$\Rightarrow \angle A+\angle B+\angle B=180^o$

$\Rightarrow 52^o+2\angle B=180^o$

$\Rightarrow 2\angle B=128^o$

$\Rightarrow \angle B=64^o=\angle C$ .........(i)

$\angle B=\angle ADE$ [Given DE||BC]
Given

$DE||BC$ and

$DC$ is the transversal.

$\Rightarrow \angle CDE=\angle DCB=52^o$ ........(ii)
Also,

$\angle ECB=64^o$ .........[From(i)]
But,

$\angle ECB=\angle DCE+\angle DCB$

$\Rightarrow 64^o=\angle DCE+52^o$

$\Rightarrow \angle DCE=64^o-52^o$

$\Rightarrow \angle DCE=12^o$

In adjoining $AP \perp BC , AD || BC,$ then find $A \left ( \Delta ABC \right ) : A \left (\Delta BCD \right )$

Given

$AP \perp BC$
AD || BC

$\therefore \dfrac{A\left ( \Delta ABC \right )}{A\left ( \Delta BCD \right )} = \dfrac{AP\times BC}{AP \times BC}$

$= \dfrac{1}{1}$
Hence , the ratio of

$A \left ( \Delta ABC \right ) and A\left ( \Delta BCD \right )$ is 1 : 1

Base of a is 9 and height is 5 . Base of another triangle is 10 and height is 6 . Find the ratio of areas of these triangle .

Let ABC and PQR be two right triangle with

$AB \perp BC$ and

$PQ \perp QR$
Given
BC = 9 , AB = 6 and QR = 10 .

$\therefore \dfrac{A\left ( \Delta ABC \right )}{A \left ( \Delta PQR \right )} = \dfrac{AB \times BC}{PQ \times QR}$

$= \dfrac{5 \times 9}{6\times 10}$

$= \dfrac{3}{4}$

In the following figure, $DE$ is parallel to $BC$ . Shows that:
Area $(\triangle BOD)=$ Area $(\triangle COE)$

We know that area of triangles on the same parallel are equal
Area (triangle

$DBC$ ) =Area (triangle

$BCE$ )
Area (triangle

$DOB$ )+ Area (triangle

$BOC$ )= Area (triangle

$BOC$ ) +Area (triangle

$COE$ )
So Area (triangle

$DOB$ )=Area (triangle

$COE$ )

In the following figure, $DE$ is parallel to $BC$ . Shows that:
Area $(\triangle ADC)=$ Area $(\triangle AEB)$

$\triangle ABC, D$ is midpoint of

$AB$ and

$E$ is the midpoint of

$AC$ .

$\dfrac {AD}{AB}=\dfrac {AE}{AC}$

$DE$ is parallel to

$BC$

$\therefore Ar.(\triangle ADC)= Ar.(\triangle BDC)=\dfrac 12 Ar.(\triangle ABC)$
Again

$\therefore Ar.(\triangle AEB)= Ar.(\triangle BEC)=\dfrac 12 Ar.(\triangle ABC)$
From the above two equation, we have
Area

$(\triangle ADC)=Area\ (\triangle AEB)$ .
Hence proved.

In the given figure $BC \perp AB$ and $AD \perp AB$ , BC = 4 , AD = 8 , then find $\dfrac{A\left ( \Delta ABC \right )}{A\left ( \Delta ADB \right )}$

Given
BC = 4
AD = 8

$\therefore \dfrac{A\left ( \Delta ABC \right )}{A\left ( \Delta ADB \right )} = \dfrac{AB\times BC}{AB \times AD}$

$= \dfrac{BC}{AD} \left [ \because BC = 4\ and\ AD = 8 \right ]$

$= \dfrac{4}{8}$

$= \dfrac{1}{2}$

In $\Delta ABC$ , ray BD bisects $\angle ABC$ and ray CE bisects $\angle ACB$ . If seg $AB\cong seg AC$ then prove that ED || BC .

Given

ray BD bisects

$\angle ABC$

ray BD bisects

$\angle ACB$

seg

$AB\cong seg AC$

$\Delta ABC , \angle ABD = \angle DBC$

$\dfrac{AD}{DC} = \dfrac{AB}{BC}$ ....(I)(By angle bisector theorem)

$\Delta ABC , \angle BEC = \angle ACE$

$\dfrac{AE}{EB} = \dfrac{AC}{BC}$ .(II)(By angle bisector theorem)

From (I)and (II)

$\dfrac{AD}{DC} = \dfrac{AE}{EB} \left ( \therefore seg AB\cong seg AC \right ) ED || BC$ (converse of basic proportional theorem)

1914539_1852103_ans_c93161d7bfd449b8b2fbb5a599cf9063.png

In $\Delta PQR ,$ PM = 15 , PQ = 25 PR = 20 , NR = 8 . State whether line NM is parallel to side RQ . Give reason .

Given
PM = 15,
PQ = 25,
PR = 20 and
NR = 8
Now , PN = PR - NR
= 20 - 8
= 12
Also , MQ = PQ - PM
25 - 15
= 10
In

$\Delta PRQ$

$\dfrac{PR}{NR} = \dfrac{12}{8}$

$= \dfrac{3}{2}$
Also,

$\dfrac{PM}{MQ} = \dfrac{15}{10}$

$= \dfrac{3}{2}$

$\therefore \dfrac{PR}{NR} = \dfrac{PM}{MQ}$
By convers of basic proportionality theorem , NM is parallel to side RQ or NM || RQ .

If $\Delta ABC \sim \Delta PQR$ and AB : PQ = 2 :3 , then fill in the blanks $\dfrac{A\left ( \Delta ABC \right )}{A\left ( \Delta PQR \right )} = \dfrac{AB^{2}}{...} = \dfrac{2^{2}}{3^{2}} = \dfrac{........}{.......}$

Given :

$\Delta ABC \sim \Delta PQR$

$AB : PQ = 2 :3$ According to theorem of areas of similar triangles "When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the square of their corresponding sides''.

$\therefore \dfrac{A\left ( \Delta ABC \right )}{A\left ( \Delta PQR \right )} = \dfrac{AB^{2}}{PQ^{2}} = \dfrac{2^{2}}{3^{2}} = \dfrac{4}{9}$

Measure ofsome angles in the figure are given . Prove that $\dfrac{AP}{PB} = \dfrac{AQ}{QC}$

Given :

$\angle APQ = 60^{o}$

$\angle ABC = 60^{o}$
Since , the corresponding angles

$\angle APQ and \angle APC$ are equal
Hence, line PQ ||BC.
In

$\Delta ABC , PQ ||BC$

$\dfrac{AP}{PB} = \dfrac{AQ}{QC}$ (By proportionality theorem)

If $\Delta ABC \sim \Delta PQR$ $A (\Delta ABC) = 80 (\Delta PQR) = 125$ then, fill in the blanks $\dfrac{A\left ( \Delta ABC \right )}{A\left ( \Delta .... \right )} = \dfrac{80}{125} \therefore \dfrac{AB}{PQ} = \dfrac{.....}{......}$

Given

$\Delta ABC \sim \Delta PQR$

$A (\Delta ABC) = 80$

$A (\Delta PQR) = 125$

According to theorem of areas of similar triangles ""When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the square of their corresponding sides'

$\therefore \dfrac{A\left ( \Delta ABC \right )}{A\left ( \Delta PQR \right )} = \dfrac{AB^{2}}{PQ^{2}}$

$\Rightarrow \dfrac{80}{125} = \dfrac{AB^{2}}{PQ^{2}}$

$\dfrac{16}{25} = \dfrac{AB^{2}}{PQ^{2}}$

$\Rightarrow \dfrac{4^{2}}{5^{2}} = \dfrac{AB^{2}}{PQ^{2}}$

$\Rightarrow \dfrac{AB}{PQ} = \dfrac{4}{5}$ Therefore,

$\dfrac{A\left ( \Delta ABC \right )}{A\left ( \Delta PQR \right )} = \dfrac{80}{125} \ and\ \dfrac{AB}{PQ} = \dfrac{4}{5}$

In adjoining figure $PQ \perp BC , AD \perp BC$ then find following ratios.
$\dfrac{A\left ( \Delta ADC \right )}{A\left ( \Delta PQC \right )}$

1959139_1851963_ans_e5ab8d837fea4b29a09c0c58b7c321a4.jpg

In adjoining figure $PQ \perp BC , AD \perp BC$ then find following ratios.
$\dfrac{A\left ( \Delta ABC \right )}{A\left ( \Delta ADC \right )}$

$\Delta ABC\ and\ \Delta ADC have\ same\ height\ AD$

$\dfrac{Area\ of\ \Delta{ABC}}{Area\ of\ \Delta{ADC}} = \dfrac{BC}{DC}$ [Triangles having equal hights]

In adjoining figure $PQ \perp BC , AD \perp BC$ then find following ratios.
$\dfrac{A\left ( \Delta PQB \right )}{A\left ( \Delta PBC \right )}$

$\Delta PQB\ and\ \Delta PBC have\ same\ height\ PQ$

$\dfrac{Area\ of\ \Delta{PQB}}{Area\ of\ \Delta{PBC}} = \dfrac{BQ}{BC}$ [Triangles having equal hights]

In the given figure , seg AC and BD intersect each other in point P and $\dfrac{AP}{CP} = \dfrac{BP}{DP}$ Prove that , $\Delta ABP \sim \Delta CDP$

Given :

$\dfrac{AP}{CP} = \dfrac{BP}{DP}$
To Prove :

$\Delta ABP \sim \Delta CDP$
Proof : In

$\Delta ABP\ and\ \Delta DCP$

$\dfrac{AP}{CP} = \dfrac{BP}{DP}$

$\angle APB = \angle DPC$ (Vertically opposite angles)
By SAS test of similarity

$\Delta ABP \sim \Delta CDP$

In the given figure , X is any point in the interior of triangle . Point X is joined to vertices of triangles .Seg PQ || seg DE , sed QR || seg EF . Fill in the blanks to prove that, sef PR || seg DF .
Proof : In $\Delta XDE , PQ || DE$
$\therefore \dfrac{XP}{[ ]} = \dfrac{[ ]}{QE}$ ...(I)(Basic proportionality theorem)
In $\Delta XEE, QR || EF$
$\therefore \dfrac{[ ]}{[ ]} = \dfrac{[ ]}{[ ]}$ .......(II)_______
$\dfrac{[ ]}{[ ]} = \dfrac{[ ]}{[ ]}$ ...From (I)and (II)
$\therefore seg PR || seg DE$ ........(converse of basic proportionality theorem)

Given
seg PQ || seg DE
seg QR || seg EF
In

$\Delta DXE , PQ || DE$

$\dfrac{XP}{PD} = \dfrac{XQ}{QE}$ .......(I)(By Basic proportionality theorem)
In

$\Delta XEF , QR || EF$ ......Given

$\therefore \dfrac{XQ}{QE} = \dfrac{XR}{RF}$ .(II)(By Basic proportionality theorem)

$\therefore \dfrac{XP}{PD} = \dfrac{XR}{RF}$ From (I)and(II)

$\therefore seg PR || seg DF$ (Converse of basic proportionality theorem)

$\Delta LMN \sim \Delta PQR , 9 \times A \left ( \Delta LMN \right ) =16 \times A (\Delta LMN )$ . IF QR = 20 then find MN.

Given

$\Delta LMN \sim \Delta PQR = 16 \times A (\Delta LMN )$

$9 \times A \left ( \Delta LMN \right )= 16 \times A (\Delta LMN )$

$\dfrac{A\left ( \Delta LMN \right )}{A\left ( \Delta PQR \right )} = \dfrac{9}{16}$

$\Rightarrow \dfrac{MN^{2}}{QR^{2}} = \dfrac{3^{2}}{4^{2}}$

$\Rightarrow \dfrac{MN}{QR} = \dfrac{3}{4}$

$\Rightarrow MN = \dfrac{3}{4}\times QR$

$\Rightarrow MN = \dfrac{3}{4} \times 20 [\therefore QR = 20]$

$\Rightarrow MN = 15$

Find the length of the side and perimeter of an equilateral triangle whose height is $\sqrt{3}cm$

Since ,

$\Delta ABC$ is an equilateral triangle

$, CD$ is the perpendicular bisector of

$AB$

Now. According to pythagoras theorem ,

$In \bigtriangleup ACD$

$AC^{2} = AD ^{2} + CD ^{2}$

$\Rightarrow (2a)^{2} = a^{2} + (\sqrt{3})^{2}$

$\Rightarrow 4a^{2} - a^{2} = 3$

$\Rightarrow 3a^{2} = 3$

$\Rightarrow a^{2} = 1$

$\Rightarrow a = \sqrt{1}= 1 cm$

Hence , the length of the side of an equilateral triangle is 2 cm.

Now, Perimeter of the triangle

$= (2 + 2 + 2 ) cm$

$=6 cm$

Hence perimeter of an equilateral triangle is 6 cm

1802054_1852368_ans_12acac944a7d481495dd496ebd0d5bc6.png

In $\square ABCD , seg AD || seg BC$ .Diagonal AC and diagonal BD intersect each other in point P .Then show that $\dfrac{AP}{PD} = \dfrac{PC}{BP}$

Given

$\square ABCD$ is a parallelogram
To prove :

$\dfrac{AP}{PD} = \dfrac{PC}{BP}$

$\angle APD = \angle CPB$

$\angle PAD = \angle PCB$
By AA test of similarity

$\Delta APD \sim \Delta CPB$

$\therefore \dfrac{AP}{PC} = \dfrac{PD}{PB}$

$\dfrac{AP}{PD} = \dfrac{PC}{BP}$
Hence proved.

In the given figure , A - D - C and B - E - C seg || side AB If AD = 5 , DC = 3 , BC = 6.4 then find BE.

1959280_1852332_ans_141f27b2749e472c85bf5f9e79434bc7.png

Do sides 7 cm , 24 cm , 25 cm form a right angled triangle ? Give reason

In the triplet

$( 7 , 24 , 25 )$

$7^{2} = 49 , 4 ^{2} = 576 , 25 ^{2} = 625 \ and \ 9 + 576 = 625$

The square of the largest number is equal to the sum of the square of the other two numbers.

$\therefore$ Sides

$7 cm , 24 cm , 25 cm$ form a right angled triangle

Pranali and Prasad started walking to the East and to the North respectively , from the same point and at the same speed. After 2 hours distance between them was $15\sqrt{2}$ km. Find their speed per hour.

It is given that , Pranali and Prasad have same speed.

Thus, they cover same distance in

$2$ hours .

i.e.

$OA = OB$

By pythagoras theorem ,

$In \bigtriangleup AOB$

$AB^{2} = AO^{2} + OB^{2}$

$\Rightarrow (15\sqrt{2})^{2} = AO^{2} + OA^{2}$

$\Rightarrow 450 = 2AO^{2}$

$\Rightarrow AO^{2} = \dfrac{450}{2}$

$\Rightarrow AO^{2} = 225$

$\Rightarrow AO = 15 km$

$\Rightarrow BO = 15 km$

$\Rightarrow Speed = \dfrac{Distance}{Time}$

$=\dfrac{15}{2}$

$= 7.5 km$ per hour

1803060_1852440_ans_04cb23a072724cd49cbeebca3ec5cf7e.png

$\Delta MNT \sim \Delta QRS$ .Length of altitude drawn from point T is 5 and length of altitude drwan from point S is 9 .Find the ratio $\dfrac{A\left ( \Delta MNT \right )}{A\left ( \Delta QRS \right )}$

$\therefore \dfrac{A\left ( \Delta MNT \right )}{A\left ( \Delta QRS \right )} = \left ( \dfrac{T}{S} \right )^{2} =(\dfrac{5}{9})^2$

$= \dfrac{25}{81}$

Areas of two similar triangles are $225 sq .cm$ and $81 sq. cm$ . If a side of the smaller triangle is $12 cm$ , then find the corresponding side of the bigger triangle.

