Quadrilaterals - Class 9 Maths - Extra Questions
Can a quadrilateral ABCD be a parallelogram for the following condition?(Enter 1 for True and 0 for False)$$AB = DC = 8$$ cm, $$AD = 4$$ cm and $$BC = 4.4$$ cm.
Can a quadrilateral ABCD be a parallelogram for the following condition?(Enter 1 for True and 0 for False)$$\angle A = 70^{0}\ and \ \angle C = 65^{0}$$.
$$ In\quad a\quad parallelogram\quad the\quad opposite\quad angles\quad must\quad be\quad equal.\\ Here\quad ABCD\quad is\quad a\quad quadrilateral.\\ So\quad \angle A\quad \& \quad \angle C\quad are\quad opposite\quad angles\quad which\quad must\quad be\quad equal\\ for\quad ABCD\quad to\quad be\quad a\quad parallelogram.\\ But\quad \angle A={ 70 }^{ o }\quad \& \quad \angle C={ 65 }^{ o }\\ i.e\quad \angle A\neq \angle C\\ \therefore \quad ABCD\quad cannot\quad be\quad a\quad parallelogram.\\ Ans-\quad 0$$
If two adjacent angles of a parallelogram are in the ratio of $$2 : 3$$ find all the angles of the parallelogram.
The angles of a quadrilateral are of the measures $$120$$$$^o$$, $$90$$$$^o$$, $$72$$$$^o$$ and $$x$$$$^o$$, then find $$x$$.
Three angles of a quadrilateral are of the measure 130$$^o$$, 82$$^o$$, 40$$^o$$. Find the measure of the fourth angle.
The angles of a quadrilateral are in the ratio $$2 : 3 : 5 : 8$$. Find the smallest angle of the quadrilateral.
In square $$ABCD$$, $$\angle A = 120^o$$. If $$\angle B = \angle C = \angle D$$, then find the measures of $$\angle B$$.
The measures of two angles of a quadrilateral are $$55$$$$^o$$ each. The third angle is $$150$$$$^o$$. What is the measure of the fourth angle?
If all the sides of a parallelogram touches a circle, show that the parallelogram is a rhombus.
Examine whether the following points taken in order form a square.
(-1, 2), (1, 0), (3, 2) and (1, 4)
Examine whether the following points taken in order form a rectangle.
(8, 3), (0, -1), (-2, 3) and (6, 7)
Examine whether the following points taken in order form a square.
(0, -1), (2, 1), (0, 3) and (-2, 1)
Let us consider the given points as $$A(0,-1) ; B(2,1) ; C(0,3) ; D(-2,1)$$.
The distance between $$AB = \sqrt {(2-0)^2+(1+1)^2} =\sqrt{4+4} = \sqrt 8$$
The distance between $$BC = \sqrt{(0-2)^2+(3-1)^2} = \sqrt {4+4}=\sqrt 8$$
The distance between $$CD = \sqrt {(-2-0)^2+(1-3)^2} =\sqrt{4+4} = \sqrt 8$$
The distance between $$DA = \sqrt{(0+2)^2+(-1-1)^2} = \sqrt {4+4}=\sqrt 8$$
The distance between $$AC = \sqrt {(0-0)^2+(-1-3)^2} =\sqrt {16} =4$$
The distance between $$BD = \sqrt{(2+2)^2+(1-1)^2} = \sqrt {16}=4$$
As the sides are equal $$AB = BC = CD = DA$$
The diagonals are also equal $$AC= BD$$.
Therefore, the given points represents square.
Hence, this is the answer.
Examine whether the following points taken in order form a square.
(5, 2), (1, 5), (-2, 1) and (2, -2)
A piece of wire is $$60\ cm$$ long. What will be the length of each side if the wire is used to form a square?
Three angles of quadrilateral are respectively equal to $${110^ \circ },{50^ \circ },{40^ \circ }$$. Find its fourth angle.
In the given figure, In $$\Delta PQR\quad XY\ \parallel QR$$, $$PX : QX=1 : 3$$ $$YR=4.5 cm$$ and $$QR=9 cm$$. Find $$XY$$. Further if $$Ar (\Delta PXY$$)=$$A { cm }^{ 2 }$$, find in terms of $$A$$ the area of $$\Delta PQR$$ and the area of trapezium $$XYRQ$$.
Quadrilateral $$MORE$$
$$MO=6\ cm$$
$$OR=4.5\ cm$$
$$\angle M={60}^{o}$$
$$\angle O={105}^{o}$$
$$\angle R={105}^{o}$$
A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?
Square is a regular quadrilateral as all sides and angles are equal.
As per the property mentioned above,
$$1).$$ All sides are equal
$$2).$$ All angles are $$90^{\circ}$$ (means equal).
$$1).$$ All sides are equal
$$2).$$ All angles are equal
Hence, the given figure can be identified as the regular quadrilateral.
Fill in the blanks using the correct word given in brackets:
All squares are .......... (similar, congruent).
A quadrant has its all four interior angles equal. What is the measure of each interior angle.
The measure of two adjacent angles of a quadrilateral are $$110^o$$ and $$50^o$$ and the other two angles are equal. Find the measure of each angle.
In the given figures, $$PQRS$$ is a parallelogram .Find the value of $$x,y$$ and $$z$$.
In the given figure, ABCD is a rhombus where $$BO = 12cm$$ and $$OC=5cm$$. find the values of $$x,y$$ and $$z$$
The angle of a quadrilateral are respectively $$100^{o}, 98^{o}, 92^{o}$$. Find the fourth angle.
Show that the point $$(5, -1, 1)(7, -4, 7) , (1, -6, 10)(-1, -3, 4)$$ are the vertices of a rhombus.
Angles of a quadrilateral are in the ratio $$3:4:4:7$$ . Find all the angles of the quadrilateral.
Prove that the sum of angles of a quadrilateral is $${360}^{o}$$
Let ABCD be any quadrilateral divided into four triangles, then using sum of angles in a $$\Delta $$ is $$180^o \Rightarrow$$
$$\angle 1 + \angle 2 + \angle 3 = 180^o$$
$$\angle 4 + \angle 5 + \angle 6 = 180^o$$
$$\angle 7 + \angle 8 + \angle 9 = 180^o$$
$$\angle 10 + \angle 11 + \angle 12 = 180^o$$
$$\therefore \angle 1 + \angle 2 + \angle 3 + ... \angle 12 = 180 \times 4$$ ...(1)
Now
$$\angle 3 + \angle 6 + \angle 7 + \angle 10 = 360^o$$
equation (1) - equation (2) $$\Rightarrow$$
$$\angle 1 + \angle 2 + \angle 4 + \angle 5 + \angle 8 + \angle 9 + \angle 11 + \angle 12 = 180 \times 4 - 360$$
$$\Rightarrow $$ Sum of angles in a quadrilateral = $$180 \times 4 - 360 = 360^o$$
Hence proved
The sum of two opposite angles of a quadrilateral is $${172}^{o}$$. The other two angles of the quadrilateral are equal. Find the equal angles.
In a quadrilateral if opposite vertices are joint it is called_____
One of the angle of quadrilatral has measure $$\dfrac{2\pi}{9}$$ radians and other three angles are in the ratio $$3: 5: 8$$, Find the measures in radian.
Find the area of trapezium whose sum of parallel sides is 28 cm and height is 10 cm.