Refer image,

Given,

Area of bigger triangle

$=225cm^2$

Area of smaller triangle

$=81cm^2$

One side of smaller triangle

$=12cm$

Let us draw a suitable diagram,

As we know, for two similar triangles,

Ratio of their area =Square of the ratios of corresponding sides

$\dfrac{Area\,of\,\Delta ABC}{Area\,of\,\Delta PQR}=\dfrac{AB^2}{PQ^2}$

$\Rightarrow \dfrac { 225 }{ 81 } =\dfrac {AB^2 }{12^2 } \Rightarrow \dfrac { 25 }{ 9 } =\dfrac {AB }{ 12 } \Rightarrow \dfrac { 25 }{ 9} \times 12=AB$

$\Rightarrow AB=\dfrac{25}{9}\times 12=\dfrac{25}{3}\times 4=\dfrac{100}{3}=33.33cm$

How the corresponding side of the bigger triangle measures

$33.33cm$ (approx)

1927562_1852249_ans_1598572698284d3f8afdc7deec6c30b2.png

In the given figure 1.66 , seg PQ || seg DE , $A\left ( \Delta PQF \right ) = 20$ units , $PF = 2 DP$ then find $A \left ( \square DPQE \right )$ by completing the following activity .

2011613_1852258_ans_1c9480023de8413598d790c99dabed63.jpg

In $\bigtriangleup$ PQR, PQ = $\sqrt{8}$ , QR = $\sqrt{5}$ and PR = $\sqrt{3}$ . Is $\bigtriangleup$ PQR a right angled triangle? If yes, which angle is of 90 $^0$ ?

Given that, in

$\bigtriangleup PQR,$

$PQ=\sqrt{8},QR=\sqrt{5},PR=\sqrt{3}$

$(\sqrt{8})^{2}=8, \ (\sqrt{5})^{2} = 5, \ (\sqrt{3})^{2} = 3$

We can see that, the square of the length of the longest side of the triangle is equal to the sum of squares of the other two sides.

i.e.,

$(\sqrt8)^2=(\sqrt5)^2+(\sqrt3)^2$

Hence, according to the converse of Pythagoras theorem,

$\bigtriangleup PQR$ is a right angled triangle.

Thus, the longest side of the triangle is

$PQ$ , which is opposite to

$\angle R$ .

Hence,

$\angle R=90^0$

Hence, the given

$\bigtriangleup PQR$ is a right angled triangle, having

$\angle R=90^0$ .

In a $\triangle ABC, D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE \parallel BC$ .
In $AD=4x-3,AE=8x-7,BD=3x-1$ and $CE=5x-3$ then find $x$ .

$\triangle ABC$ ,

$DE\parallel BC$

$\cfrac{AD}{DB}=\cfrac{AE}{CE}$
(By B.P theorem)

$\cfrac{4x-3}{3x-1}=\cfrac{8x-7}{5x-3}$

$(4x-3)(5x-3)=(8x-7)(3x-1)$

$\Rightarrow$

$20{x}^{2}-12x-15x+9=24{x}^{2}-8x-21x+7$

$\Rightarrow$

$24{x}^{2}-20{x}^{2}-29x+27x-9+7=0$

$\Rightarrow$

$4{x}^{2}-2x-2=0$

$\Rightarrow$

$2{x}^{2}-x-1=0$

$\Rightarrow$

$2{x}^{2}-2x+x+1=0$

$\Rightarrow$

$2x(x-1)+1(x-1)=0$

$\Rightarrow$

$(x-1)(2x+1)=0$

$\Rightarrow$

$x-1=0$

$\Rightarrow$

$x=1$
or

$2x+1=0$

$2x=-1$

$x=-\cfrac{1}{2}$
Thus,

$x=1,-\cfrac{1}{2}$
1862131_1875924_ans_1479869fa3034105b5fb3ccbf3b18753.png

Points $D$ and $E$ lie on sides $AB$ and $AC$ of $\triangle ABC$ , respectively. In the following questions check whether $DE\parallel BC$ or not.
$AB=5.6cm$ , $AD=1.4cm$ , $AC=9.0cm$ and $AE=1.8cm$

$\cfrac{AD}{BD}=\cfrac{AD}{AB-AD}=\cfrac{1.4}{5.6-1.4}$

$\cfrac{AD}{BD}=\cfrac{1.4}{4.2}=\cfrac{1}{3}$

$\cfrac{AE}{CE}=\cfrac{AE}{AC-CE}=\cfrac{1.8}{9.0-1.8}$

$\cfrac{AE}{CE}=\cfrac{1.8}{7.2}=\cfrac{1}{4}$

$\cfrac{AD}{BD}\ne \cfrac{AE}{CE}$
Hence

$DE\ne BC$

Points $D$ and $E$ lie on sides $AB$ and $AC$ of $\triangle ABC$ , respectively. In the following questions check whether $DE\parallel BC$ or not.
$AD=10.5cm$ , $BD=4.5cm$ , $AC=4.8cm$ and $AE=2.8cm$

$\cfrac{AD}{BD}=\cfrac{10.5}{4.5}=\cfrac{7}{3}$

$\cfrac{AE}{CE}=\cfrac{AE}{AC-AE}=\cfrac{2.8}{4.8-2.8}$

$\cfrac{AE}{CE}=\cfrac{2.8}{2}=\cfrac{7}{5}$

$\cfrac{AD}{BD}\ne \cfrac{AE}{CE}$
Hence

$DE\ne BC$

Points $D$ and $E$ lie on sides $AB$ and $AC$ of $\triangle ABC$ , respectively. In the following questions check whether $DE\parallel BC$ or not.
$AB=12cm$ , $AD=8cm$ , $AE=12cm$ and $AC=18cm$

$\cfrac{AD}{BD}=\cfrac{AD}{AB-AD}=\cfrac{8}{12-8}$

$\cfrac{AD}{BD}=\cfrac{8}{4}=\cfrac{2}{1}$

$\cfrac{AE}{CE}=\cfrac{AE}{AC-AE}=\cfrac{12}{18-12}$

$\cfrac{AE}{CE}=\cfrac{12}{6}=\cfrac{2}{1}$

$\cfrac{AD}{BD}=\cfrac{AE}{CE}=\cfrac{2}{1}$
Thus

$DE\parallel BC$

1862134_1875925_ans_c296882cae7c40d2bb3519e53ec71b26.png

In a $\triangle ABC, D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE \parallel BC$ .
If $\cfrac { AD }{ DB } =\cfrac { 4 }{ 13 }$ and $AC=20.4cm$ . Then find $EC$ .

$\triangle ABC$ ,

$DE\parallel BC$

$\cfrac{AD}{DB}=\cfrac{AE}{CE}$
(By B.P theorem)

$\cfrac{4}{13}=\cfrac{AC-CE}{CE}$

$(AE+CE=AC)$

$\cfrac{4}{13}=\cfrac{20.4-CE}{CE}$

$4\times CE=13(20.4-CE)$

$4\times CE+13CE=13\times 20.4$

$17CE=13\times 20.4$

$CE=\cfrac{13\times 20.4}{17}=13\times 1.2$

$CE=15.6cm$
1862123_1875921_ans_1b66b5aa880b4ab9a505906ea2db5f74.png

In a $\triangle ABC, D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE \parallel BC$ .
If $\cfrac { AD }{ DB } =\cfrac { 7 }{ 4 }$ and $AE=6.3cm$ then find $AC$ .

$\triangle$ ,

$DE\parallel BC$

$\cfrac{AD}{DB}=\cfrac{AE}{CE}$
(By B.P theorem)

$\cfrac{7}{4}=\cfrac{6.3}{AC-AE}$

$\cfrac{7}{4}=\cfrac{6.3}{AC-6.3}$

$7(AC-6.3)=4\times 6.3$

$AC-6.3=\cfrac{4\times 6.3}{7}$

$AC=6.3+3.6=9.9cm$
1862128_1875922_ans_66126633349b406a9959481e18922026.png

In a triangle ABC , AB = AC . Suppose P is a point on AB and Q is point on AC such that AP = AQ Prove that $\bigtriangleup APC \cong \bigtriangleup AQB$

$\bigtriangleup APC$ and

$\bigtriangleup AQB$

$AP = AQ$ [ data]

$AC = AB$ [data]

$\angle PAC = \angle BAQ$ [ common angle ]

$\therefore \bigtriangleup APC \cong \bigtriangleup AQB$ [ SAS postulate ]

1995878_1871306_ans_5c69ea048f014a27a512b6dc7bf84a15.png

Draw the shape of any two examples of similar figures.

Two quadrilaterals are similar because their corresponding angles are equal and corresponding sides are proportional.
1862115_1875916_ans_a5fc0377609f4d1ba93b11cf77922e16.png

1862115_1875916_ans_a5fc0377609f4d1ba93b11cf77922e16.png

In given figure $L,M$ and $N$ are points on $OA,OB$ and $OC$ such that $LM\parallel AB$ and $MN\parallel BC$ , then show that $LN\parallel AC$

Given: In

$\triangle ABC$ , point

$L,M,N$ respectively lie

$OA,OB$ and

$OC$ such that

$LM\parallel AB$ and

$MN\parallel BC$
To prove:

$LN\parallel AC$
Proof: In

$\triangle ABO$

$LM\parallel AB$ (given)

$\cfrac{OL}{LA}=\cfrac{OM}{MB}$ ...(ii) (By basic prop. theorem)
Again, In

$\triangle BCO$

$MN\parallel BC$

$\cfrac{OM}{MB}=\cfrac{ON}{NC}$ ...(ii) (By basic prop. theorem)
From equation (i) and (ii)

$\Rightarrow$

$\cfrac{OL}{LA}=\cfrac{OM}{MB}=\cfrac{ON}{NC}$

$\cfrac{OL}{LA}=\cfrac{ON}{NC}$ (By converse of B.P theorem)

$\Rightarrow$

$LN\parallel AC$

Points $D$ and $E$ lie on sides $AB$ and $AC$ of $\triangle ABC$ , respectively. In the following questions check whether $DE\parallel BC$ or not.
$AD=5.7cm$ , $BD=9.5cm$ , $AE=3.3cm$ and $EC=5.5cm$

$\cfrac{AD}{BD}=\cfrac{5.7}{9.5}=\cfrac{3}{5}$

$\cfrac{AE}{CE}=\cfrac{3.3}{5.5}=\cfrac{3}{5}$

$\cfrac{AD}{CD}=\cfrac{AE}{CE}=\cfrac{3}{5}$
By B.P Theorem

$\Rightarrow$

$DE\parallel BC$
1862142_1875928_ans_d6aa8d76d5d14f9c9180a228bdf1ddab.png

Solve the following questions:
In $\triangle ABC$ and $DE\parallel BC$ and $AD:DB=2:3$ then find the ratio of areas of $\triangle ADE$ and $\triangle ABC$

$\triangle ABC$

$BC\parallel DE$ and

$\cfrac{AD}{DB}=\cfrac{2}{3}$ (given)
In

$\triangle ABC$ and

$\triangle DEA$

$\angle B=\angle D$

$\angle C=\angle E$
AA similarity cirterion

$\cfrac { ar.\left( \triangle ADE \right) }{ ar.\left( \triangle ABC \right) } =\cfrac { { \left( AD \right) }^{ 2 } }{ { \left( AB \right) }^{ 2 } } =\cfrac { { \left( AD \right) }^{ 2 } }{ { \left( AD+DB \right) }^{ 2 } }$

$=\cfrac { { \left( \cfrac { AD }{ BD } \right) }^{ 2 } }{ { \left( \cfrac { AD }{ BD } +1 \right) }^{ 2 } } ={ \left( \cfrac { AD }{ BD } \right) }^{ 2 }\times \cfrac { 1 }{ { \left( \cfrac { AD }{ BD } +1 \right) }^{ 2 } } ={ \left( \cfrac { 2 }{ 3 } \right) }^{ 2 }\times \cfrac { 1 }{ { \left( \cfrac { 2 }{ 3 } +1 \right) }^{ 2 } } =\cfrac { 4 }{ 9 } \times \cfrac { 9 }{ 25 } =\cfrac { 4 }{ 25 }$
1862176_1876003_ans_9919ac17541944e6876a1d7d21868786.png

1862176_1876003_ans_9919ac17541944e6876a1d7d21868786.png

In $\triangle ABC$ and $\triangle DEF$ ; $\angle A=\angle D$ , $\angle B=\angle F$ . Is $\triangle ABC\sim \triangle DEF$ ?Given reason for your answer.

No, because when

$\angle A=\angle D$ ,

$\angle B=\angle F$ , then

$\triangle ABC\sim \triangle DEF$

In $\triangle ABC$ , $D$ and $E$ points lie on sides $AB$ and $AC$ such that $BD=CE$ . If $\angle B=\angle C$ then show that $DE\parallel BC$ .

$\triangle ABC,D$ and

$E$ lie on sides

$BD=CE$ and

$\angle B=\angle C$
To prove:

$DE\parallel BC$
Proof: In

$\triangle ABC$ ,

$\angle B=\angle C$

$AB=AC$ (opposite sides of equals angles are same)

$\Rightarrow$

$AD+DB=AE+CE$ (

$\because DB=CE$ )

$\Rightarrow$

$AD=AE$

$\angle D=\angle E$ (opposite angles of equal sides are same)

$\Rightarrow$

$\triangle ADE$ is an isosceles triangle
So,

$\triangle ABC$ and

$\triangle ADE$ are isosceles triangle in which vertex angle

$\angle A$ is common in both triangles.

$\angle B=\angle C=\angle D=\angle E$

$\angle B=\angle D$ (corresponding angle)

$\Rightarrow$

$DE\parallel BC$
1862154_1875931_ans_a7a119a97d7e40d3a1eca416b3834daf.png

In figure $PQ$ and $RS$ are parallel then prove that: $\triangle POQ\sim \triangle SOR$ .

$\triangle POQ$ and

$\triangle SOR$

$PQ\parallel RS$ (Given)

$\angle OPQ=\angle OSR$ (alternative angle)

$\angle POQ=\angle ROS$ (Vertically opposite angles)
and

$\angle OQP=\angle ORS$ (alternate angles)
By AAA similarity law

$\triangle POQ\sim \triangle SOR$
1862194_1876023_ans_15b3c6e7ea5847a6977a112885785c08.png

Solve the following questions:
In figure, find $x$ in terms of $a,b$ and $c$

$\angle A=\angle B={50}^{o}$ (corresponding angle)

$AE\parallel BD$
In

$\triangle BDC$ and

$\triangle AEC$

$\angle A=\angle B={50}^{o}$ (Given)

$\angle E=\angle D$ (corresponding angles)
By AA similarity criterion

$\triangle BDC\sim \triangle AEC$

$\cfrac{BC}{AC}=\cfrac{BD}{AE}$

$\Rightarrow$

$\cfrac{c}{b+c}=\cfrac{x}{a}$

$\Rightarrow$

$x=\cfrac{ac}{b+c}$
1862182_1876006_ans_49cef64ebdb942dcba57f3c68b938f49.png

In $\triangle ABC,\angle B={90}^{o}$ and $BD$ is perpendicular to hypotenuse $AC$ then prove that
$\triangle ADB\sim \triangle BDC$

$\triangle ABC$

$\angle A+\angle C={90}^{o}$ (

$\because \angle B={90}^{o}$ )...(i)
In

$\triangle BDC$

$\angle DBC+\angle C={90}^{o}$ (

$\because \angle BDC={90}^{o}$ )...(ii)
from equation (i) and (ii)

$\angle A+\angle C=\angle DBC+\angle C$

$\angle A=\angle DBC$ ...(iii)
Now In

$\triangle ADB$ and

$\triangle BDC$

$\angle ADB=\angle BDC={90}^{o}$ (given)

$\angle DAB=\angle DBC$ (by eq. (ii))
By AA similarity cirterion

$\triangle ADB\sim \triangle BDC$
1862186_1876007_ans_6b0b5112436346a8af83d9fb7ffcefb9.png

$D$ is a mid-point of side $BC$ of $\triangle ABC$ . A line is drawn through $B$ and bisects $AD$ at point $E$ and $A$ at $C$ , then prove that $\cfrac{EX}{BE}=\cfrac{1}{3}$

Given:

$D$ is mid point of side

$BC$ of

$\triangle ABC$

$BD=DC$
and

$AE=ED$
To prove:

$\cfrac{EX}{BE}=\cfrac{1}{3}$
Construction: Draw

$DY\parallel BX$ meeting

$AC$ at

$Y$
Proof: In

$\triangle BCX$ and

$\triangle DCY$

$\angle BCX=\angle DCY$ (common)

$\angle BXC=\angle DYC$ (corresponding angle)
So,

$\triangle BCX\sim \triangle DCY$ (AA similarity)

$\Rightarrow$

$\cfrac{BX}{DY}=\cfrac{BC}{DC}=\cfrac{CX}{CY}$ (corresponding sides are proportional)

$\Rightarrow$

$\cfrac{BX}{DY}=\cfrac{2CD}{CD}$ (

$D$ is the mid point of

$BC$ )

$\Rightarrow$

$\cfrac{BX}{DY}=2$ ....(i)
Similarly,

$\triangle AEX\sim \triangle ADY$

$\cfrac{EX}{DY}=\cfrac{AE}{ED}=\cfrac{1}{2}$ ....(ii) (

$E$ is the midpoint of

$AD$ )
from (i) and (ii) on dividing (i) by (ii) we get

$\cfrac{BX}{EX}=4$

$\Rightarrow$

$BE+EX=4\times EX$

$\Rightarrow$

$3EX=BE$

$\Rightarrow$

$\cfrac{EX}{BE}=\cfrac{1}{3}$
1864268_1876032_ans_547d4b2f5e474df1bf3e9c8c5c5d33ba.png

In figure, $EF\parallel DC\parallel AB$ . Then prove that: $\cfrac{AE}{ED}=\cfrac{BF}{FC}$ .

Given:

$EF\parallel DC\parallel AB$
To prove:

$\cfrac{AE}{ED}=\cfrac{BF}{FC}$
Construction: Join

$AC$
Proof: In

$\triangle ADC$

$GE\parallel DC$

$\cfrac{AE}{ED}=\cfrac{AG}{GC}$ ...(i) (Basic proportionality theorem)
Now in

$\triangle ABC$

$AB\parallel GF$

$\cfrac {CF}{BF}=\cfrac{CG}{AG}$
or

$\cfrac{BF}{CF}=\cfrac{AG}{GC}$ ....(ii)
From (i) and (ii)

$\cfrac{AE}{ED}=\cfrac{BF}{FC}$
1863635_1876350_ans_288414a292e4404ab44b22fe9c48d1fa.png

In figure if $EF\parallel BC$ and $GE\parallel DC$ then, prove that $\cfrac{AG}{AD}=\cfrac{AF}{AB}$

$\triangle ADC$

$GE\parallel DC$

$\cfrac{AG}{AD}=\cfrac{AE}{AC}$ (By basic prop. theorem)
Thus, in

$\triangle ABC$

$FE\parallel BC$

$\cfrac{AE}{AC}=\cfrac{AF}{AB}$
From equation (i) and (ii)

$\cfrac{AG}{AD}=\cfrac{AF}{AB}$

In figure, $DE\parallel BC$ and $CD\parallel EF$ then prove that: ${AD}^{2}=AB\times AF$

Given:

$\triangle ABC$ in which

$DEBC$ and

$CDEF$
To prove:

${AD}^{2}=AB\times AF$
Proof: In

$\triangle ABC$

$DE\parallel BC$

$\cfrac{AD}{DB}=\cfrac{AE}{EC}$

$\cfrac{DB}{AD}=\cfrac{CE}{AE}$
(inverse proportionality law)

$\Rightarrow$

$\cfrac{DB}{AD}+1=\cfrac{CE}{AE}+1$

$\Rightarrow$

$\cfrac{DB+AD}{AD}=\cfrac{CE+AE}{AE}$

$\Rightarrow$

$\cfrac{AB}{AD}=\cfrac{AC}{AE}$ ...(i)
Again in

$\triangle ADC$ ,

$EF\parallel DC$

$\cfrac{AF}{FD}=\cfrac{AE}{CE}$
(By basic Pro. theorem)

$\Rightarrow$

$\cfrac{FD}{AF}=\cfrac{CE}{AE}$

$\Rightarrow$

$\cfrac{FD}{AF}+1=\cfrac{CE}{AE}+1$

$\cfrac{FD+AF}{AF}=\cfrac{CE+AE}{AE}$

$\Rightarrow$

$\cfrac{AD}{AF}=\cfrac{AC}{AE}$

$\Rightarrow$

$\cfrac{AF}{AD}=\cfrac{AE}{AC}$ ....(ii)
From (i) and (ii)

$\Rightarrow$

$\cfrac{AF}{AD}=\cfrac{AD}{AB}$

${AD}^{2}=AF\times AB$

Point $D$ and $E$ lie on side $AB$ of $\triangle ABC$ such that $AD=BE$ . If $DP\parallel BC$ and $EQ\parallel AC$ then prove that $PQ\parallel AB$ .

Given:

$\triangle ABC$ in which

$AD=BE$

$DP\parallel BC$ and

$EQ\parallel AC$
To prove

$PQ\parallel AB$
Proof: In

$\triangle ABC$

$DP\parallel BC$

$\cfrac{AD}{BD}=\cfrac{AP}{CP}$ ....(i) (By B.P theorem)
and

$QE\parallel AC$

$\cfrac{BE}{AE}=\cfrac{BQ}{QC}$ .....(ii)

$AD=BE$ (given)....(iii)

$AD+DE=BE+DE$

$AE=BD$ ....(iv)
putting values from equation (iii) and (iv) in (ii)

$\cfrac{BE}{AE}=\cfrac{AD}{BD}=\cfrac{BQ}{QC}$
From equation (i) and (v)

$\cfrac{AP}{DB}=\cfrac{AP}{CP}=\cfrac{BQ}{QC}$

$\cfrac{AP}{CP}=\cfrac{BQ}{QC}$

$\cfrac{CP}{AP}=\cfrac{QC}{BQ}$

$PQ\parallel AB$
1863662_1876374_ans_6c141958c666457bb4cecab5466c8664.PNG

In fig, $O$ is center of circle and $AB || CD$ , if $\angle ADC = 25^{0}$ , then find $\angle AEB$ .

Join

$CA$ and

$BD$ , we know that angle subtended by arc at center of the circle is double the angle at remaining part by same arc.
Thus

$\angle AOC = 2\angle ADC$

$= 2 \times 25^{0}$

$\angle AOC = 50^{0}$
Now

$AB || CD$ and

$AO$ is transversal. We know that line

$AO$ , intersects

$CD$ and

$AB$ at

$O$ and

$A$ respectively such that

$\angle OAB = \angle AOC$ and

$\angle OAB = 50^{0}$
In

$\Delta OAB$

$OA = OB$ (radius of circle)

$\angle OAB = \angle OBA$ (In a triangle angles opposite to equal sides arc equal)

$\angle OBA = \angle OAB$

$\angle OBA = 50^{0}$
In

$\Delta OAB$

$\angle OAB + \angle OBA + \angle AOB = 180^{0}$ [

$\because$ Sum of triangle angles of a triangle

$\Delta$ is

$180^{0}$ ]

$50^{0} + 50^{0} + \angle AOB = 180^{0}$

$100^{0} + \angle AOB = 180^{0}$

$\angle AOB = 180^{0}- 100^{0}$

$\angle AOB = 80^{0}$
Again, we know that angle subtended by an arc at center is double the angle at remaining part of circle by same arc.

$\therefore \angle AOB = 2\angle AEB$

$80^{0} = 2\angle AEB$

$\angle AEB = \dfrac{80^{0}}{2}= 40^{0}$

In given figure, line segment $XY$ is parallel to side $AC$ of $\triangle ABC$ and divides the triangle in two equal parts. Find ratio $\cfrac{AX}{AB}$ .

$XY\parallel AC$
In

$\triangle BXY$ and

$\triangle BAC$

$\angle BXY=\angle BAC$

$\angle BYX=\angle BCA$ (corresponding angle)

$\triangle BXY\sim \triangle BAC$ (AA similarity criterion)

$\cfrac { ar.\quad \triangle BXY }{ ar.\quad \triangle BAC } =\cfrac { { XB }^{ 2 } }{ { AB }^{ 2 } }$

$\cfrac { ar.\quad \triangle BXY }{ ar.\quad \triangle BAC } ={ \left( \cfrac { XB }{ AB } \right) }^{ 2 }$ ...(i)
But

$(ar.\triangle BAC)=2ar(\triangle XBY)$ (given)

$\Rightarrow$

$\cfrac { ar.\quad \triangle BXY }{ ar.\quad \triangle BAC } =\cfrac{1}{2}$ ....(ii)
From equation (i) and (ii)

${ \left( \cfrac { XB }{ AB } \right) }^{ 2 }=\cfrac{1}{2}$

$\Rightarrow$

$\cfrac{XB}{AB}=\cfrac{1}{\sqrt 2}$
(Taking square root of both sides)

$\Rightarrow$

$1-\cfrac{XB}{AB}=1-\cfrac{1}{\sqrt 2}$

$\Rightarrow$

$\cfrac{AB-XB}{AB}=\cfrac{\sqrt{2}-1}{\sqrt{2}}$

$\Rightarrow$

$\cfrac{AX}{AB}=\cfrac {\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}\times \sqrt{2}}$
(multiplying numerator and denominator by

$\sqrt 2$ )

$\Rightarrow$

$\cfrac{AX}{AB}=\cfrac{2-\sqrt{2}}{2}$
Thus,

$\cfrac{AX}{AB}=\cfrac{2-\sqrt{2}}{2}$

Points $E$ and $F$ lie on sides $PQ$ and $PR$ of any $\triangle PQR$ . From the following, for each case find is $EF\parallel QR$
(i) $PE=3.9cm,EQ=3cm.PF=3.6cm$ and $FR=2.4cm$
(ii) $PE=4cm,QE=4cm, PF=8cm$ and $RF=9cm$
(iii) $PQ=1.28cm,PR=2.56cm,PE=0.18cm$ and $PF=0.36cm$

$\triangle PQR$ , two point

$E$ and

$F$ lie on sides

$PQ$ and

$PR$ respectively
(i)

$\cfrac{PE}{EQ}=\cfrac{3.9}{2}=1.3cm$

$\cfrac{PF}{FR}=\cfrac{3.6}{2.4}=\cfrac{36}{24}=\cfrac{3}{2}=1.5cm$

$\cfrac{PE}{EQ}\ne \cfrac{PF}{FR}$

$EF$ is not parallel to

$QR$
(ii)

$PE=4cm$ ,

$QE=4.5cm$ ,

$PF=8cm$ and

$RF=9cm$

$\cfrac{PE}{QE}=\cfrac{4}{4.5}=\cfrac{40}{45}=\cfrac{8}{9}$ ...(i)
and

$\cfrac{PF}{RF}=\cfrac{8}{9}$
From equation (i) and (ii)

$\cfrac{PE}{QE}=\cfrac{PF}{RF}$ (by converse of basic prop. theorem)

$EF\parallel QR$
(iii)

$PQ=1.28cm,PR=2.56cm,PE=0.18cm$ and

$PF=0.36cm$

$=1.28-0.18=1.10cm$

$FR=PR-PF$

$2.56-0.36=2.20cm$

$\cfrac{PE}{EQ}=\cfrac{0.18}{1.10}=\cfrac{18}{110}=\cfrac{9}{55}$ ...(i)
and

$\cfrac{PF}{FR}=\cfrac{0.36}{2.20}=\cfrac{36}{220}=\cfrac{9}{55}$ ...(ii)
from equation (i) and (ii)

$\cfrac{PE}{EQ}=\cfrac{PF}{FR}$ (by converse of basic prop. theorem)

$EF\parallel QR$
1864096_1876669_ans_a4e9f4b0dcdf42d4a7bd4feefc418411.png

In figure, if $AB || CD$ and $E$ is the mid-point of $AC$ , then show that $E$ is the midpoint of $BD$ .

$\Delta AEB$ and

$\Delta DEC$

$AB || CD$

$\angle 1 = \angle 2$ (alternate angles)

$\angle 3 = \angle 4$ (vertically opposite angles)
and

$AE = EC$ (

$\because E$ is the mid-point of

$AC$ )
Similarly,

$\Delta AEB \cong \Delta DEC$
(by ASA congruency property)
Hence

$DE = EB$ (by c.p.c.t)
Similarly

$E$ is the mid-point of

$BD$ .

In right triangle $ABC$ , right angled at $C, M$ is the mid-point of hypotenuse $AB$ . $C$ is joined to $M$ and produced to a point $D$ such that $DM = CM$ . Point $D$ is joined to point $B$ (see figure).
Show that:
(i) $\Delta AMC \cong \Delta BMD$
(ii) $\angle DBC$ is a right angle.
(iii) $\Delta DBC \cong \Delta ACB$
(iv) $CM = \frac{1}{2}AB$

(i) In

$\Delta AMC$ and

$\Delta BMD$

$DM = CM$ (given)

$\angle 3 = \angle 4$ (vertically opposite angles)

$AM = MB$ (as

$M$ is the mid-point of

$AB$ )

$\Delta AMC \cong \Delta BMD$ (by SAS congruency rule)
(ii)

$\because \Delta AMC \cong \Delta BMD$

$\angle 1 = \angle 2$ (by c.p.c.t)
But they are alternate angles,
so

$AC || BD$

$\angle ACB + \angle DBC = 180^{0}$
(sum of the interior angles on the same side of a transversal is equal to

$180^{0}$ )
But

$\angle ACB = 90^{0}$ (given)

$\angle DBC = 180^{0} = 90^{0}$
(iii) In

$\Delta DBC$ and

$\Delta ACB$

$\angle DBC = \angle ACB = 90^{0}$

$BC = BC$ (common side)

$BD = CA$

$(\because \Delta AMC \cong \Delta BMD)$

$\therefore \Delta DBC \cong \Delta ACB$ (by SAS congruency rule)
(iv)

$\because \Delta DBC \cong \Delta ACB$

$DC = AB$
But

$M$ mid-point of

$DC$

$2CM = CD$

$2CM = AB [\because DC = AB]$

$CM =\frac{1}{2} AB$ .
Hence proved.
1869972_1878463_ans_bb1253cac94244a28ad9c91e97d956d8.png

Prove that ratio of areas of two similar triangles is equal to ratio of their corresponding medians.