Diagonals of a quadrilateral JOSH intersect at P in such a way that ar(JPH) = ar(OPS).Prove that JOSH is a trapezium.
ar.(JPH) = ar (OPS)
$$\dfrac{1}{2} JH \times PM = \dfrac{1}{2} OS \times PN$$
Area JOSH = ar. $$\Delta OHJ + ar \, \Delta OHS$$
$$= \dfrac{1}{2} \times JH \times MN + \dfrac{1}{2} \times os \times MN$$
$$= \dfrac{1}{2} MN (JH + Os)$$
$$\therefore$$ Ar. g traperiam = $$\dfrac{1}{2} \times 1 ar \times sam g$$. let side
$$\dfrac{1}{2} JH \times PM = \dfrac{1}{2} OS \times PN$$
Area JOSH = ar. $$\Delta OHJ + ar \, \Delta OHS$$
$$= \dfrac{1}{2} \times JH \times MN + \dfrac{1}{2} \times os \times MN$$
$$= \dfrac{1}{2} MN (JH + Os)$$
$$\therefore$$ Ar. g traperiam = $$\dfrac{1}{2} \times 1 ar \times sam g$$. let side
The four angles of a quadrilateral are equal. Find each of them.
A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.
If the measures of internal angles of a quadrilateral are in the ratio of $$3:6:4:5$$. what are the measures of its angle?
$$ { ABCD }$$ is a parallelogram where $${ A } ( x , y ) , { B } ( 5,8 ) ,{ C } ( 4,7 )$$ and $${ D } ( 2 , - 4 )$$ . Find
(i) the coordinates of $$A$$.
(ii) the equation of the diagonal $$BD$$.
Find the value of $$x$$
Find the area and perimeter of the rectangle in the figure.
ABCD is a rectangle with length AB$$=\sqrt{8}$$cm and breadth AD$$=\sqrt{2}$$cm
Area of a rectangle$$=$$length$$\times$$breadth
$$=(\sqrt{8}\times \sqrt{2})cm^2$$
$$=\sqrt{16}cm^2$$
$$=4cm^2$$
Perimeter of rectangle$$=2$$(length$$+$$breadth)
$$=2(\sqrt{8}+\sqrt{2})$$cm
$$(\sqrt{8}=\sqrt{4\times 2}=2\sqrt{2})=2(2\sqrt{2}+\sqrt{2})$$cm
$$=2(3\sqrt{2})$$cm
$$=6\sqrt{2}$$cm.
Find the length of a diagonal of a quadrilateral whose area is $$195\ m^{2}$$ and the lengths of the perpendicular to the diagonal vertices are $$8.5\ m$$ and $$11\ m$$ respectively.
Write each properties of trapezium, parallelogram, rectangle, rhombus, square.
If the four angles of a quadrilateral are equal, what will be the measure of each angle ? What is the special name of this figure ?
The ratio of the angles of a quadrilateral is $$3 : 5 : 7 : 9.$$ Find the angles.
Find x in the picture:
The angle subtended by an arc at the centre is double the angle subtended by it at any on the remaining part of the circle.
$$\therefore \angle AOC=90^o$$
& $$\angle AOC^{\ast}$$(outside) $$=360^o-90^o$$
$$=270^o$$
$$\angle ABC=x=\dfrac{1}{2}(\angle AOC^{\ast})$$
$$x=\dfrac{1}{2}\times 270=135^o$$.
Prove that a cyclic parallelogram is rectangle.
Given : $$ABCD$$ be a cyclic parallelogram.
To prove :$$ABCD$$ is a rectangle.
A rectangle is a parallelogram with one angle $$90^0$$
So, we have to prove angle $$90^0$$
Since $$ABCD$$ is a paralleogram
$$\angle A = \angle C$$ (opposite angle of parallelogram are equal)......................(i)
In cyclic parallelogram $$ABCD$$
$$\angle A + \angle C=180^0$$ (Sum of opposite angles of a cyclic quadrilateral is $$180^0$$)
$$\angle A + \angle A=180^0$$ from(i)
$$ 2 \angle A =180^0$$
$$\angle A = \dfrac{{{{180}^0}}}{2} = {90^0}$$
Hence,
$$\angle A = \angle C= 90^o$$
Similarly we can prove other two angles are also equal to $$90^o$$
Thus, a cyclic parallelogram is rectangle.
In the figure , 0 is the center of the circle and $$BA = AC$$.If$$\angle ABC = {50^.}$$,find$$\angle BOC$$ and $$\angle BDC$$
Ref. Image.
$$ BA=AC$$
$$ \angle ABC = 50^{\circ}$$
Since BA = AC
$$ \angle ABC = \angle ACB = 50^{\circ}$$
so $$ \angle A = 180^{\circ} - (50^{\circ}+50^{\circ})$$
$$ = 80^{\circ}$$
We know $$ \angle BOC = 2(\angle BAC)$$
$$ \angle BOC = 2 \times 80= 160^{\circ} $$
ABCD is a cyclic Quadrilateral
So $$ \angle A + \angle D = 180^{\circ}$$
$$ 80^{\circ} + \angle D = 180^{\circ} \Rightarrow \angle BDC = 100^{\circ}$$
The angles of a quadrilateral are in the ratio $$2:5:5:6$$. Find the greatest angle.
Using information given in the adjacent figure , find the value of $$x$$.
$${\textbf{Step - 1: Finding angle y}}$$
$${\text{We know that, sum of interior angle of a quadrilateral is 360}}^\circ $$
$$\because {\text{ ABCD is a quadrilateral}}$$
$$\therefore {\text{ }}\angle {\text{DAB + }}\angle {\text{ABC + y + }}\angle {\text{CDA = 360}}^\circ $$
$$ \Rightarrow {\text{ 46 + 38 + 26 + y = 360}}$$
$$ \Rightarrow {\text{ y = 250}}^\circ $$
$${\textbf{Step - 2: Finding angle x}}$$
$${\text{x + y = 360}}^\circ $$
$$ \Rightarrow {\text{ x + 250 = 360}}$$
$$ \Rightarrow {\text{ x = 360 - 250}}$$
$$ \Rightarrow {\text{ x = 110}}^\circ $$
$${\textbf{Hence, The value of angle x is 110}}^\circ $$
Find the value of $$a$$ in the following figures.
on a line segment AB = 4 cm, construct a square,
ABCD is a quadrilateral such that diagonal AC bisects the angles $$\angle A$$ an $$\angle C$$. Prove that $$AB=AD$$ and $$CB=CD$$.
In $$\Delta ABC$$ and $$\Delta ADC$$,
$$\angle BAC=\angle DAC$$ (given)
$$\angle ACB=\angle ACD$$ (given)
$$AC=AC$$ (common)
$$\therefore \Delta ABC\cong \Delta ADC$$ (ASA axiom)
$$\therefore AB=AD$$ and $$CB=CD$$ by CPCT.
$$AO$$ and $$DO$$ are the bisectors of $$\angle A$$ and $$\angle D$$ of the quadrilateral $$ABCD$$. Prove that $$\angle AOD=\dfrac {1}{2}(\angle B+\angle C)$$
$$\begin{array}{l} \angle AOD=\dfrac { 1 }{ 2 } \left( { \angle B+\angle C } \right) \\ proof:\, \angle A+\angle B+\angle C+\angle D={ 360^{ 0 } }\, \, \, \left[ { angle\, \, of\, \, a\, \, quard. } \right] \\ =\dfrac { 1 }{ 2 } \angle A+\dfrac { 1 }{ 2 } \angle B+\dfrac { 1 }{ 2 } \angle C+\dfrac { 1 }{ 2 } \angle D={ 180^{ 0 } } \\ =\angle OAD+\angle DAD={ 180^{ 0 } }-\dfrac { 1 }{ 2 } \left( { \angle B+\angle C } \right) .............\left( 1 \right) \\ \because \angle ODA+\angle DAD+\angle AOD={ 180^{ 0 } }-\angle AOD \\ Putting\, \, this\, \, value\, \, in\, \, \left( 1 \right) \\ { 180^{ 0 } }-\angle AOD={ 180^{ 0 } }-\dfrac { 1 }{ 2 } \left( { \angle B+\angle C } \right) \\ \angle AOD=\dfrac { 1 }{ 2 } \left( { \angle B+\angle C } \right) \end{array}$$
Hence, proved.
Find the value of $$x$$ in each of the following figures.
What is the name given to a parallelogram whose all sides are equal ?
In $$\Box ABCD$$, side $$BC\parallel \ AD$$. Diagonal $$AC$$ and diagonal $$BD$$ Intersect in point $$Q$$. If $$AB=CD$$ .Then show that $$\Delta ABC$$ is congruent to $$\Delta DBC$$
In the following figure,
If $$\angle BAD=96^{o}$$, find $$\angle BCD$$ and $$\angle BFE$$
In $$\Box ABCD$$, side $$BC <$$ side $$ AD$$
side $$BC \parallel $$ side $$AD$$ and if
side $$BA \cong$$ side $$CD$$
then prove that $$\angle ABC \cong \angle DCB$$.
Find area of trapezium whose parallel sides $$9cm$$ and $$5cm$$ respectively and the distance between these sides is $$8cm$$.
Given: Length of parallel sides are $$9$$cm and $$5$$cm
Height $$=8$$cm
Area of trapezium$$=\dfrac{1}{2}\times$$ Height $$\times$$(Sum of parallel sides)
$$=\dfrac{1}{2}\times 8cm\times (9+5)cm$$
$$=\dfrac{8+14}{2}cm^2$$
$$=56cm^2$$.
The angles of a qudarilateral are in the ratio 3 : 5 : 9 :Find all the angles of the quadrilateral.
Fill in the blanks:
(i) A quadrilateral with all sides and all angles equal is a .............
(ii) A quadrilateral with four equal sides and no right angles can be called a...................
(iii) A quadrilateral with exactly two sides parallel is a......................
(iv) The diagonals of this quadrilateral are equal but not perpendicular. The quadrilateral is a.............
(v) All rectangles, squares and rhombus are ............,but a trapezium is not.
What is trapezium ?
The angles of quadrilateral rae $$x,x+20,x+30,x+60$$ Find $$x$$
.Use the information given in the following figure to find :
$$(1)$$ x
$$(2)$$ $$\angle B$$ and $$\angle C$$
EFGH is a parallelogram with diagonals EG and FH meeting at a point X. Show that $$ar(\Delta GXF)=ar(\Delta EXH)=ar(\Delta GXH)=ar(\Delta XEF)$$
X is the midpoint of both the diagonals (diagonals of a parallelogram bisect each other)
'
In triangle $$EFG$$,
XF is the median.
Therefore $$area (\triangle EXF)=area(\triangle GXF)$$ (since median of a triangle divides the triangle into two triangles of equal area) ..........1
In triangle $$FGH$$,
GX is the median.
Therefore, $$area(\triangle HXG)=area(\triangle GXF)$$ (reason same as above) .......2
And in Triangle $$EHG,$$
HX is the median.
Therefore, $$area(\triangle EXH)=area(\triangle HXG)$$ (reason same as above) ..............3
From eq 1 , 2 and 3 it can be concluded that
$$ar(\Delta GXF)=ar(\Delta EXH)=ar(\Delta GXH) = ar(\Delta EXF)$$
The angles of a quadrilateral are in the ratio of $$2:3:4:5$$, then find the angles.
Let the angles be $$2k, 3k, 4k, 5k$$
As sum of angles of Quadrilateral $$=360^o$$
$$\Rightarrow 14k=360$$
$$\Rightarrow k=\dfrac{180}{7}$$
$$\Rightarrow$$ Angles $$=\dfrac{360^o}{7}, \dfrac{540^o}{7}, \dfrac{720^o}{7}, \dfrac{900^o}{7}$$.
In quadrilateral PQRS.
$$\angle P:\angle Q:\angle R:\angle S=3:5:9:7$$. Find all the angles of quadrilateral PQRS.
Let the angles be $$3x, 5x, 9x, 7x$$
Sum of all angles of quadrilateral is $$360^o$$
So $$3x+5x+9x+7x=360^o$$
$$24x=360^o$$
$$x=15^o$$
Angles $$=45, 75, 135, 105$$.
If a triangle and a parallelogram are on the same base and between the same parallels, then prove that area of triangle is half the area of parallelogram.
Draw $$AL \perp BC$$ and $$DM \perp BC$$, meeting BC produced in M.
Since, A, E and D are collinear and $$BC\parallel AD$$
Therefore,
$$AL=DM$$
Now,
$$ar(\triangle ABC)=\dfrac{1}{2}(BC\times AL)$$
$$ar(\triangle ABC) =\dfrac{1}{2}(BC\times DM)$$
$$ar(\triangle ABC)=\dfrac{1}{2}ar({\parallel}^{gm} BCDE)$$
Give the difference between a square and a rhombus.
A square has all angle $$90^o$$ while rhombus does not and diagonals of square are equal in length but of rhombus not.
2 angles of quadrilateral are $${ 70 }^{ o }$$ and $${ 80 }^{ o }$$ third angle is supplementary of $${ 70 }^{ o }$$ find the $${ 4 }^{ th }$$ angle.
Angles $$=70^o, 80^o$$
$$3^{rd}$$ Angle $$=180-70^o=110$$
$$4^{th}$$ angle $$=x$$
$$\Rightarrow 70+80+110+x=360^o$$
$$x=360^o-260^o$$
$$x=100^o$$.
Do the angles $$120,55,150,35$$ make a qudrilateral give reasons.
Find the area of the rectangles whose sides are:
$$3\ cm$$ and $$4\ cm$$
The angles of quadrilateral are $$x,2x,2x,5x+10$$ find the value of $$x$$
(i) How many pairs of adjacent sides does a quadrilateral have?
(ii) How many pairs of opposite sides does a quadrilateral have?
(iii) How many pairs of adjacent angles does a quadrilateral have?
(iv) How many pairs of opposite angles does a quadrilateral have?
(v) How many diagonals does a quadrilateral have?
In the figure, given below, $$ABCD$$ is a cyclic quadrilateral in which $$\angle BAD=75^{o}; \angle ABD=58^{o}$$ and $$\angle ADC=77^{o}$$. Find:
$$\angle BDC$$
The angles in quadrilateral are $$2x,3x-50,4x,5x-10$$ find $$x$$
Show that sum of all sides of a quadrilateral is greater than the sum of the diagonals
ABCD is a quadrilaterial Prove that $$ A B+B C+C D+D A>A C+B D ? $$
The angle of a quadrilateral are in the ratio 3:5:9:Find all the angles of the quadrilateral.
$$3:5:9:13$$
$$\Rightarrow 3x+5x+9x+13x$$
=360
50x=360
x=12
angle = $$36^{\circ}$$
$$60^{\circ}$$
$$108^{\circ}$$
$$156^{\circ}$$
Angles of a quadrilateral are $$\left( 4x \right) ^{ \circ },5\left( x+2 \right) ^{ \circ },\left( 7x-20 \right) ^{ \circ }$$ and $$6\left( x+3 \right) ^{ \circ }.$$ Find each angle of the quadrilateral.