Proof:

$\triangle ABC\sim \triangle DEF$

$\cfrac{AB}{DE}=\cfrac{BC}{EF}$

$\Rightarrow$

$\cfrac{AB}{DE}=\cfrac{2BX}{2EY}$
(

$\because AX$ and

$DY$ medians

$BC=2BX,EF=2EY$ )

$\Rightarrow$

$\cfrac{AB}{DE}=\cfrac{BX}{EY}$ ...(i)
In

$\triangle ABX$ and

$\triangle DEY$

$\angle B=\angle E$ (

$\because \triangle ABC\sim \triangle DEF$ )

$\cfrac{AB}{DE}=\cfrac{BX}{EY}$
SAS similarity criterion

$\triangle ABX\sim \triangle DEY$

$\cfrac{AB}{DE}=\cfrac{AX}{DY}$ ...(ii)
Ratio of areas of two similar triangles is equal to ratio of squares of their corresponding sides

$\cfrac { ar.\quad \triangle ABC }{ ar.\quad \triangle DEF } =\cfrac { { AB }^{ 2 } }{ { DE }^{ 2 } } =\cfrac { { AX }^{ 2 } }{ { DY }^{ 2 } }$
1864098_1876671_ans_cb678bbcc8a24b54a6c2b7211927b453.png

In figure $LM\parallel CB$ and $LN\parallel CD$ . Prove that $\cfrac{AM}{MB}=\cfrac{AN}{AD}$

$\triangle ABC$

$ML\parallel BC$ (given)

$\cfrac{AM}{AB}=\cfrac{AL}{LC}$ ...(i) Basic prop. Theorem
Again in

$\triangle ADC$

$LN\parallel DC$ (given)

$\cfrac{AN}{ND}=\cfrac{AL}{LC}$ ...(ii)
From equation (i) and (ii)

$\cfrac{AM}{MB}=\cfrac{AN}{ND}$
or

$\cfrac{MB}{AM}=\cfrac{ND}{AN}$

$\Rightarrow$

$\cfrac{MB}{AM}+1=\cfrac{ND}{AN}+1$

$\Rightarrow$

$\cfrac{MB+AM}{AM}=\cfrac{ND+AN}{AN}$

$\Rightarrow$

$\cfrac{AB}{AM}=\cfrac{AD}{AN}$
Thus,

$\cfrac{AM}{AB}=\cfrac{AN}{AD}$

Sides $AB$ and $AC$ and median $AD$ of a triangle and respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle. Show that $\triangle ABC\sim \triangle PQR$

Given:

$\cfrac{AB}{PQ}=\cfrac{AC}{PR}=\cfrac{AD}{PM}$ ...(i)
Construction: Produce

$AD$ upto

$E$ such that

$AD=DE$ . Join

$BE$ and

$CE$ and produce

$PM$ up to

$N$ such that

$PM=MN$ . Join

$QN$ and

$NR$
Proof: Diagonals

$AE$ and

$BC$ of quadrilateral

$ABEC$ bisect each other at point

$D$ .
Quadrilateral

$ABEC$ is a parallelogram

$BE=AC$ ...(ii)
Similarly

$PQNR$ is a parallelograom

$QN=PR$ ...(iii)
Dividing equation (ii) by (iii)

$\cfrac{BE}{QN}=\cfrac{AC}{PR}$ ...(iv)
Now,

$\cfrac{AD}{PM}=\cfrac{2AD}{2PM}=\cfrac{AD}{PM}=\cfrac{AE}{PN}$ ....(v)
In equation (i), (iv), (v)

$\cfrac{AB}{PQ}=\cfrac{BE}{QN}=\cfrac{AE}{PN}$

$\triangle ABE\sim \triangle PQN$ (By SSS)

$\angle BAE=\angle QPN$ ...(vi)
Similarly

$\triangle AEC\sim \triangle PNR$

$\angle EAC=\angle NPR$ ...(vii)
Adding equation (vi) and (vii)

$\angle BAE+\angle EAC=\angle QPN+\angle NPR$

$\Rightarrow$

$\angle BAC=\angle QPR$
Now, in

$\triangle ABC$ and

$\angle PQR$

$\cfrac{AB}{PQ}=\cfrac{AC}{PR}$ (from equation (i))

$\angle A=\angle P$
By SAS similarity criterion

$\triangle ABC\sim \triangle PQR$
1864322_1876674_ans_47ef81252acd4503af1491dde8e1c321.png

If in $\triangle ABCD$ is any point on $BC$ such that $\cfrac{AB}{AC}=\cfrac{BD}{DC}$ and $\angle B={70}^{o}$ , $\angle C={50}^{o}$ , then find $\angle BAD$

$\triangle ABC$

$\angle B={70}^{o}$ ,

$\angle C={50}^{o}$

$\angle A+\angle B+\angle C={180}^{o}$

$\Rightarrow$

$\angle A={180}^{o}-(\angle B+\angle C)$

$\Rightarrow$

$\angle A={180}^{o}-({70}^{o}+{50}^{o})$

$\Rightarrow$

$\angle A={180}^{o}-{120}^{o}$

$\Rightarrow$

$\angle A={60}^{o}$
Thus,

$\cfrac{AB}{AC}=\cfrac{BD}{DC}$ (given)

$\triangle ABD\sim \triangle ADC$

$\Rightarrow$

$\angle BAD$ and

$\angle DAC$ will be equal to each other

$2\angle BAD={60}^{o}$

$\angle BAD=\cfrac{60}{2}$

$\angle BAD={30}^{o}$

1862588_1877587_ans_1efb511cedb345ebbe5fcf77559ff331.png

In $\triangle ABC$ , $DE\parallel BC$ and $AD=6cm,DB=9cm$ and $AE=8cm$ , then find $AC$ .

$\triangle ABC$ ,

$DE\parallel BC$

$\cfrac{AD}{DB}=\cfrac{AE}{EC}$ (by basic proportionality theorem)

$\Rightarrow$

$\cfrac{6}{9}=\cfrac{8}{CE}$

$\Rightarrow$

$6\times CE=9\times 8$

$\Rightarrow$

$CE=\cfrac{8\times 9}{6}$

$\Rightarrow$

$CE=12cm$

$AC=AE+CE$

$=(8+12)cm=20cm$
Thus,

$AC=20cm$

1862622_1877590_ans_c3a9c7f942ae475c9ce36b8b2cc6e276.png

In figure, $D, E$ and $F$ are the mid-points of the sides $BC, CA$ and $AB$ respectively. If $AB= 4.3 \ cm; BC= 5.6 \ cm$ and $AC= 3.9 \ cm$ , then find the perimeter of the following:
(i) $ADEF$ and
(ii) quad. $BDEF$

(i) Perimeter of

$\triangle DEF$

$= DE+ FE + DF$

$= 2.15 + 2.8 + 1.95 = 6.9 \ cm$

(ii) Perimeter of quad.

$BDEF$

$= BD+ DE+ EF+ FB$

$= 2.8+ 2.15+ 2.8+ 2.15$

$= 9.9 \ cm$

In figure, find $\angle x, \angle y$ and $\angle A C D$ where line $B A \| F E .$

$\mathrm{BA} \| \mathrm{FE}$

$\angle \mathrm{x}=42^{\circ} \quad$ (alternate angle)

$\angle \mathrm{ACD}=\angle \mathrm{x}+66^{\circ}$ (ext. angle is equal to sum of opposite interior angles)

$\angle \mathrm{ACD}=42^{\circ}+66^{\circ}=108^{\circ}$
But

$\angle y+\angle A C D=180^{\circ}$

$\angle y=180^{\circ}-108^{\circ}$ (linear pair axiom)

$\angle y=72^{\circ}$

$\angle \mathrm{x}=42^{\circ}$ and

$\angle \mathrm{y}=72^{\circ}$

In figure, $X$ and $Y$ are the points on equal sides of $AB$ and $AC$ such that $AX = AY$ . Show that $XC = BY$ .

$\Delta ABC$

$AB = AC$ (given)
and

$AX = AY$ (given)

$AB – AX = AC – AY$

$BX = CY$ ….(i)
Now in

$\Delta BXC$ and

$\Delta BYC$

$BX = CY$ [from (i)]

$BC = BC$ (common)

$\angle B = \angle C$ (angle opp. to equal sides are equal)

$\Delta BXC = \Delta BYC$ (by SAS property)

$XC = BY$ (by c.p.c.t)
Hence proved.

In $\Delta ABC, D$ is the mid-point of side $AC$ such that $BD =\frac{1}{2} AC$ . Show that $\angle ABC = 90^{0}$ .

$D$ is mid-point of

$AC$

$AD = DC =\frac{1}{2} AC$
But

$BD = \frac{1}{2}AC$ (given)

$AD = DC = BD$
In

$\Delta ABD$

$AD = BD$

$\angle 2 = \angle 1 = x$ (angles opposite to equal sides are equal)
Also

$BD = DC$

$\angle 3 = \angle 4 = x$
In

$\Delta ABC$

$\angle 1 + (\angle 2 + \angle 3) + \angle 4 = 180^{0}$

$x + (x + x) + x = 180^{0}$

$4x = 180^{0}$

$x = 45^{0}$

$\angle ABC = x + x = 45^{0} + 45^{0} = 90^{0}$
Hence proved.
1869986_1878691_ans_42c481179b064461b086e239121d95ad.png

In the figure, $ABC$ and $DBC$ are two triangles on the same base $BC$ . If $AD$ intersects $BC$ at $O$ , show that $\dfrac{ar(ABC)}{ar(DBC)}=\dfrac{AO}{DO}$

$\textbf{Step 1: Check }\boldsymbol{\triangle ABD\ \&\ \triangle DOM}\textbf{ are similar triangles.}$

$\text{Draw }AM\perp OB$

$DM\perp OC$

$\text{In }\Delta AMO$ and

$\Delta DNO$

$\angle AMO=\angle DNO=90^\circ$

$\angle AOM=\angle DON$ (VOA)

$\Delta ABD\sim \Delta DOM$ (by AA)

$\text{So, }\dfrac{AM}{DN}=\dfrac{AO}{DO}=\dfrac{MO}{NO}$

$...........(1)$

$\textbf{Step 2: Find ratio of area of two triangles and use equation (1)}$

$\dfrac{ar\Delta (ABC)}{er\Delta(BDC)}=\dfrac{\dfrac{1}{2}\times BC\times AM}{\dfrac{1}{2}\times BC\times DN}$

$\dfrac{ar\Delta ABC}{ar\Delta BDC}=\dfrac{AM}{ON}$

$\dfrac{ar\Delta ABC}{er\Delta BDC}=\dfrac{AO}{DO}$

$\textbf{Hence, Proved.}$

Find the height of a cone of diameter $30m$ and slant height $25m$ .

$\textbf{ Step 1: Use Pythagoras theorem to find the height of the cone.}$

$\text{Given,}$

$\text{l = 25 m and d = 30 m}$

$\text{Now,}$

$r=\dfrac{30}{2}=15m$

$\left[\because\boldsymbol{r=\dfrac{d}{2}}\right]$

$\text{By using Pythagoras theorem}$

$h=\sqrt{l^2-r^2}$

$h=\sqrt{25^2-15^2}$

$\left[\because\boldsymbol{h=\sqrt{l^2-r^2}}\right]$

$\,=\sqrt{625-225}$

$\,=\sqrt{400}$

$h=20 \text{ m}$

$\textbf{Hence, height of cone is 20 }\boldsymbol{m.}$

Through $A, B$ and $C$ lines $RQ, PR$ and $QP$ have been drawn respectively parallel to sides $BC, CA$ and $AB$ of a $\triangle ABC$ as shown in figure. Show that $BC= \dfrac12 QR$ .

2017235_1879045_ans_7a979bb592164d5eac52faa6e373fc38.png

In figure, $D$ and $E$ are the mid-points of the sides $AB$ and $AC$ respectively of $\triangle ABC$ . If $BC= 6.4\ cm$ , find $DE$ .

From mid-point theorem,

$DE= \dfrac12 BC$

$\implies DE= \dfrac12 \times 6.4 = 3.2$

$\implies DE= 3.2 \ cm$ .

Derive the formula for height and area of an equilateral triangle.

Let

$△ABC$ be the triangle where

$AP⊥BC,AB=AC=a,BP=PC=\cfrac {a}{2}$ and

$AP=h$ as shown in the above figure:

$△APC$ ,

$∠P=90^{ 0 }$

Using pythagoras theorem, we have

$AC^{ 2 }=AP^{ 2 }+PC^{ 2 }\\ \Rightarrow a^{ 2 }=h^{ 2 }+\left( \cfrac { a }{ 2 } \right) ^{ 2 }\quad \\ \Rightarrow a^{ 2 }=h^{ 2 }+\cfrac { a^{ 2 } }{ 4 } \quad \quad \\ \Rightarrow h^{ 2 }=a^{ 2 }-\cfrac { a^{ 2 } }{ 4 } \\ \Rightarrow h^{ 2 }=\cfrac { 4a^{ 2 }-a^{ 2 } }{ 4 } \\ \Rightarrow h^{ 2 }=\cfrac { 3a^{ 2 } }{ 4 } \\ \Rightarrow h=\sqrt { \cfrac { 3a^{ 2 } }{ 4 } } \\ \Rightarrow h=\cfrac { \sqrt { 3 } a }{ 2 }$

Now, the area of the triangle is

$A=\cfrac { 1 }{ 2 } bh$ therefore,

$A=\cfrac { 1 }{ 2 } \times b\times h=\cfrac { 1 }{ 2 } \times a\times \cfrac { \sqrt { 3 } a }{ 2 } =\cfrac { \sqrt { 3 } a^{ 2 } }{ 4 }$

Hence, height of an equilateral triangle is

$h=\cfrac { \sqrt { 3 } a }{ 2 }$ and area is

$A=\cfrac { \sqrt { 3 } a^{ 2 } }{ 4 }$ .

In $\triangle ABC$ , $a + b = 18$ units, $b + c = 25$ units and $c + a = 17$ units. What type of triangle is $ABC$ ? Give reason.