The diagram shows a regular pentagon and a kite.
Complete the following statement.
The kite has rotational symmetry of order _________.
Find the value of $$y$$
Find the area of the rectangles whose sides are:
$$12\ cm$$ and $$21\ cm$$
Find the area of the rectangles whose sides are:
$$2\ cm$$ and $$3\ cm$$
In the given figure
$$b=2a+15^{\circ}$$ and
$$c=3a+5^{\circ}$$, find the values of b and c.
The angles of quadrilateral are $$3x,4x,5x,6x.$$ Find the value of $$x.$$
Angles of a quadrilateral are $$\left( 4x \right) ^{ \circ },5\left( x+2 \right) ^{ \circ },\left( 7x-20 \right) ^{ \circ }$$ and $$6\left( x+3 \right) ^{ \circ }.$$ Find
the value of $$x$$.
The sum of the angles of a quadrilateral is_______
The sum of the angles of a quadrilateral is $$ 360^o $$
If we draw one diagonal to quadrilateral we will get two triangles i.e. $$ \triangle ABC $$ and $$ \triangle ACD $$
But we know that sum of angles of triangle is $$ 180^o $$
Sum of angles of quadrilateral ABCD= $$ \triangle ABC+ \triangle ACD $$
=$$ 180^o + 180^o $$
=$$ 360^o $$
If we draw one diagonal to quadrilateral we will get two triangles i.e. $$ \triangle ABC $$ and $$ \triangle ACD $$
But we know that sum of angles of triangle is $$ 180^o $$
Sum of angles of quadrilateral ABCD= $$ \triangle ABC+ \triangle ACD $$
=$$ 180^o + 180^o $$
=$$ 360^o $$
A quadrilateral has ...... diagonals.
The tree angles of a quadrilateral are $$ 76^o, 54^o $$ and $$ 108^o $$. Find the measure of the fourth angle.
Solve the following :
Of the four quadrilaterals - square , rectangle , rhombus and trapezium - one is somewhat different from the others because of its design. Find it and give justifications.
In the given figure, $$ABCD$$ is a quadrilateral shaped field in which diagonal $$BD$$ is $$36\ m$$, $$AL \bot BD$$ and $$CM \bot BD$$ such that $$AL=19\ m$$ and $$CM=11 \ m$$. Find the area
Solve the following :
The point of intersection of diagonals of a quadrilateral divides one diagonal in the ratio $$ 1 : 2 $$. Can it be a parallelogram ? Why or why not ?
A quadrilateral has_______ angles.
Four
Prove that the sum of the angles of a quadrilateral is $$ 360^o $$
If we draw one diagonal to quadrilateral we will get two triangles i.e.
$$ \triangle ABC $$ and $$ \triangle ACD $$ and we know that sum of
the angles of a triangle is $$ 180^o $$
Sum of angles of quadrilateral ABCD=sum of angles of $$ \triangle ABC $$ + sum of the angles $$ \triangle ACD $$
= $$ 180^o + 180^o $$
= $$ 360^o $$
Sum of angles of quadrilateral ABCD=sum of angles of $$ \triangle ABC $$ + sum of the angles $$ \triangle ACD $$
= $$ 180^o + 180^o $$
= $$ 360^o $$
The figure given below, shows a trapezium $$ABCD. M$$ and $$N$$ are the mid-points of the non-parallel sides $$AD$$ and $$BC$$ respectively. Find:
$$MN$$, if $$AB=11\ cm$$ and $$DC=8\ cm$$
Let we draw a diagonal $$AC$$ as shown in the figure below,
Given that $$AB=11\ cm, CD=8\ cm$$
From triangle $$ABC$$
From triangle $$ACD$$
$$OM=\dfrac{1}{2}CD=\dfrac{1}{2}\times 8=4\ cm$$
Hence $$MN=OM+ON=(4+5.5)=9.5\ cm$$
Given that $$AB=11\ cm, CD=8\ cm$$
From triangle $$ABC$$
From triangle $$ACD$$
$$OM=\dfrac{1}{2}CD=\dfrac{1}{2}\times 8=4\ cm$$
Hence $$MN=OM+ON=(4+5.5)=9.5\ cm$$
The diagonal of a quadrilateral intersect at right angles. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is a rectangle.
The figure is shown below
Let $$ABCD$$ be a quadrilateral where $$P, Q, R, S$$ are the midpoint of $$AB, BC, CD, DA$$. Diagonal $$AC$$ and $$BD$$ intersects at right angle at point $$O$$. We need to show that $$PQRS$$ is a rectangle
Proof:
From $$\triangle ABC$$ and $$\triangle ADC$$
$$2PQ=AC$$ and $$PQ\parallel AC$$ .......(1)
$$2RS=AC$$ and $$RS\parallel AC$$ .......(2)
From $$(1)$$ and $$(2)$$ we get,
$$PQ=RS$$ and $$PQ\parallel RS$$
Similarly we can show that $$PS=PQ$$ and $$PS\parallel RQ$$
Therefore $$PQRS$$ is a parallelogram.
Now $$PQ\parallel AC$$, therefore $$\angle AOD=\angle PXO=90^{o}$$ [Corresponding angle]
Again $$BD\parallel RQ$$, therefore $$\angle PXO=\angle RQX=90^{o}$$ [Corresponding angle]
Similarly $$\angle QRS=\angle RSP=\angle SPQ=90^{o}$$
Therefore $$PQRS$$ is a rectangle.
Hence proved
Let $$ABCD$$ be a quadrilateral where $$P, Q, R, S$$ are the midpoint of $$AB, BC, CD, DA$$. Diagonal $$AC$$ and $$BD$$ intersects at right angle at point $$O$$. We need to show that $$PQRS$$ is a rectangle
Proof:
From $$\triangle ABC$$ and $$\triangle ADC$$
$$2PQ=AC$$ and $$PQ\parallel AC$$ .......(1)
$$2RS=AC$$ and $$RS\parallel AC$$ .......(2)
From $$(1)$$ and $$(2)$$ we get,
$$PQ=RS$$ and $$PQ\parallel RS$$
Similarly we can show that $$PS=PQ$$ and $$PS\parallel RQ$$
Therefore $$PQRS$$ is a parallelogram.
Now $$PQ\parallel AC$$, therefore $$\angle AOD=\angle PXO=90^{o}$$ [Corresponding angle]
Again $$BD\parallel RQ$$, therefore $$\angle PXO=\angle RQX=90^{o}$$ [Corresponding angle]
Similarly $$\angle QRS=\angle RSP=\angle SPQ=90^{o}$$
Therefore $$PQRS$$ is a rectangle.
Hence proved
Prove that the figure obtained by joining the mid-point of the adjacent sides of a rectangle is a rhombus.