$△ABC$ , the followings are given:

$a+b=18 ........(1)$

$b+c=25 ........(2)$

$c+a=17 ........(3)$

Subtracting equation

$(1)$ from equation

$(2)$ :

$(b+c)-(a+b)=25-18$

$\Rightarrow (c-a)+(b-b)=25-18$

$\Rightarrow$

$c-a=7 .......(4)$

Now, adding equation

$(3)$ and

$(4)$ :

$(c+a)+(c-a)=17+7$

$\Rightarrow$

$c+a+c-a=24$

$\Rightarrow$

$2c=24$

$\Rightarrow$

$c=12$

Substituting

$c=12$ in equation

$(3)$ :

$12+a=17$

$\Rightarrow$

$a=17-12=5$

Now substitute

$a=5$ in equation

$(1)$ :

$5+b=18$

$\Rightarrow$

$b=18-5=13$

Finally consider

$AB^2+BC^2$ as shown below:

$AB^{ 2 }+BC^{ 2 }$

$=c^{ 2 }+a^{ 2 }$

$=12^{ 2 }+5^{ 2 }$

$=144+25$

$=169$

$=(13)^{ 2 }$

$=b^{ 2 }$

$=AC^{ 2 }$

Since

$AB^2+BC^2=AC^2$ .

Hence,

$△ABC$ is a right angled triangle.

Using basic proportionality theorem, prove that a line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side.

In $\triangle MNS$ , line KL $\parallel$ side NS .....(given)
$\therefore \dfrac {MK}{KN} = \dfrac {ML}{LS}$ ..... (By BPT)

But $MK = KN$
$\therefore \dfrac {MK}{KN}=1$

$\therefore \dfrac {ML}{LS} = 1$

$\therefore ML=LS$
This means that line KL bisects side MS.

Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.

Given triangle ABC is a right angle triangle at B

Draw BD perpendicular to AC

In $\Delta ABC$ and $\Delta ADB$

$\angle ABC=\angle ADB$

$\angle A=\angle A$

$\therefore \Delta ABC\sim \Delta ADB$

$\therefore \dfrac{AB}{AD}=\dfrac{BC}{DB}=\dfrac{AC}{AB}$

$\Rightarrow AB^{2}=AC\times AD$ .............................................(1)

In $\Delta ABC$ and $\Delta BDC$

$\angle ABC=\angle bDC$

$\angle C=\angle C$

$\therefore \Delta ABC\sim \Delta BDC$

$\therefore \dfrac{AB}{BD}=\dfrac{BC}{DC}=\dfrac{AC}{BC}$

$\Rightarrow BC^{2}=AC\times DC$

Add (1) and (2) we get

$AB^{2}+BC^{2}=AC\times AD+AC\times DC$

$=AC(AD+DC)=AC\times AC$

$=AC^{2}$

Hence proved.

The corresponding altitudes of two similar triangles are 3 cm and 5 cm respectively. Find the ratio between their areas.

As shown in the above figure, let

$△ABC$ and

$△DEF$ be the similar triangles with altitudes

$AM$ and

$DN$ and with

$AM⊥BC$ and

$DN⊥EF$ .

We know that the arc of similar triangles are proportional to squares of their corresponding altitude, therefore with the given altitudes

$3$ cm and

$5$ cm, we have,

$\dfrac { { A }r(△ABC) }{ { A }r(△DEF) } =\dfrac { AM^{ 2 } }{ DN^{ 2 } } \\=\dfrac { 3^{ 2 } }{ 5^{ 2 } } \\=\dfrac { 9 }{ 25 }$

Hence the ratio between the areas is

$9:25$ .

Two isosceles triangles are having equal vertical angles and their areas are in the ratio 9 :Find the ratio of their corresponding altitudes.

As shown in the above figure, let

$△ABC$ and

$△DEF$ are the isosceles triangles with altitudes

$AM$ and

$DN$ and with

$∠A=∠D$ , therefore,

$∠B=∠C$ and

$∠E=∠F$ that is

$∠B=∠C=∠E=∠F$

Thus,

$△ABC∼△DEF$

We know that the arc of similar triangles are proportional to squares of their corresponding altitude, therefore with the given ratio of area of triangles

$9:16$ , we have,

$\frac { { A }r(△ABC) }{ { A }r(△DEF) } =\frac { AM^{ 2 } }{ DN^{ 2 } } \\ \Rightarrow \frac { 9 }{ 16 } =\frac { AM^{ 2 } }{ DN^{ 2 } } \\ \Rightarrow \sqrt { \frac { AM^{ 2 } }{ DN^{ 2 } } } =\sqrt { \frac { 9 }{ 16 } } \\ \Rightarrow \frac { AM }{ DN } =\frac { 3 }{ 4 } \\ \Rightarrow AM:DN=3:4$

Hence the ratio of their altitudes is

$3:4$ .

$\triangle BAC$ and $\triangle$ BDC are two right-angled triangles with common hypotenuse BC. The sides $AC$ and $BD$ intersect at $'P'$
Prove that $AP. PC = DP. PB$

In the given figure,

$∆BAC$ and

$∆CDB$ are right angled triangles.

$∆BAP$ and

$∆CDP$ ,

$∠BAP=∠CDP=90^{ 0 }$

$∠APB = ∠DPC$ (Vertically opposite angles)

Therefore,

$∠ABP = ∠DCP$ (Remaining angles)

Thus

$\Delta BAP$ and

$\Delta CDP$ are similar by AA criterion.

$\Rightarrow\dfrac { AB }{ CD } =\dfrac { AP }{ PD } =\dfrac { PB }{ PC }$

$\Rightarrow \dfrac { AP }{ PD } =\dfrac { PB }{ PC } \\ \Rightarrow AP\times PC=DP\times PB$

Hence

$AP\times PC=DP\times PB$ .

The shortest distance $AP$ from a point ' $A$ ' to $QR$ is $12 cm$ . $Q$ and $R$ are $15 cm$ and $20 cm$ from ' $A$ ' and on opposite sides of $AP$ . Prove that $\angle QAR = {90}^{o}$ .

$△APR$ ,

$∠P=90^{ 0 }$ ,

$AR=20$ cm and

$AP=12$ cm.

Using pythagoras theorem, we have

$AP^{ 2 }+PR^{ 2 }=AR^{ 2 }\\ \Rightarrow (12)^{ 2 }+PR^{ 2 }=(20)^{ 2 }\quad \quad \\ \Rightarrow 144+PR^{ 2 }=400\\ \Rightarrow PR^{ 2 }=400-144\\ \Rightarrow PR^{ 2 }=256\\ \Rightarrow PR=\sqrt { 256 } \\ \Rightarrow PR=\sqrt { 16^{ 2 } } \\ \Rightarrow PR=16$

$△AQP$ ,

$∠P=90^{ 0 }$ ,

$AQ=15$ cm and

$AP=12$ cm.

Again using pythagoras theorem, we have

$AP^{ 2 }+PQ^{ 2 }=AQ^{ 2 }\\ \Rightarrow (12)^{ 2 }+PQ^{ 2 }=(15)^{ 2 }\quad \quad \\ \Rightarrow 144+PQ^{ 2 }=225\\ \Rightarrow PQ^{ 2 }=225-144\\ \Rightarrow PQ^{ 2 }=81\\ \Rightarrow PQ=\sqrt { 81 } \\ \Rightarrow PQ=\sqrt { 9^{ 2 } } \\ \Rightarrow PQ=9$

From the figure, we observe that

$QR=PQ+PR$ , therefore,

$QR=PQ+PR=16+9=25$

Now, in

$△AQR$ , consider

$AQ^2+AR^2$ that is:

$AQ^{ 2 }+AR^{ 2 }=15^{ 2 }+20^{ 2 }=225+400=625=(25)^{ 2 }=QR^{ 2 }\quad$

Since

$AQ^2+AR^2=QR^2$ , therefore

$△AQR$ is a right angled triangle.

Hence,

$∠QAR=90^{ 0 }$ .

In the figure shown above, AC || BD and CE || DF. If OA = 12 cm, AB = 9 cm, OC = 8 cm and EF = 4.5 cm, find OE.

Given that, in

$\Delta OBD, \ AC||BD,\ OA=12\ cm, \ AB=9\ cm,\ and \ OC=8\ cm$ .

To find out: The length of

$OE$ .

According to the basic proportionality theorem, if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points, then it divides the two sides in the same ratio.

Hence, using the basic proportionality theorem in

$\Delta OBD$ , we have:

$\dfrac { OA }{ BA } =\dfrac { OC }{ DC }$

$\Rightarrow \dfrac { 12 }{ 9 } =\dfrac { 8 }{ DC }$

$\Rightarrow 12DC=8\times 9$

$\Rightarrow 12DC=72$

$\Rightarrow DC=\dfrac { 72 }{ 12 }$

$\therefore \ DC=6\ cm$

Now, in

$\Delta ODF,\ CE||DF,\ EF=4.5\ cm, \ DC=6\ cm\ and \ OC=8\ cm$ .

Using the basic proportionality theorem in

$\Delta ODF$ , we have

$\dfrac { OC }{ DC } =\dfrac { OE }{ EF }$

$\Rightarrow \dfrac { 8 }{ 6 } =\dfrac { OE }{ 4.5 }$

$\Rightarrow 6OE=8\times 4.5$

$\Rightarrow 6OE=36$

$\Rightarrow OE=\dfrac { 36 }{ 6 }$

$\therefore \ OE =6$

Hence, the required length of

$OE$ is

$6\ cm$ .

, $BM = x + 2, AM = x + 7, CM = x$ , find $x$

It is given that

$∠ABC=90^{ 0 }$ and

$BM⊥AC$ as shown in the above figure:

Now with

$BM=(x+2)$ ,

$CM=x$ and

$AM=(x+7)$ , consider,

$BM^{ 2 }=CM\times AM\\ \Rightarrow (x+2)^{ 2 }=x\times (x+7)\\ \Rightarrow x^{ 2 }+4x+4=x^{ 2 }+7x\\ \Rightarrow x^{ 2 }+4x+4-x^{ 2 }-7x=0\\ \Rightarrow -3x+4=0\\ \Rightarrow -3x=-4\\ \Rightarrow x=\frac { 4 }{ 3 }$

Hence,

$x=\frac { 4 }{ 3 }$ .

In the isosceles $\triangle ABC$ , $AB = AC$ , $BC = 18 cm$ , $AD \perp BC$ , $AD = 12 cm$ , $BC$ is produced to ' $E$ ' and $AE = 20 cm$ . Prove that $\angle BAE = {90}^{o}$ .

$△ABD$ ,

$∠D=90^{ 0 }$ ,

$BD=9$ cm and

$AD=12$ cm.

Using pythagoras theorem, we have

$AB^{ 2 }=AD^{ 2 }+BD^{ 2 }\\ \Rightarrow AB^{ 2 }=9^{ 2 }+12^{ 2 }\quad \quad \\ \Rightarrow AB^{ 2 }=81+144\quad \\ \Rightarrow AB^{ 2 }=225\\ \Rightarrow AB=\sqrt { 225 } \\ \Rightarrow AB=\sqrt { 15^{ 2 } } \\ \Rightarrow AB=15$

$△ADE$ ,

$∠D=90^{ 0 }$ ,

$AE=20$ cm and

$AD=12$ cm.

Again using pythagoras theorem, we have

$AE^{ 2 }=AD^{ 2 }+DE^{ 2 }\\ \Rightarrow 20^{ 2 }=12^{ 2 }+DE^{ 2 }\quad \\ \Rightarrow 400=144+DE^{ 2 }\quad \\ \Rightarrow DE^{ 2 }=400-144\\ \Rightarrow DE^{ 2 }=256\\ \Rightarrow DE=\sqrt { 256 } \\ \Rightarrow DE=\sqrt { 16^{ 2 } } \\ \Rightarrow DE=16$

From the figure, we observe that

$BE=BD+DE$ , therefore,

$BE=BD+DE=9+16=25$

Now, in

$△ABE$ , consider

$AB^2+AE^2$ that is:

$AB^{ 2 }+AE^{ 2 }=15^{ 2 }+20^{ 2 }=225+400=625=(25)^{ 2 }=BE^{ 2 }\quad \quad$

Since

$AB^2+AE^2=BE^2$ , therefore

$△ABE$ is a right angled triangle.

Hence,

$∠BAE=90^{ 0 }$ .

$\triangle$ ABC and $\triangle$ BDE are two equilateral triangles and BD = DC. Find the ratio between areas of $\triangle$ ABC and $\triangle$ BDE.

In the given figure, it is given that

$△ABC$ and

$△BDF$ are equilateral triangles that is

$△ABC∼△BED$ and

$BD=DC$ .

We know that the area of similar triangles are proportional to squares of their corresponding altitude, therefore,

$\dfrac { { A }r(△ABC) }{ { A }r(△BDE) } =\dfrac { BC^{ 2 } }{ BD^{ 2 } }$

$=\dfrac { (2BD)^{ 2 } }{ BD^{ 2 } } =\dfrac { 4BD^{ 2 } }{ BD^{ 2 } } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (\because \ BC=BD+DC,\quad BD=DC)$

$=4$

Hence the ratio between areas of

$△ABC$ and

$△BDE$ is

$4:1$ .

In the quadrilateral $ABCD$ , $\angle ADC = {90}^{o}$ , $AB = 9 cm$ , $BC = AD = 6 cm$ and $CD = 3 cm$ . Prove that $\angle ACB = {90}^{o}$

$△ADC$ ,

$∠D=90^{ 0 }$ ,

$AD=6$ cm and

$CD=3$ cm.

Using pythagoras theorem, we have

$AC^{ 2 }=AD^{ 2 }+CD^{ 2 }\\ \Rightarrow AC^{ 2 }=6^{ 2 }+3^{ 2 }\quad \\ \Rightarrow AC^{ 2 }=36+9\quad \quad \\ \Rightarrow AC^{ 2 }=45\quad \\ \Rightarrow AC=\sqrt { 45 } \\ \Rightarrow AC=\sqrt { 3^{ 2 }\times 5 } \\ \Rightarrow AC=3\sqrt { 5 }$

Now, in

$△ACB$ , consider

$AC^2+BC^2$ that is:

$AC^{ 2 }+BC^{ 2 }=(3\sqrt { 5 } )^{ 2 }+6^{ 2 }\\=(9\times 5)+36\\=45+36\\=81\\=(9)^{ 2 }\\=AB^{ 2 }\quad$

Since

$AC^2+BC^2= AB^2$ , therefore

$△ACB$ is a right angled triangle.

Hence,

$∠ACB=90^{ 0 }$ .

In the adjoining figure, DE || AB, AD = 7 cm, CD = 5 cm and BC = 18 cm. Find BE and CE.

It is given that

$DE||AB, AD=7$ cm,

$CD=5$ cm and

$BC=18$ cm.

Using the basic proportionality theorem, we have

$\dfrac { CD }{ CA } =\dfrac { DE }{ AB } =\dfrac { CE }{ CB }$

$\Rightarrow \dfrac { CD }{ CA } =\dfrac { CE }{ CB }$

$\Rightarrow \dfrac { 5 }{ 12 } =\dfrac { CE }{ 18 }$

$\Rightarrow 18\times 5=12CE$

$\Rightarrow 12CE=90$

$\Rightarrow CE=\dfrac { 90 }{ 12 } =7.5$

Since

$CE=7.5$ cm, therefore,

$CB=CE+EB$

$\Rightarrow 18=7.5+EB$

$\Rightarrow EB=18-7.5=10.5$

Hence,

$BE=10.5$ cm and

$CE=7.5$ cm.

In the given figure, $XY||AC$ , if $2AX=3BX$ and $XY=9$ find the value of $AC$

1184069_1128278_ans_979be1235c9e44508e07f4161df78788.png

In $\triangle ABC$ $DE||BC$ and $\dfrac{AD}{DB}=\dfrac{3}{5}$ $AC =5.6$ find $AE$ ?

given that

$\dfrac{AD}{DB}$ =

$\dfrac { 3 }{ 5 }$

$AC=5.6$ cm ; therefore

$AC-AE=EC$

Since

$DE||BC$

$\dfrac { AB }{ DB } =\dfrac { AE }{ EC }$

reversing

$\dfrac { DB }{ AD } =\dfrac { AC }{ AE }$

adding 1 then cross multiplying

$\dfrac { AB+AD }{ AD } =\dfrac { EC+AE }{ AE }$

$\dfrac { AB }{ AD } =\dfrac { AC }{ AE }$

$\dfrac { 8 }{ 3 } =\dfrac { 5.6 }{ AE }$

$AE=5.6\times \dfrac { 3 }{ 8 }$

$AE=2.1$ cm

$EC=3.5$ cm

Therefore

$5.6 - 2.1 = 3.5$ cm

In a fig. a crescent is formed by two circles which touch at A. C is the center of a larger circle. The width of the crescent at BD is 9 cm and at EF it is 5 cm. Find the radii of two circles.

Let the radii of the larger and smaller circles be R and r respectively. Then,

$BD = 9 cm$

$\Rightarrow \,$

$2R - 2r = 9$

$\Rightarrow \,$

$R - r = 4.5$

Join

$AE$ and

$DE$ . Let

$\angle CAE \, = \, \theta$ Then,

$\angle AEC \, = \, 90^\circ \, - \, \theta$

Now,

$\angle AED \, = \, 90^\circ \, \Rightarrow \, \angle AEC \, + \, \angle DEC \, = \, 90^\circ \, \Rightarrow \, \angle DEC \, = \, 90^\circ \, - \, (90^\circ \, - \, \theta) \, = \, \theta$

Thus, in

$\Delta$ ACE and DCE, we have

$\angle CAE \, = \, \angle CED \, = \, \theta \, and \, \angle ACE \, = \, \angle ECD \, = \, 90^\circ$

So, by AA similarly criterion, we have

$\Delta \, ACE \, ~ \, \Delta \, ECD$

$\Rightarrow \, \dfrac{AC}{EC} \, = \, \dfrac{CE}{CD}$

$\Rightarrow \, \dfrac{AC}{CF \, - \, EF} \, = \, \dfrac{CF \, - \, EF}{BC \, - \, BD}$

$\Rightarrow \, \dfrac{R}{R \, - \, 5} \, = \, \dfrac{R \, - \, 5}{R \, - \, 9}$

$\Rightarrow \, R(R \, - \, 9) \, = \, (R \, - \, 5)^2$

$\Rightarrow \, 0 \, = \, -R \, + \, 25 \, \Rightarrow \, R \, = \, 25 \, cm$

Substituting the value of

$R$ in

$(i),$ we get

$25 - r = 4.5$

$\Rightarrow$

$r = 20.5 cm$

In a triangle $ABC,AB=AC$ and $D$ is a point on side $AC$ such that $BC^{2}=AC \times CD$ prove that $BD=BC$

As $B{C^2} = AC \times CD$ , then,

$\dfrac{{BC}}{{AC}} = \dfrac{{CD}}{{BC}}$

In $\Delta ABC$ and $\Delta BDC$ ,

$\dfrac{{BC}}{{AC}} = \dfrac{{CD}}{{BC}}$

And,

$\angle C = \angle C$

Therefore, by SAS property,

$\Delta ABC \sim \Delta BDC$

So,

$\dfrac{{AB}}{{AC}} = \dfrac{{BD}}{{BC}}$ [C.S.C.T.]

$\dfrac{{AB}}{{AB}} = \dfrac{{BD}}{{BC}}$ [ $\because AB=AC$ ]

$1 = \dfrac{{BD}}{{BC}}$

$BD = BC$ $[hence \, proved]$

In given figure, find $\angle F$

In triangles ABC and DEF, we have

$\dfrac { AB }{ DF } =\dfrac { BC }{ FE } =\dfrac { CA }{ ED } =\dfrac { 1 }{ 2 }$

Therefore, by SSS-criterion of similarity, we have

$\triangle ABC\sim \triangle DEF$

$\Rightarrow$

$\angle A=\angle D,\angle B=\angle F$ and

$\angle C=\angle E$

$\Rightarrow$

$\angle D=80^0 ,\angle F=60^0$

Hence,

$\angle F=60^0$

In $\Delta ABC, \bar{PQ}\parallel \bar{AB}$ , P,Q are on BC and CA respectively if CQ: QA=1:3 are CP=4 then BC=......

We have,

$In\,\Delta ABC$

P and Q are mid point of side BC and CA respectively.

Given that,

$CQ:QA=1:3$ and $CP=4$

Then, BC=?

So,

We know that,

Similar triangle theorem,

$In\,\Delta ABC$

$\dfrac{CQ}{QA}=\dfrac{CP}{PB}$

$\dfrac{1}{3}=\dfrac{4}{PB}$

$PB=12$

Now,

$BC=BP+PB$

$BC=12+4$

$BC=16$

Hence, this is the answer.

Let $\triangle ABC\sim \triangle DEF$ and there areas be, respectively $64cm^{2}$ and $121cm^{2}$ if $EF=15.4cm$ find $BC$ ?

$BC=1.07cm$

Given two similar triangles

$∆ABC~ ∆DEF$ and also their are as are respectively

${ 64cm }^{ 2 }$ and

${ 121cm }^{ 2 }\quad$ Also if one of the sidei .e

$EF=15.4cm$ then we have to find the side

$BC.$

$\dfrac { area\left( ABC \right) }{ area\left( DEF \right) } =\dfrac { { BC }^{ 2 } }{ { EF }^{ 2 } }$

$\dfrac { 64 }{ 121 } =\dfrac { { BC }^{ 2 } }{ { 15.4 }^{ 2 } }$

$\dfrac { 8 }{ 11 } =\dfrac { { BC } }{ 3.92 }$

$BC=1.07cm$

In the given fig, the line segment $xy$ in parallel to $AC$ of $\triangle ABC$ and its dividend the triangle into two points of equal areas. Prove that $\dfrac {Ax}{AB}=\dfrac {\sqrt {2-1}}{\sqrt {2}}$

Given in

$\Delta ABC$ ,

$xy$ is parallel to

$AC$ and

$area\:of\:\Delta XBY=area\:of\:AXYC$

$\Delta ABC$ and

$\Delta XBY$ ,

$\angle ABC=\angle XBY$ (common angle)

$\angle ACB=\angle XYB$ (corresponding angles)

$\Delta ABC \sim \Delta XBY$ (AA similarity)

We know that in similar triangles, ratio of area of triangle is equal to ratio of square of corresponding sides.

$\dfrac{Area\:of\:\Delta ABC}{Area\:of\:\Delta XBY}=\left(\dfrac{AB}{XB}\right)^2$

$\Rightarrow \dfrac{Area\:of\:\Delta XBY+Area\:of\:AXYC}{Area\:of\:\Delta XBY}=\left(\dfrac{AB}{XB}\right)^2$

$\Rightarrow \dfrac{Area\:of\:\Delta XBY+Area\:of\:\Delta XBY}{Area\:of\:\Delta XBY}=\left(\dfrac{AB}{XB}\right)^2$ (given)

$\Rightarrow \dfrac{2\times Area\:of\:\Delta XBY}{Area\:of\:\Delta XBY}=\left(\dfrac{AB}{XB}\right)^2$

$\Rightarrow 2=\left(\dfrac{AB}{XB}\right)^2$

$\Rightarrow \dfrac{AB}{XB}=\sqrt{2}$

$\Rightarrow XB=\dfrac{AB}{\sqrt{2}}$

Now let us find

$\dfrac{AX}{AB}$

Since

$AX+XB=AB$

$\Rightarrow AX+\dfrac{AB}{\sqrt{2}}=AB$

$\Rightarrow AX=AB-\dfrac{AB}{\sqrt{2}}$

$\Rightarrow AX=\dfrac{AB(\sqrt{2}-1)}{\sqrt{2}}$

$\therefore \dfrac{AX}{AB}=\dfrac{\sqrt{2}-1}{\sqrt{2}}$

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR.
Show that $\Delta ABC\, \sim \,\Delta PQR$

2041586_1058293_ans_991d23bbf4204fc89531ad3670a10a22.jpg

In the trapezium ABCD, AB || DC, AB = 2CD and ar ( $\triangle$ AOB) = 84 cm $^2$ , find the area of $\triangle$ COD

It is given that

$ABCD$ is a trapezium and

$AB||DC$ .

$△AOB$ and △COD

$∠ABO=∠CDO$ (Alternate angles)

$∠BAO=∠DCO$ (Alternate angles)

$∠AOB=∠COD$ (Vertically opposite angles)

Therefore, △ABC∼△COD

We know that the arc of similar triangles are proportional to squares of their corresponding altitude, therefore with

${ A }r(△AOB)=84$ cm

$^2$ and

$AB=2CD$ , we have,

$\frac { { A }r(△AOB) }{ { A }r(△COD) } =\frac { AB^{ 2 } }{ CD^{ 2 } } \\ \Rightarrow \frac { 84 }{ { A }r(△COD) } =\frac { 4CD^{ 2 } }{ CD^{ 2 } } \\ \Rightarrow \frac { 84 }{ { A }r(△COD) } =4\\ \Rightarrow { A }r(△COD)=\frac { 84 }{ 4 } \\ \Rightarrow { A }r(△COD)=21$

Hence, area of

$△COD$ is

$21$ cm

$^2$ .

Prove that the ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding medians.

Given,

$\triangle ABC\sim \triangle PQR$

$BD=DC$ &

$QS=SR$

$\cfrac{AB}{PQ}=\cfrac{BC}{QR}=\cfrac{\cfrac{1}{2}BC}{\cfrac{1}{2}QR}=\cfrac{BD}{QS}$

Also,

$\angle B=\angle Q$

$\therefore \triangle ABD$ &

$\triangle PQS$ are similar.

$\therefore \cfrac{AB}{PQ}=\cfrac{AD}{PS}$

Also,

$\cfrac{Area \,of\,\triangle ABC}{Area\,of\,\triangle PQR}=(\cfrac{AB}{PQ})^{2}$

$\therefore \cfrac{Area \,of\,\triangle ABC}{Area\,of\,\triangle PQR}=(\cfrac{AD^{2}}{PS^{2}})$

State and prove that basic proportionality theorem.

2039492_1288600_ans_11ff714f53904bcd92c71c17f2e7cbf5.jpg

In the given figure, $3SR = 2SP,\,\,\left. {ST} \right|\left| {PM} \right.$ and $ar\left( {\triangle PMR} \right) = 50\,\,c{m^2}.$
Find the -
1. $ar\left( {\triangle RST} \right)$
2. $ar\left( {PMTS} \right)$

2070431_1408223_ans_370fa31522a844f2af3b5dee195c4813.jpg

In figure, two triangles ABC and DBC lie on the same side of base BC.P is a point on BC such that PQ || BA and PQ || BA and PR || BD. prove that QR || AD.

2040782_1327626_ans_b7590bdd6d11427aba7651edb12c452f.jpg

Prove that $\triangle ABC\sim \triangle EDC$

$\begin{array}{l} In\, \, \Delta ABC\, \, and\, \, \Delta EDC \\ \Rightarrow \angle ABC={ 90^{ 0 } }\, \, \, \, \, \left( { given } \right) \\ \Rightarrow \angle EDC={ 90^{ 0 } }\, \, \, \, \, \, \left( { given } \right) \\ \therefore BC=CD \\ and\, \, AB=ED\, \, \, \left( { altitudes } \right) \\ then,\, \, \Delta ABC\sim \Delta EDC\, \, \, \\ By\, \, S.A.S\, \, criteria \end{array}$

Hence, proved.

$\Delta ABC\sim \Delta DEF$ and their areas are respectively $100{ cm }^{ 2 }$ and $49{ cm }^{ 2 }$ . If the altitude of $\Delta ABC$ is $5 \ cm$ , find the corresponding altitude of $\Delta DEF$ .

Given

$\triangle ABC \cong \triangle DEF$

$\dfrac {Area_{1}}{Area_{2}}=\dfrac {(Side_{1})^{2}}{(Side_{2})^{2}}$

Given

$Area_{1}=100cm^{2}, Area_{2}=49cm^{2}$

$altitude_{1}=5cm, altitude_{2}=x$

$\dfrac {100}{49}=\dfrac {5^{2}}{X^{2}}$

$X^{2}=\dfrac {25\times 49}{100}$

$X=\dfrac{5\times 7}{100}$

$x=3.5\ cm$

In the figure, $\ ED\parallel QO$ and $DF\parallel OR$ . Prove that,
$EF \parallel QR$

$\Delta PQR , DE \parallel OQ$

$\Delta PRO,DF \parallel OR$

Since if line deacon parallel to one side of triangle,

intersects the other two sides in distrinct points, then it divides the other two side in the same ratio.

$\Rightarrow \dfrac{PE}{EQ}=\dfrac{PD}{DO}\,\cdots (1)$

$\Rightarrow \dfrac{PF}{FR}=\dfrac{PD}{DO}\,\cdots (2)$

From (1) and (2)

$\dfrac{PF}{FR}=\dfrac{PE}{EQ}$

i.e., line EF divides the triangle PQR in the same ratio.

Using the fact that if a line divides any two sides of a triangle in the same ratio then the line is parallel to the third side

$\therefore EF \parallel QR .$

Hence proved.

Triangles ABC and DBC have side BC common, AB $=$ BD and AC $=$ CD. Are the two triangles congruent? State in symbolic form. which congruence condition do you use? Does $\angle$ ABD equal $\angle$ ACD? Why or why not?

Triangles ABC and DBC have side BC common, AB

$=$ BD and AC

$=$ CD.

So,

$BC=BC\cdots\cdots[\text{common side}]$

$AB=BD\cdots\cdots[\text{given}]$

$AC=CD\cdots\cdots[\text{given}]$

So, by

$SSS$ test we can say that,

$\Delta ABC\cong\Delta DBC$

Congruence condition used is

$SSS$

Suppose congruent triangles are as shown as in the figure.

Then,

$\angle ABD\neq\angle ACD$

Then they will not be equal.

Because ,

$\angle ABD=\angle ABC\pm\angle DBC$

and

$\angle ACD=\angle ACB\pm\angle DCB$

From congruency,

$\angle ABC\cong\angle DBC$ and

$\angle ACB\cong\angle DCB$

So, we cannot say that

$\angle ABD$ is equal to

$\angle ACD$

1682415_1678072_ans_3f00ea5f701c4c1e80a885a47cd59d4f.png

The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas.

According to theorem of areas of similar triangles:

''When two triangles are similar , the ratio of areas of those triangles is equal to the ratio of the square their corresponding sides''

Ratio

$= \dfrac{3^{2}}{5^{2}}$

$= \dfrac{9}{25}$

$ABC$ is a right angle triangle such that $AB = AC$ and bisector of angle $C$ intersects the side $AB$ at $D$ P.T. $AC + AD = BC$ .