The figure is shown below:
Let $$ABCD$$ be a rectangle where $$P, Q, R, S$$ are the midpoint of $$AB, BC, CD, DA$$. We need to show that $$PQRS$$ is a rhombus
For help we draw two diagonal $$BD$$ and $$AC$$ as shown in figure Where $$BD=AC$$ (Since diagonal of rectangle are equal)
Proof:
From $$\triangle ABD$$ and $$\triangle BCD$$
$$PS=\dfrac{1}{2}BD=QR$$ and $$PS\parallel BD\parallel QR$$
$$2PS=2QR=BD$$ and $$PS\parallel QR$$ .......(1)
Similarly $$2PQ=2SR=AC$$ and $$PQ\parallel SR$$ ........(2)
From (1) and (2) we get
$$PQ=QR=RS=PS$$
Therefore $$PQRS$$ is a rhomus.
Hence proved
Let $$ABCD$$ be a rectangle where $$P, Q, R, S$$ are the midpoint of $$AB, BC, CD, DA$$. We need to show that $$PQRS$$ is a rhombus
For help we draw two diagonal $$BD$$ and $$AC$$ as shown in figure Where $$BD=AC$$ (Since diagonal of rectangle are equal)
Proof:
From $$\triangle ABD$$ and $$\triangle BCD$$
$$PS=\dfrac{1}{2}BD=QR$$ and $$PS\parallel BD\parallel QR$$
$$2PS=2QR=BD$$ and $$PS\parallel QR$$ .......(1)
Similarly $$2PQ=2SR=AC$$ and $$PQ\parallel SR$$ ........(2)
From (1) and (2) we get
$$PQ=QR=RS=PS$$
Therefore $$PQRS$$ is a rhomus.
Hence proved
In triangle $$ABC, M$$ is mid-point of $$AB$$ and a straight line through $$M$$ and parallel to $$BC$$ cuts $$AC$$ at $$N$$. Find the lengths of $$AN$$ and $$MN$$, if $$BC=7\ cm$$ and $$AC=5\ cm$$
Solve the following :
Two sticks each of length $$ 5\,\text{cm} $$ are crossing each other such that they bisect each other. What shape is formed by joining their endpoints? Give reason.
D, E and F are mid-point of the sides BC, CA and AB respectively of a $$ \triangle $$ ABC. Prove that:
(i) FDCE is a parallelogram
(ii) area of $$ \Delta $$ DEF $$ =1 / 4 $$ area of $$ \Delta \mathrm{ABC} $$
(iii) area of $$ \| \mathrm{gm} $$ FDCE $$ =1 / 2 $$ area of $$ \Delta \mathrm{ABC} $$
Given:D, E and F are mid-point of the sides BC, CA and AB respectively of a $$ \Delta $$ ABC.
To prove:
(i) FDCE is a parallelogram
(ii) area of $$ \Delta \mathrm{DEF}=1 / 4 $$ area of $$ \Delta \mathrm{ABC} $$
(iii) area of $$ \| \mathrm{gm} \mathrm{FDCE}=1 / 2 $$ area of $$ \Delta \mathrm{ABC} $$
Proof:
(i) $$ \mathrm{F} $$ and $$ \mathrm{E} $$ are mid-points of $$ \mathrm{AB} $$ and $$ \mathrm{AC} $$.
So, $$ \mathrm{FE} \| \mathrm{BC} $$ and $$ \mathrm{FE}=1 / 2 \mathrm{BC} \ldots \ldots(1) $$
Also, $$ \mathrm{D} $$ is mid-point of $$ \mathrm{BC} $$
$$ \mathrm{CD}=1 / 2 \mathrm{BC} \ldots \ldots $$(2)
From ( 1 ) and (2) $$ \mathrm{FE} \| \mathrm{BC} $$ and $$ \mathrm{FE}=\mathrm{CD} $$
$$ \mathrm{FE} \| \mathrm{CD} $$ and $$ \mathrm{FE}=\mathrm{CD} \ldots \ldots $$(3)
Similarly, $$ \mathrm{D} $$ and $$ \mathrm{F} $$ are mid-points of $$ \mathrm{BC} $$ and $$ \mathrm{AB} $$.
So, DF $$ \| \mathrm{EC} $$ is a parallelogram.
Hence proved.
(ii) we know that, FDCE is a parallelogram. And DE is a diagonal of $$ \| \mathrm{gm} $$ FDCE
So, ar $$ (\Delta \mathrm{DEF})=\operatorname{ar}(\Delta \mathrm{DEC}) \ldots \ldots(4) $$
Similarly, we know BDEF and DEAF are $$ \| \mathrm{gm} $$
So, ar $$ (\Delta \mathrm{DEF})=\operatorname{ar}(\Delta \mathrm{BDF})=\operatorname{ar}(\Delta \mathrm{AFE}) \ldots \ldots(5) $$
From (4) and (5) $$ \operatorname{ar}(\Delta \mathrm{DEF})=\operatorname{ar}(\Delta \mathrm{DEC})=\operatorname{ar}(\Delta \mathrm{BDF})=\operatorname{ar}(\Delta \mathrm{AFE}) $$
Now, ar $$ (\Delta \mathrm{ABC})=\operatorname{ar}(\Delta \mathrm{DEF})+\operatorname{ar}(\Delta \mathrm{DEF})+\operatorname{ar}(\Delta \mathrm{DEF})+\operatorname{ar}(\Delta \mathrm{DEF}) $$
$$ =4 \operatorname{ar}(\Delta \mathrm{DEF}) $$
$$ \operatorname{ar}(\Delta \mathrm{DEF})=1 / 4 \operatorname{ar}(\Delta \mathrm{ABC}) \ldots \ldots(6) $$
Hence proved.
(iii) ar of $$ \| \mathrm{gm} \mathrm{FDCE}=\operatorname{ar}(\Delta \mathrm{DEF})+\operatorname{ar}(\Delta \mathrm{DEC}) $$
$$\begin{array}{l}=\operatorname{ar}(\Delta \mathrm{DEF})+\operatorname{ar}(\Delta \mathrm{DEF}) \\=2 \operatorname{ar}(\Delta \mathrm{DEF})[\mathrm{From}(4)]\end{array}$$
$$ =2[1 / 4 \operatorname{ar}(\Delta \mathrm{ABC})][\mathrm{From}(6)] $$
ar of $$ \| \mathrm{gm} \mathrm{FDCE}=1 / 2 $$ ar of $$ \Delta \mathrm{ABC} $$
Hence proved.
To prove:
(i) FDCE is a parallelogram
(ii) area of $$ \Delta \mathrm{DEF}=1 / 4 $$ area of $$ \Delta \mathrm{ABC} $$
(iii) area of $$ \| \mathrm{gm} \mathrm{FDCE}=1 / 2 $$ area of $$ \Delta \mathrm{ABC} $$
Proof:
(i) $$ \mathrm{F} $$ and $$ \mathrm{E} $$ are mid-points of $$ \mathrm{AB} $$ and $$ \mathrm{AC} $$.
So, $$ \mathrm{FE} \| \mathrm{BC} $$ and $$ \mathrm{FE}=1 / 2 \mathrm{BC} \ldots \ldots(1) $$
Also, $$ \mathrm{D} $$ is mid-point of $$ \mathrm{BC} $$
$$ \mathrm{CD}=1 / 2 \mathrm{BC} \ldots \ldots $$(2)
From ( 1 ) and (2) $$ \mathrm{FE} \| \mathrm{BC} $$ and $$ \mathrm{FE}=\mathrm{CD} $$
$$ \mathrm{FE} \| \mathrm{CD} $$ and $$ \mathrm{FE}=\mathrm{CD} \ldots \ldots $$(3)
Similarly, $$ \mathrm{D} $$ and $$ \mathrm{F} $$ are mid-points of $$ \mathrm{BC} $$ and $$ \mathrm{AB} $$.