$\textbf{Step 1: Use pythagoras theorem}$

$\text{Let AB = AC = a and AD = b}$

$\Delta$

$\text{ABC is right angled at A}$

$AB^2 + AC^2 = BC^2$

$a^2 + a^2 = BC^2 \Rightarrow BC = a\sqrt{2}$

$\textbf{Step 2: Find relation between a and b}$

$\dfrac{AD}{BD} = \dfrac{AC}{BC}$

$\text{(Angle bisector theorem)}$

$\dfrac{b}{a - b} = \dfrac{a}{a\sqrt{2}} = \dfrac{1}{\sqrt{2}}$

$b = \dfrac{a}{1 + \sqrt{2}} = a(\sqrt{2} - 1)$

$b = a(\sqrt{2} - 1)\Rightarrow a + b = a\sqrt{2}$

$....(1)$

$\textbf{Step 3: Compare values of AC, AD and BC}$

$\text{AC = a, AD = b, BC =}$

$a\sqrt{2}$

$\text{Now}$

$\text{AC + AD = BC}$

$a + b = a\sqrt{2}$

$\text{(from equation 1)}$

$\textbf{Hence, required result is proved.}$

In the adjoining figure, the diagonals of a parallelogram intersect at $O. OE$ is drawn parallel to $CB$ to meet $AB$ at $E$ . Find the following ratio:
Area of $\triangle AOE$ : Area of parallelogram $ABCD$ .

In the figure, the four triangles formed have equal area.
So, Area of

$\triangle OAB = \dfrac {1}{4}$ Area of parallelogram

$ABCD$

Then,

$O$ is midpoint of

$AC$ of

$\triangle ABC$ and

$DE \parallel CB$

$E$ is also midpoint of

$AB$
Therefore,

$OE$ is the median of

$\triangle AOB$ .
Area of

$\triangle AOE = \dfrac {1}{2}$ area of

$\triangle AOB$

$= \dfrac {1}{2} \times \dfrac {1}{4}$ area of parallelogram

$ABCD$

$= \dfrac{1}{8}$ area of parallelogram

$ABCD$
So, area of

$\triangle AOE :$ area of parallelogram

$ABCD = 1 : 8$

The sides of certain triangle are given. Determine whether it is right triangle.
$a=9$ cm, $b=16$ cm and $c=18$ cm.

2039873_1677884_ans_7b943b4c96634cfeaa4385d2ca8af4d7.png

$\operatorname{In} \Delta A B C, D E \| B C$ . If $A D=5 \mathrm{cm}, B D=7 \mathrm{cm}$ and $A C=18 \mathrm{cm},$ find the length of $A E .$

Let

$A E=x$

$ΔADE$ and

$ΔABC$

$∠ADE = ∠ABC$ ( Corresponding angles are equal)

$∠A = ∠A$ ( Common angles are equal)

So, By

$AA$ postulate of similarity of triangles,

$ΔADE ~ ΔABC$

Since, The corresponding sides of the similar triangles are proportional to each other

$\dfrac{A D}{B D}=\dfrac{A E}{C E} \\$

$\dfrac{5}{7}=\dfrac{x}{18-x}\\$

$90-5 x=7 x \\$

$12 x=90 \\$

$x=\dfrac{90}{12}=7.5$

Triangles - Class 10 Maths - Extra Questions

In the given figure you find two triangles. Indicate whether the triangles are similar. Give reasons in support of your answer.

In fig PQ=5,XY=7.5,QR=4,YZ=PR=6,XZ=9.PQ = 5, XY = 7.5, QR = 4,YZ =PR = 6, XZ = 9. Are ΔPQR\Delta PQR and ΔXYZ\Delta XYZ similar? If yes, state the test and the correspondence of similarity.

Without drawing exact triangles, state, giving reasons, whether the given pairs of triangle are congruent or not :In ΔABCandΔPBC;AB=BP,AC=PC \Delta ABC \,\,\, and \,\,\, \Delta PBC; AB=BP, AC=PC

BP = 2ACIf the above statement is true then mention answer as 1, else mention 0 if false

In the figure above, straight lines ABAB and CDCD intersect at PP, and AC || BDAC \ || \ BD. Then ΔAPC\Delta\,APC and ΔBPD\Delta\,BPD are similar.If true enter 1 else 0.

A vertical stick 1212m long casts a shadow 88m long on the ground. At the same time a tower casts the shadow of length 4040m on the ground. Determine the height of the tower.

In the figure DE∥BCDE\parallel BC , DE=4cmDE=4\,cm, BC=8cmBC=8\,cm, A(△ADE)=25cm2A\displaystyle(\triangle ADE)=25\,{ cm }^{ 2 }, find A(△ABC)A\displaystyle(\triangle ABC)

If △ABC∼△DEF\triangle ABC\sim \triangle DEF and if A△ABC=9cm2A\,\triangle ABC=\,9\,{ cm }^{ 2 }, A△DEF=64cm2A\,\triangle DEF=\,64\,{ cm }^{ 2 }, DE=5.6cmDE=5.6\,cm, then find ABAB.

In ΔABC\Delta ABC, the bisector of ∠A\angle A intersects the base BCBC at the point DD. Prove that AB×AC=BD×DC+AD2AB\times AC=BD\times DC+AD^2.

In the given figure if, line PYPY \parallel\parallel side BCBC, AP=3AP=3, PB=6PB=6, AY=6AY=6 and YC=xYC=x, then find xx.

Write the conditions that are required for "Two polygons to be similar to each other".

In \triangle PQR \ 2PM = 3PN\triangle PQR \ 2PM = 3PN and 2PQ = 3PR2PQ = 3PR. Prove that MQRN is a trapezium

In the following figure drawn, line PYPY \parallel\parallel side BCBC, AP=5, PB=10, AY=8, YC=xAP=5, PB=10, AY=8, YC=x, then find xx.

In \triangle ABC, \angle ABC = \angle DAC, AB = 8\ cm, AC = 4\ cm, AD = 5\ cm\triangle ABC, \angle ABC = \angle DAC, AB = 8\ cm, AC = 4\ cm, AD = 5\ cm.Find \dfrac{\text{area of}\, \triangle ABC}{\text{area of}\, \triangle ACD}\dfrac{\text{area of}\, \triangle ABC}{\text{area of}\, \triangle ACD}

What is similar triangles?

Prove that "In a trapezium, the line joining the mid points of non-parallel sides is (i) parallel to the parallel sides and (ii) Half of the sum of the parallel sides"

In the figure, line PQPQ is parallel to side BCBC. Find the value of 'xx'. If AP=3AP=3, PB=6PB=6, AQ=5AQ=5, QC=xQC=x.

In \triangle ABC\triangle ABC, line PQ \parallelPQ \parallel side BCBC. Find the value of xx from the information given in the following figure.

In the figure, it is given that BR = PCBR = PC and \angle ACB = \angle QPR\angle ACB = \angle QPR and AB \parallel PQAB \parallel PQ. Prove that AC= QR AC= QR.

State the converse of Pythagorean Theorem.

Prove that"If two triangles are equiangular then their corresponding sides are in proportion"

In the following figure, DE \parallel ABDE \parallel AB. If AD = 7\ cm, CD = 5\ cmAD = 7\ cm, CD = 5\ cm and BC = 18\ cmBC = 18\ cm, find CECE.

\triangle {ABC}\sim \triangle{DEF}\triangle {ABC}\sim \triangle{DEF}. Their areas are 64\ {cm}^{2}64\ {cm}^{2} and 121\ {cm}^{2}121\ {cm}^{2}.Then ratio of corresponding sides ==

In \triangle ABC, DE\parallel BC\triangle ABC, DE\parallel BC and \dfrac {AD}{DB} = \dfrac {2}{3}\dfrac {AD}{DB} = \dfrac {2}{3}. If AE = 3.7\ cmAE = 3.7\ cm, find ECEC

State Thale's theorem.

If the ratio of corresponding medians of two similar triangles are 9:169:16, then find the ratio of their areas.

Fill in the blanks using the correct word given in brackets:All .......... triangles are similar (isosceles, equilateral).

The corresponding sides of two similar triangles are in the ratio 1 :If the area of the smaller triangle in 40 \text{ cm }^{ 2 }\text{ cm }^{ 2 }, find the area of the larger triangle.

Prove that, in a right-angled triangle, the square of hypotenuse is equal to the sum of the square of remaining two sides.

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively find the ratio of their areas.

What values of x will make DE \parallel\parallel AB in the given figure?

The area of two similar triangles ABCABC and PQRPQR are in the ratio 9 : 169 : 16. If BC = 4.5 cmBC = 4.5 cm, find the length of QRQR.

In \Delta\DeltaPQR, MN is parallel to QR and \displaystyle\frac{PM}{MQ}=\frac{2}{3}\displaystyle\frac{PM}{MQ}=\frac{2}{3}. Find, Area of \Delta\DeltaOMN:Area of \Delta\DeltaORQ.

Given \Delta\Delta ABC \sim\sim \Delta\Delta PQR, if \displaystyle\frac{AB}{PQ}=\frac{1}{3}\displaystyle\frac{AB}{PQ}=\frac{1}{3}, then find \displaystyle\frac{ar \Delta ABC}{ar \Delta PQR}\displaystyle\frac{ar \Delta ABC}{ar \Delta PQR}.

ABCABC is a triangle in which \angle A= 90^{o}\angle A= 90^{o}, AN \bot BCAN \bot BC, BC=12\ cmBC=12\ cm and AC=5\ cmAC=5\ cm. Find the ratio of the areas of \triangle ANC\triangle ANC and \triangle ABC\triangle ABC.

In the given Figure, \triangle ABC \sim\triangle\triangle ABC \sim\triangle PQRPQR and quad ABCDABCD \sim\sim quad PQRSPQRS. Determine the values of x, y, zx, y, z in each case.

Give two different examples of pair of (ii) non-similar figures.

Give two different examples of pair of (i) similar figures

In the figure, DE \parallel BCDE \parallel BC(i) Prove that \triangle ADE\triangle ADE and \triangle ABC\triangle ABC are similar

In the figure PS=3,SQ=6.QR=5,PT=xPS=3,SQ=6.QR=5,PT=x and TR=yTR=y.Give any two pairs of value of xx and yy such that line ST||QRST||QR

Area of similar triangles are in the ratio 25:3625:36 then ratio of their similar sides is _________?

Write the properties of similar triangles.

In figure if AD=6\ cm, DB=9\ cm, AE=8\ cmAD=6\ cm, DB=9\ cm, AE=8\ cm and EC=12\ cmEC=12\ cm and \angle ADE=48^{o}\angle ADE=48^{o}. Find \angle ABC\angle ABC.

Given that \Delta ABC \sim \Delta DEF,BC = 3\,cm,EF = 4\,cm\Delta ABC \sim \Delta DEF,BC = 3\,cm,EF = 4\,cm ans the area of \Delta ABC\Delta ABC is 54\,sq.cm.54\,sq.cm.Then find the area of \Delta DEF.\Delta DEF.

In figure \angle 1=\angle 2\angle 1=\angle 2 and \Delta NSQ\cong \Delta MTR\Delta NSQ\cong \Delta MTR, then prove that \Delta PTS\sim \Delta PRQ\Delta PTS\sim \Delta PRQ.

DA,CB,OMDA,CB,OM are each perpendicular to line segment ABAB where OO is the point of intersection of ACAC and DBDB.If OA=2.4OA=2.4cm,OC=3.6OC=3.6cm then find the ratio of AM:BMAM:BM

In \Delta ABC, \bar{PQ} || \bar{BC}\Delta ABC, \bar{PQ} || \bar{BC} and AP = 3x - 19, PB = x - 5, AQ = x - 3, QC = 3 \,cmAP = 3x - 19, PB = x - 5, AQ = x - 3, QC = 3 \,cm. Find xx.

Ler \Delta A B C \sim \Delta D E F\Delta A B C \sim \Delta D E F and their areas be respectively, 64\mathrm { cm } ^ { 2 }\mathrm { cm } ^ { 2 } and 121\mathrm { cm } ^ { 2 } .\mathrm { cm } ^ { 2 } . If EF =15.4 =15.4 find BC

In figure, AC\parallel BDAC\parallel BD and CE\parallel DFCE\parallel DF Find CD,where OD=7,OA=10.5,AB=4.5

Prove that the ratio of the area of two similar triangle is equal to the square of the ratio of their corresponding sides.

If D and E are points on sides AB and AC respectively of ABC and AB- 12 cm, AD= 8 cm,AE = 12 cm, AC =18 cm, then prove that DE \parallel BCDE \parallel BC.

State whether the following pair of triangle is similar or not.

In the adjoining figure,DE \parallel BCDE \parallel BC,find x.

State whether the following pair of triangle is similar or not. Write the similarity criterion used and write it in symbolic form.

In the adjoining figure, PP and QQ are points on sides ABAB and ACAC respectively of mix such that PQ \parallel BCPQ \parallel BC and $$AP = 8\ cm, AB =12\ cm,AQ = 3x\ cm, QC = (x+2)\ cm.$$Find x.

P and Q are points on sides AB and AC respectivelyof \triangle \triangle ABC. For each of the following cases, statewhether PQ\parallel \parallel BC.AB = 5 cm, AC =10 cm, AP: 4 cm, AQ = 8 cm.

State whether the following pair of triangle is similar or not. Write the similarity criterion used and write it in symbolic form.

If DE has been drawn parallel to side BC of ABC cutting AB and AC at points D and E respectively , such that \frac{AD}{DB} = \frac {3}{4}\frac{AD}{DB} = \frac {3}{4} then and the value of \frac {AE}{EC}\frac {AE}{EC}

P and Q are points on sides AB and AC respectivelyof \triangle \triangle ABC. For each of the following cases, statewhether PQ\parallel \parallel BC.AP: 8 cm, PB = 3 cm, AC = 22 cm and AQ =16 cm.

State whether the following pair of triangle is similar or not. Write the similarity criterion used and write it in symbolic form.

Give two different examples of pair of similar figures.

In the following figure , ABCDABCD is a trapezium with AB || DC. If AB = 9 AB = 9 cmcm, DC = 18DC = 18 cmcm, CF = 13.5CF = 13.5 cmcm, AP = 6AP = 6 cmcm and BE = 15BE = 15 cmcm. Find the value of AFAF

In the given figure, XY \parallel BCXY \parallel BC. Find the length of XY, given BC = 6 cm

In the figure given below, AB EF CD. If AB = 22.5\,cmAB = 22.5\,cm, EP = 7.5\,cmEP = 7.5\,cm,PC = 15\,cmPC = 15\,cm and DC = 27\,cmDC = 27\,cm. Calculate EF.

In a \Delta ABC , BD \Delta ABC , BD is the median to the side AC , BD is produce to E such that BD = DEProve that AE is parallel to BC.

The following figure shows a triangle ABC in which AB = AC . M is a point on AB and N is a point AC such that BM = CNProve thatBN = CM

Statements in the List I have to be matched with statements in the List II.