So, DF $$ \| \mathrm{EC} $$ is a parallelogram.
Hence proved.
(ii) we know that, FDCE is a parallelogram. And DE is a diagonal of $$ \| \mathrm{gm} $$ FDCE
So, ar $$ (\Delta \mathrm{DEF})=\operatorname{ar}(\Delta \mathrm{DEC}) \ldots \ldots(4) $$
Similarly, we know BDEF and DEAF are $$ \| \mathrm{gm} $$
So, ar $$ (\Delta \mathrm{DEF})=\operatorname{ar}(\Delta \mathrm{BDF})=\operatorname{ar}(\Delta \mathrm{AFE}) \ldots \ldots(5) $$
From (4) and (5) $$ \operatorname{ar}(\Delta \mathrm{DEF})=\operatorname{ar}(\Delta \mathrm{DEC})=\operatorname{ar}(\Delta \mathrm{BDF})=\operatorname{ar}(\Delta \mathrm{AFE}) $$
Now, ar $$ (\Delta \mathrm{ABC})=\operatorname{ar}(\Delta \mathrm{DEF})+\operatorname{ar}(\Delta \mathrm{DEF})+\operatorname{ar}(\Delta \mathrm{DEF})+\operatorname{ar}(\Delta \mathrm{DEF}) $$
$$ =4 \operatorname{ar}(\Delta \mathrm{DEF}) $$
$$ \operatorname{ar}(\Delta \mathrm{DEF})=1 / 4 \operatorname{ar}(\Delta \mathrm{ABC}) \ldots \ldots(6) $$
Hence proved.
(iii) ar of $$ \| \mathrm{gm} \mathrm{FDCE}=\operatorname{ar}(\Delta \mathrm{DEF})+\operatorname{ar}(\Delta \mathrm{DEC}) $$
$$\begin{array}{l}=\operatorname{ar}(\Delta \mathrm{DEF})+\operatorname{ar}(\Delta \mathrm{DEF}) \\=2 \operatorname{ar}(\Delta \mathrm{DEF})[\mathrm{From}(4)]\end{array}$$
$$ =2[1 / 4 \operatorname{ar}(\Delta \mathrm{ABC})][\mathrm{From}(6)] $$
ar of $$ \| \mathrm{gm} \mathrm{FDCE}=1 / 2 $$ ar of $$ \Delta \mathrm{ABC} $$
Hence proved.
In the quadrilateral $$ABCD$$; prove that
$$AB+BC+CD > DA$$
Construction:
Join $$AC$$ and $$BD$$
In $$\triangle ABC$$,
$$AB +BC > AC....(i)$$ [Sum of two sides is greater than the third side]
In $$\triangle ACD$$,
$$AC +CD> DA.....(ii)$$ [Sum of two sides is greater than the third side]
Adding (i) and (ii)
$$AB+BC+AC+CD > AC +DA$$
$$AB +BC+CD > AC +DA-AC$$
$$AB+BC + CD > DA ....(iii)$$
Join $$AC$$ and $$BD$$
In $$\triangle ABC$$,
$$AB +BC > AC....(i)$$ [Sum of two sides is greater than the third side]
In $$\triangle ACD$$,
$$AC +CD> DA.....(ii)$$ [Sum of two sides is greater than the third side]
Adding (i) and (ii)
$$AB+BC+AC+CD > AC +DA$$
$$AB +BC+CD > AC +DA-AC$$
$$AB+BC + CD > DA ....(iii)$$
The figure given below, shows a trapezium $$ABCD. M$$ and $$N$$ are the mid-points of the non-parallel sides $$AD$$ and $$BC$$ respectively. Find:
$$AB$$, if $$DC=20\ cm$$ and $$MN=27\ cm$$
Let we draw a diagonal $$AC$$ as shown in the figure below,
Given that $$CD=20\ cm, MN=27\ cm$$
From triangle $$ACD$$
$$OM=\dfrac{1}{2}CD=\dfrac{1}{2}\times 20=10\ cm$$
Therefore $$ON=27-10=17\ cm$$
From triangle $$ABC$$
$$AB=2ON=2\times 17=34\ cm$$
Given that $$CD=20\ cm, MN=27\ cm$$
From triangle $$ACD$$
$$OM=\dfrac{1}{2}CD=\dfrac{1}{2}\times 20=10\ cm$$
Therefore $$ON=27-10=17\ cm$$
From triangle $$ABC$$
$$AB=2ON=2\times 17=34\ cm$$
The figure given below, shows a trapezium $$ABCD. M$$ and $$N$$ are the mid-points of the non-parallel sides $$AD$$ and $$BC$$ respectively. Find:
$$DC$$, if $$MN=15\ cm$$ and $$AB=23\ cm$$
Let us draw a diagonal $$AC$$ as shown in the figure below,
Given that $$AB=23\ cm, MN=15\ cm$$
From triangle $$ABC$$
$$ON=\dfrac{1}{2}AB=\dfrac{1}{2}\times 23=11.5\ cm$$
Therefore $$OM=15-11.5=3.5\ cm$$
From triangle $$ACD$$
$$CD=2OM=2\times 3.5=7\ cm$$
Given that $$AB=23\ cm, MN=15\ cm$$
From triangle $$ABC$$
$$ON=\dfrac{1}{2}AB=\dfrac{1}{2}\times 23=11.5\ cm$$
Therefore $$OM=15-11.5=3.5\ cm$$
From triangle $$ACD$$
$$CD=2OM=2\times 3.5=7\ cm$$
The following figure shows a trapezium $$ABCD$$ in which $$AB\parallel DC. P$$ is the mid-point of $$AD$$ and $$PR\parallel AB$$. Prove that:
$$PR=\dfrac{1}{2}(AB+CD)$$
$$D, E$$ and $$F$$ are the mid-point of the sides $$AB, BC$$ and $$CA$$ respectively of triangle $$ABC, AE$$ meets $$DF$$ at $$O. P$$ and $$Q$$ are the mid-points of $$OB$$ and $$OC$$ respectively. Prove that $$DPQF$$ is a parallelogram.
The required figure is shown below
For triangle $$ABC$$ and $$OBC$$
$$2DE=BC$$ and $$2PQ=BC$$, therefore $$DE=PQ ......(1)$$
For triangle $$ABO$$ and $$ACO$$
$$2PD=AO$$ and $$2FQ=AO$$, therefore $$PD=FQ ......(2)$$
From $$(1), (2)$$ we get that $$PQFD$$ is a parallelogram.
Hence proved
For triangle $$ABC$$ and $$OBC$$
$$2DE=BC$$ and $$2PQ=BC$$, therefore $$DE=PQ ......(1)$$
For triangle $$ABO$$ and $$ACO$$
$$2PD=AO$$ and $$2FQ=AO$$, therefore $$PD=FQ ......(2)$$
From $$(1), (2)$$ we get that $$PQFD$$ is a parallelogram.
Hence proved
A parallelogram $$ABCD$$ has $$P$$ the mid-point of $$DC$$ and $$Q$$ a point of $$AC$$ such that $$CQ=\dfrac{1}{4}AC. PQ$$ produced meets $$BC$$ at $$R$$
Prove that :
$$R$$ is the mid-point of $$BC$$.
For help we draw the diagonal $$BD$$ as shown below
The diagonal $$AC$$ and $$BD$$ cuts at point $$X$$.
We know that the diagonal of a parallelogram intersects equally each other. Therefore
$$AX=CX$$ and $$BX=DX$$
Given,
$$CQ=\dfrac{1}{4}AC$$
$$CQ=\dfrac{1}{4}\times 2CX$$
$$CQ=\dfrac{1}{2}CX$$
Therefore $$Q$$ is the midpoint of $$CX$$.
For triangle $$CDX:\ PQ\parallel DX$$ or $$PR\parallel BD$$
Since for triangle $$CBX$$
$$Q$$ is the midpoint of $$CX$$ and $$QR\parallel BX$$. Therefore $$R$$ is the midpoint of $$BC$$.
The diagonal $$AC$$ and $$BD$$ cuts at point $$X$$.
We know that the diagonal of a parallelogram intersects equally each other. Therefore
$$AX=CX$$ and $$BX=DX$$
Given,
$$CQ=\dfrac{1}{4}AC$$
$$CQ=\dfrac{1}{4}\times 2CX$$
$$CQ=\dfrac{1}{2}CX$$
Therefore $$Q$$ is the midpoint of $$CX$$.
For triangle $$CDX:\ PQ\parallel DX$$ or $$PR\parallel BD$$
Since for triangle $$CBX$$
$$Q$$ is the midpoint of $$CX$$ and $$QR\parallel BX$$. Therefore $$R$$ is the midpoint of $$BC$$.
In trapezium $$ABCD, AB$$ in parallel to $$DC; P$$ and $$Q$$ the mid-points of $$AD$$ and $$BC$$ respectively. $$BP$$ produced meets $$CD$$ produced at point $$E$$. Prove that:
Point $$P$$ bisects $$BE$$
From $$\triangle PED$$ and $$\triangle ABP$$
$$PD=AP$$ [ $$P$$ is the midpoint of $$AD$$]
$$\angle DPB=\angle APB$$ [Opposite angle]
$$\angle PED=\angle PBA$$ $$[Ab\parallel CB]$$
$$\therefore \triangle PED\cong \triangle ABP$$ [$$ASA$$ postulate]
$$\therefore EP=BP$$
In the quadrilateral $$ABCD$$; prove that
$$AB+BC+CD + DA > 2BD$$
In $$\triangle ABD$$
$$AB+DA > BD....(v)$$ [Sum of two sides is greater than the third side]
In $$\triangle BCD$$
$$BC+CD > BD....(vi)$$ [Sum of two sides is greater than the third side]
Adding $$AB+DA+BC+CD > BD +BD$$
$$AB+DA+BC+CD > 2BD$$
$$AB+DA > BD....(v)$$ [Sum of two sides is greater than the third side]
In $$\triangle BCD$$
$$BC+CD > BD....(vi)$$ [Sum of two sides is greater than the third side]
Adding $$AB+DA+BC+CD > BD +BD$$
$$AB+DA+BC+CD > 2BD$$
A parallelogram $$ABCD$$ has $$P$$ the mid-point of $$DC$$ and $$Q$$ a point of $$AC$$ such that $$CQ=\dfrac{1}{4}AC. PQ$$ produced meets $$BC$$ at $$R$$
Prove that :
$$PR=\dfrac{1}{2}DB$$
For help we draw the diagonal $$BD$$ as shown below
The diagonal $$AC$$ and $$BD$$ cuts at point $$X$$.
We know that the diagonal of a parallelogram intersects equally each other. Therefore
$$AX=CX$$ and $$BX=DX$$
Given,
$$CQ=\dfrac{1}{4}AC$$
$$CQ=\dfrac{1}{4}\times 2CX$$
$$CQ=\dfrac{1}{2}CX$$
Therefore $$Q$$ is the midpoint of $$CX$$.
For triangle $$BCD$$
As $$P$$ and $$R$$ are the midpoint of $$CD$$ and $$BC$$, therefore $$PR=\dfrac{1}{2}DB$$
The diagonal $$AC$$ and $$BD$$ cuts at point $$X$$.
We know that the diagonal of a parallelogram intersects equally each other. Therefore
$$AX=CX$$ and $$BX=DX$$
Given,
$$CQ=\dfrac{1}{4}AC$$
$$CQ=\dfrac{1}{4}\times 2CX$$
$$CQ=\dfrac{1}{2}CX$$
Therefore $$Q$$ is the midpoint of $$CX$$.
For triangle $$BCD$$
As $$P$$ and $$R$$ are the midpoint of $$CD$$ and $$BC$$, therefore $$PR=\dfrac{1}{2}DB$$
In trapezium $$ABCD, AB$$ in parallel to $$DC; P$$ and $$Q$$ the mid-points of $$AD$$ and $$BC$$ respectively. $$BP$$ produced meets $$CD$$ produced at point $$E$$. Prove that:
$$PQ$$ is parallel to $$AB$$.
For triangle $$ECB:\ PQ\parallel CE$$
Again $$CE\parallel AB$$
Therefore $$PQ\parallel AB$$
Hence proved
Again $$CE\parallel AB$$
Therefore $$PQ\parallel AB$$
Hence proved
In a triangle $$ABC, P$$ is the mid-point of side $$BC$$. A line through $$P$$ and parallel to $$CA$$ meets $$AB$$ at point $$Q$$; and a line through $$Q$$ and parallel to $$BC$$ meets median $$AP$$ at point $$R$$. Prove that:
$$BC=4QR$$
The required figure is shown below
From the figure it is seen that $$P$$ is the midpoint of $$BC$$ and $$PQ\parallel AC$$ and $$QR\parallel BC$$
Therefore $$Q$$ is the midpoint of $$AB$$ and $$R$$ is the midpoint of $$AP$$.
$$\angle PQR-=angle ARS$$ (opposite angle)
$$PR=AR$$
$$PQ=AS\left[PQ=AS=\dfrac{1}{2}AC\right]$$
$$\triangle PQR\cong \triangle ARS$$ (SAS Postulate)
Therefore $$QR=RS$$
Now
$$BC=2QS$$
$$BC=2\times 2QR$$
$$BC=4QR$$
Hence proved
Here we increase $$QR$$ so that it cuts $$AC$$ at $$S$$ as shown in the figure.
From the figure it is seen that $$P$$ is the midpoint of $$BC$$ and $$PQ\parallel AC$$ and $$QR\parallel BC$$
Therefore $$Q$$ is the midpoint of $$AB$$ and $$R$$ is the midpoint of $$AP$$.
$$\angle PQR-=angle ARS$$ (opposite angle)
$$PR=AR$$
$$PQ=AS\left[PQ=AS=\dfrac{1}{2}AC\right]$$
$$\triangle PQR\cong \triangle ARS$$ (SAS Postulate)
Therefore $$QR=RS$$
Now
$$BC=2QS$$
$$BC=2\times 2QR$$
$$BC=4QR$$
Hence proved
Here we increase $$QR$$ so that it cuts $$AC$$ at $$S$$ as shown in the figure.