In trapezium ABCD ABCD , sides AB AB and DC DC are parallel to each other. F F is mid-point of AD AD and E E is mid-point of BC BC . Can it be concluded that AB + DC= 2EF AB + DC= 2EF ?If the above statement is true then mention answer as 1, else mention 0 if false

In figure, \Delta\DeltaUVW is right angled triangle and \angle\angleUVW == 90^o^o, UV == 6cm, and UW == 8 cm, \sin\sin W = p:mp:m, then p+mp+m =

In the given figure; \displaystyle \angle BAC= 90^{\circ}\displaystyle \angle BAC= 90^{\circ} and \displaystyle AD\perp BC.\displaystyle AD\perp BC. Show that \displaystyle \Delta ADC\displaystyle \Delta ADC and \displaystyle \Delta BAC\displaystyle \Delta BAC are similar.

In a triangle,\displaystyle \angle BAC= 90^{\circ}\displaystyle \angle BAC= 90^{\circ} and \displaystyle AD\perp BC.\displaystyle AD\perp BC.If AC=13.5 cm and CD =4.5cm; calculate the length of 2BC

Find whether the sides of the triangle as given below form a right-triangled or not: 99 cm, 1212 cm and 1515 cm

If a line parallel to BCBC intersects side ABAB and ACAC of triangle ABCABC at PP and Q,Q, then prove that \dfrac {AP}{AB}=\dfrac {AQ}{AC}.\dfrac {AP}{AB}=\dfrac {AQ}{AC}.

In the given figure you find two triangles. Indicate whether the triangles are similar. Give reasons in support of your answer.

In the following figure, \displaystyle \angle ABC=90^{\circ},AB =\left ( x+8 \right )\ cm, BC=\left ( x+1 \right )\ cm\displaystyle \angle ABC=90^{\circ},AB =\left ( x+8 \right )\ cm, BC=\left ( x+1 \right )\ cm and \displaystyle AC=(x + 15)\ cm\displaystyle AC=(x + 15)\ cm.Find the value of xx.

\displaystyle \Delta ABD\equiv \Delta ADC\displaystyle \Delta ABD\equiv \Delta ADCIf the above statement is true then mention answer as 1, else mention 0 if false

O is mid-point of APIf the above statement is true then mention answer as 1, else mention 0 if false

In the given figure, DE//BCDE//BC, AEAE = 1515 cm, ECEC = 99 cm, NCNC = 66 cm and BNBN = 2424 cm.Find lengths of ME(in\ cm)ME(in\ cm).

Two angles of one triangle are 85^{\circ}85^{\circ} and 65^{\circ}65^{\circ} is equal to angles of the other. Are they similar? Prove it.

In fig $PQ = 5, XY = 7.5, QR = 4,YZ =PR = 6, XZ = 9.$ Are $\Delta PQR$ and $\Delta XYZ$ similar? If yes, state the test and the correspondence of similarity.

Without drawing exact triangles, state, giving reasons, whether the given pairs of triangle are congruent or not :
In $\Delta ABC \,\,\, and \,\,\, \Delta PBC; AB=BP, AC=PC$

BP = 2AC
If the above statement is true then mention answer as 1, else mention 0 if false

In the figure above, straight lines $AB$ and $CD$ intersect at $P$ , and $AC \ || \ BD$ . Then $\Delta\,APC$ and $\Delta\,BPD$ are similar.If true enter 1 else 0.

A vertical stick $12$ m long casts a shadow $8$ m long on the ground. At the same time a tower casts the shadow of length $40$ m on the ground. Determine the height of the tower.

$DE=4\,cm$ , $BC=8\,cm$ , $A\displaystyle(\triangle ADE)=25\,{ cm }^{ 2 }$ , find $A\displaystyle(\triangle ABC)$

f $A\,\triangle ABC=\,9\,{ cm }^{ 2 }$ , $A\,\triangle DEF=\,64\,{ cm }^{ 2 }$ , $DE=5.6\,cm$ , then find $AB$ .

In $\Delta ABC$ , the bisector of $\angle A$ intersects the base $BC$ at the point $D$ . Prove that $AB\times AC=BD\times DC+AD^2$ .

In the given figure if, line $PY$ $\parallel$ side $BC$ , $AP=3$ , $PB=6$ , $AY=6$ and $YC=x$ , then find $x$ .

In $\triangle PQR \ 2PM = 3PN$ and $2PQ = 3PR$ . Prove that MQRN is a trapezium

In the following figure drawn, line $PY$ $\parallel$ side $BC$ , $AP=5, PB=10, AY=8, YC=x$ , then find $x$ .

In $\triangle ABC, \angle ABC = \angle DAC, AB = 8\ cm, AC = 4\ cm, AD = 5\ cm$ .
Find $\dfrac{\text{area of}\, \triangle ABC}{\text{area of}\, \triangle ACD}$

In the figure, line $PQ$ is parallel to side $BC$ . Find the value of ' $x$ '. If $AP=3$ , $PB=6$ , $AQ=5$ , $QC=x$ .

In $\triangle ABC$ , line $PQ \parallel$ side $BC$ . Find the value of $x$ from the information given in the following figure.

In the figure, it is given that $BR = PC$ and $\angle ACB = \angle QPR$ and $AB \parallel PQ$ . Prove that $AC= QR$ .

Prove that
"If two triangles are equiangular then their corresponding sides are in proportion"

In the following figure, $DE \parallel AB$ . If $AD = 7\ cm, CD = 5\ cm$ and $BC = 18\ cm$ , find $CE$ .

$\triangle {ABC}\sim \triangle{DEF}$ . Their areas are $64\ {cm}^{2}$ and $121\ {cm}^{2}$ .
Then ratio of corresponding sides $=$

In $\triangle ABC, DE\parallel BC$ and $\dfrac {AD}{DB} = \dfrac {2}{3}$ . If $AE = 3.7\ cm$ , find $EC$

If the ratio of corresponding medians of two similar triangles are $9:16$ , then find the ratio of their areas.

Fill in the blanks using the correct word given in brackets:
All .......... triangles are similar (isosceles, equilateral).

The corresponding sides of two similar triangles are in the ratio 1 :If the area of the smaller triangle in 40 $\text{ cm }^{ 2 }$ , find the area of the larger triangle.

What values of x will make DE $\parallel$ AB in the given figure?

The area of two similar triangles $ABC$ and $PQR$ are in the ratio $9 : 16$ . If $BC = 4.5 cm$ , find the length of $QR$ .

In $\Delta$ PQR, MN is parallel to QR and $\displaystyle\frac{PM}{MQ}=\frac{2}{3}$ . Find, Area of $\Delta$ OMN:Area of $\Delta$ ORQ.

Given $\Delta$ ABC $\sim$ $\Delta$ PQR, if $\displaystyle\frac{AB}{PQ}=\frac{1}{3}$ , then find $\displaystyle\frac{ar \Delta ABC}{ar \Delta PQR}$ .

$ABC$ is a triangle in which $\angle A= 90^{o}$ , $AN \bot BC$ , $BC=12\ cm$ and $AC=5\ cm$ . Find the ratio of the areas of $\triangle ANC$ and $\triangle ABC$ .

In the given Figure, $\triangle ABC \sim\triangle$ $PQR$ and quad $ABCD$ $\sim$ quad $PQRS$ . Determine the values of $x, y, z$ in each case.

Give two different examples of pair of
(ii) non-similar figures.

In the figure, $DE \parallel BC$
(i) Prove that $\triangle ADE$ and $\triangle ABC$ are similar

In the figure $PS=3,SQ=6.QR=5,PT=x$ and $TR=y$ .Give any two pairs of value of $x$ and $y$ such that line $ST||QR$

Area of similar triangles are in the ratio $25:36$ then ratio of their similar sides is _________?

In figure if $AD=6\ cm, DB=9\ cm, AE=8\ cm$ and $EC=12\ cm$ and $\angle ADE=48^{o}$ . Find $\angle ABC$ .

Given that $\Delta ABC \sim \Delta DEF,BC = 3\,cm,EF = 4\,cm$ ans the area of $\Delta ABC$ is $54\,sq.cm.$ Then find the area of $\Delta DEF.$

In figure $\angle 1=\angle 2$ and $\Delta NSQ\cong \Delta MTR$ , then prove that $\Delta PTS\sim \Delta PRQ$ .

$DA,CB,OM$ are each perpendicular to line segment $AB$ where $O$ is the point of intersection of $AC$ and $DB$ .If $OA=2.4$ cm, $OC=3.6$ cm then find the ratio of $AM:BM$

In $\Delta ABC, \bar{PQ} || \bar{BC}$ and $AP = 3x - 19, PB = x - 5, AQ = x - 3, QC = 3 \,cm$ . Find $x$ .

Ler $\Delta A B C \sim \Delta D E F$ and their areas be respectively, 64 $\mathrm { cm } ^ { 2 }$ and 121 $\mathrm { cm } ^ { 2 } .$ If EF $=15.4$ find BC

In figure, $AC\parallel BD$ and $CE\parallel DF$ Find CD,where OD=7,OA=10.5,AB=4.5

If D and E are points on sides AB and AC respectively of ABC and AB- 12 cm, AD= 8 cm,AE = 12 cm, AC =18 cm, then prove that $DE \parallel BC$ .

In the adjoining figure, $DE \parallel BC$ ,find x.

In the adjoining figure, $P$ and $Q$ are points on sides $AB$ and $AC$ respectively of mix such that $PQ \parallel BC$ and $$AP = 8\ cm, AB =12\ cm,AQ = 3x\ cm, QC = (x+2)\ cm.$$
Find x.

P and Q are points on sides AB and AC respectivelyof $\triangle$ ABC. For each of the following cases, statewhether PQ $\parallel$ BC.
AB = 5 cm, AC =10 cm, AP: 4 cm, AQ = 8 cm.

If DE has been drawn parallel to side BC of ABC cutting AB and AC at points D and E respectively , such that $\frac{AD}{DB} = \frac {3}{4}$ then and the value of $\frac {AE}{EC}$

P and Q are points on sides AB and AC respectivelyof $\triangle$ ABC. For each of the following cases, statewhether PQ $\parallel$ BC.
AP: 8 cm, PB = 3 cm, AC = 22 cm and AQ =16 cm.

In the following figure , $ABCD$ is a trapezium with AB || DC. If $AB = 9$ $cm$ , $DC = 18$ $cm$ , $CF = 13.5$ $cm$ , $AP = 6$ $cm$ and $BE = 15$ $cm$ . Find the value of $AF$

In the given figure, $XY \parallel BC$ . Find the length of XY, given BC = 6 cm

In the figure given below, AB EF CD. If $AB = 22.5\,cm$ , $EP = 7.5\,cm$ , $PC = 15\,cm$ and $DC = 27\,cm$ . Calculate EF.

In a $\Delta ABC , BD$ is the median to the side AC , BD is produce to E such that BD = DE
Prove that AE is parallel to BC.

The following figure shows a triangle ABC in which AB = AC . M is a point on AB and N is a point AC such that BM = CN
Prove that
BN = CM

In trapezium $ABCD$ , sides $AB$ and $DC$ are parallel to each other. $F$ is mid-point of $AD$ and $E$ is mid-point of $BC$ . Can it be concluded that $AB + DC= 2EF$ ?
If the above statement is true then mention answer as 1, else mention 0 if false

In figure, $\Delta$ UVW is right angled triangle and $\angle$ UVW $=$ 90 $^o$ , UV $=$ 6cm, and UW $=$ 8 cm, $\sin$ W = $p:m$ , then $p+m$ =

Show that $\displaystyle \Delta ADC$ and $\displaystyle \Delta BAC$ are similar.

In a triangle, $\displaystyle \angle BAC= 90^{\circ}$ and $\displaystyle AD\perp BC.$
If AC=13.5 cm and CD =4.5cm; calculate the length of 2BC

$9$ cm, $12$ cm and $15$ cm

If a line parallel to $BC$ intersects side $AB$ and $AC$ of triangle $ABC$ at $P$ and $Q,$ then prove that $\dfrac {AP}{AB}=\dfrac {AQ}{AC}.$

In the following figure, $\displaystyle \angle ABC=90^{\circ},AB =\left ( x+8 \right )\ cm, BC=\left ( x+1 \right )\ cm$ and $\displaystyle AC=(x + 15)\ cm$ .
Find the value of $x$ .

$\displaystyle \Delta ABD\equiv \Delta ADC$
If the above statement is true then mention answer as 1, else mention 0 if false

O is mid-point of AP
If the above statement is true then mention answer as 1, else mention 0 if false

Find lengths of $ME(in\ cm)$ .

Two angles of one triangle are $85^{\circ}$ and $65^{\circ}$ is equal to angles of the other. Are they similar? Prove it.

$\Delta\,APB$ is similar to $\Delta\,CPD$ .
Enter 1 if true , else 0

In the figure given below, straight lines $AB$ and $CD$ intersect at $P$ , and $AC || BD$ . If $BD = 2.4\ cm$ , $AC=4.8\ cm$ ; find the value of $\cfrac { AP }{ PB }$ .

$\angle APB$ .

$\displaystyle\,\dfrac{CP}{PA}=\dfrac m{3}$ .Find $m$

$DE$ is parallel to $FB$

$DEBF$ is a parallelogram.

$\displaystyle\,\frac{AE}{EC}$

In the given figure, AD = AE and $AD^{2}\,=\,BD\,\times\,EC$ .
Hence , $\triangle ABD$ $\sim$ $\triangle CAE$ .
If the above statement is true then mention answer as 1, else mention 0 if false

In the following figure, $XY$ is parallel to $BC$ , $AX = 9\ cm$ . $XB = 4.5 cm$ and $BC = 18\ cm$ .

What is the value of $\displaystyle\,\dfrac{AY}{YC}$ ?

$\triangle ABC \sim \triangle PQR$ . If AD and PM are corresponding medians of the two triangles , then $\frac{AB}{PQ}\,=\,\frac{AD}{PM}$ .

In triangle $ABC$ , the bisector of angle $BAC$ meets opposite side $BC$ at point $D$ . If $BD = CD$ , prove that $\Delta ABC$ is isosceles.

What is the value of $EC$ in mm?

Find $XY$ in cm.

In the following figure , $ABCD$ is a trapezium with $AB \parallel DC.$ If $AB = 9$ $cm$ , $DC = 18$ $cm$ , $CF = 13.5$ $cm$ , $AP = 6$ $cm$ and $BE = 15$ $cm$ . Find the value of $PE.$

In the given figure, $\Delta \,ABC\,\sim\,\Delta\,ADE$ . If
$AE : EC = 4 : 7$ and $DE = 6.6\ cm,$ find $BC$ . If x be the length of the perpendicular from $A$ to $DE$ , find the length of perpendicular from $A$ and $BC$ in terms of $x.$

ABCD a parallelogram. $E$ is a point on $AD$ and $CE$ is produced to meet $BA$ at $F$ . If $AE = 4$ $cm$ , $AF = 8$ $cm$ and $AB = 12$ $cm$ . Find the perimeter (in $cm$ ) of parallelogram $ABCD$ .

Calculate $AF$ .

Area $(\Delta AGB) = \frac{2}{3} \times\, Area (\Delta ADB)$

In the figure, given alongside, $\angle \,APQ\,=\,\angle\,C$ . Hence, If AQ = 9 cm, BP = 3 cm and AP = 15 cm; find AC in cm