In a triangle $$ABC, P$$ is the mid-point of side $$BC$$. A line through $$P$$ and parallel to $$CA$$ meets $$AB$$ at point $$Q$$; and a line through $$Q$$ and parallel to $$BC$$ meets median $$AP$$ at point $$R$$. Prove that:
$$AP=2AR$$
The required figure is shown below
From the figure it is seen that $$P$$ is the midpoint of $$BC$$ and $$PQ\parallel AC$$ and $$QR\parallel BC$$
Therefore $$Q$$ is the midpoint of $$AB$$ and $$R$$ is the midpoint of $$AP$$.
Therefore $$AP=2AR$$
From the figure it is seen that $$P$$ is the midpoint of $$BC$$ and $$PQ\parallel AC$$ and $$QR\parallel BC$$
Therefore $$Q$$ is the midpoint of $$AB$$ and $$R$$ is the midpoint of $$AP$$.
Therefore $$AP=2AR$$
In the quadrilateral $$ABCD$$; prove that
$$AB+BC+CD + DA > 2AC$$
In $$\triangle ACD$$
$$AB+BC+CD+DA > AC+AC$$
$$AB+BC+CD+DA > 2AC$$
$$CD+DA > AC....(i)$$ [Sum of two sides is greater than the third side]
Similarly in $$\triangle ABC$$
$$AB+BC> AC....(i)$$ [Sum of two sides is greater than the third side]
Adding $$AB+BC+CD+DA > AC+AC$$
$$AB+BC+CD+DA > 2AC$$
In quadrilateral $$ABCD$$, side $$AB$$ is the longest and side $$DC$$ is the shortest. Prove that:
$$\angle D > \angle B$$
In the quad $$ABCD$$
Since $$AB$$ is the longest side and $$DC$$ is the shortest side:
(ii) $$\angle 5 > \angle 6 [AB > AD]$$
$$\angle 3 > \angle 8 [BC > CD]$$
$$\therefore \angle 5+\angle 3 > \angle 6+\angle 8$$
$$\Rightarrow \angle D > \angle D$$
Since $$AB$$ is the longest side and $$DC$$ is the shortest side:
(ii) $$\angle 5 > \angle 6 [AB > AD]$$
$$\angle 3 > \angle 8 [BC > CD]$$
$$\therefore \angle 5+\angle 3 > \angle 6+\angle 8$$
$$\Rightarrow \angle D > \angle D$$
$$D$$ and $$F$$ are mid-points of sides $$AB$$ and $$AC$$ of a triangle $$ABC$$. A line through $$F$$ and parallel to $$AB$$ meets $$BC$$ at point $$E$$.
Prove that $$BDFE$$ is a parallelogram.
Since $$F$$ is the midpoint and $$EF\parallel AB$$.
Therefore $$E$$ is the midpoint of $$BC$$
So $$BE=\dfrac{1}{2}BC$$ and $$BF=\dfrac{1}{2}AB$$ ......(1)
Since $$D$$ and $$F$$ are the midpoint of $$AB$$ and $$AC$$
Therefore $$DE\parallel BC$$.
So $$DF=\dfrac{1}{2}BC$$ and $$DB=\dfrac{1}{2}AB$$ ......(2)
From $$(1), (2)$$ we get
$$BE=DF$$ and $$BD=EF$$
Hence $$BDEF$$ is a parallelogram.
Therefore $$E$$ is the midpoint of $$BC$$
So $$BE=\dfrac{1}{2}BC$$ and $$BF=\dfrac{1}{2}AB$$ ......(1)
Since $$D$$ and $$F$$ are the midpoint of $$AB$$ and $$AC$$
Therefore $$DE\parallel BC$$.
So $$DF=\dfrac{1}{2}BC$$ and $$DB=\dfrac{1}{2}AB$$ ......(2)
From $$(1), (2)$$ we get
$$BE=DF$$ and $$BD=EF$$
Hence $$BDEF$$ is a parallelogram.
In triangle $$ABC$$, and $$B$$ is obtuse. $$D$$ and $$E$$ are mid-point of sides $$AB$$ and $$BC$$ respectively and $$F$$ is a point on side $$AC$$ such that $$EF$$ is parallel to $$AB$$. Show that $$BEFD$$ is a parallelogram.
In parallelogram $$ABCD, E$$ and $$F$$ are mid-points of the sides $$AB$$ and $$CD$$ respectively. The line segments $$AF$$ and $$BF$$ meet the line segments $$ED$$ and $$EC$$ at points $$G$$ and $$H$$ respectively.
Prove that:
$$GEHF$$ is a parallelogram.
$$D$$ and $$F$$ are mid-points of sides $$AB$$ and $$AC$$ of a triangle $$ABC$$. A line through $$F$$ and parallel to $$AB$$ meets $$BC$$ at point $$E$$.
Find $$AB$$, if $$EF=4.8\ cm$$.
Since $$F$$ is the midpoint and $$EF\parallel AB$$.
Therefore $$E$$ is the midpoint of $$BC$$
So $$BE=\dfrac{1}{2}BC$$ and $$BF=\dfrac{1}{2}AB$$ ......(1)
Since $$D$$ and $$F$$ are the midpoint of $$AB$$ and $$AC$$
Therefore $$DE\parallel BC$$.
So $$DF=\dfrac{1}{2}BC$$ and $$DB=\dfrac{1}{2}AB$$ ......(2)
From $$(1), (2)$$ we get
$$BD=EF$$ ...(3)
So, from (1) and (3)
Therefore $$E$$ is the midpoint of $$BC$$
So $$BE=\dfrac{1}{2}BC$$ and $$BF=\dfrac{1}{2}AB$$ ......(1)
Since $$D$$ and $$F$$ are the midpoint of $$AB$$ and $$AC$$
Therefore $$DE\parallel BC$$.
So $$DF=\dfrac{1}{2}BC$$ and $$DB=\dfrac{1}{2}AB$$ ......(2)
From $$(1), (2)$$ we get
$$BD=EF$$ ...(3)
So, from (1) and (3)
$$AB=2EF$$
Since
$$AB=2EF$$$$=2\times 4.8$$
$$=9.6\ cm$$
In a triangle $$ABC, AD$$ is a median and $$E$$ is mid-point of median $$AD$$. A line through $$B$$ and $$E$$ meets $$AC$$ at point $$F$$.
Prove that: $$AC=3AF$$
The required figure is shown below
For help we draw a line $$DG\parallel BF$$
Now from triangle $$ADG, DG\parallel BF$$ and $$E$$ is the midpoint of $$AD$$
Therefore $$F$$ is the midpoint of $$AG$$, ie. $$AF=GF$$ ......(1)
From triangle $$BCF, DG\parallel BF$$ and $$D$$ is the midpoint of $$BC$$
Therefore $$G$$ is the midpoint of $$CF$$, ie $$GF=CF$$ .....(2)
$$AC=AF+GF+CF$$
$$AC=3AF$$ (From $$(1)$$ and $$(2)$$)
Hence proved
For help we draw a line $$DG\parallel BF$$
Now from triangle $$ADG, DG\parallel BF$$ and $$E$$ is the midpoint of $$AD$$
Therefore $$F$$ is the midpoint of $$AG$$, ie. $$AF=GF$$ ......(1)
From triangle $$BCF, DG\parallel BF$$ and $$D$$ is the midpoint of $$BC$$
Therefore $$G$$ is the midpoint of $$CF$$, ie $$GF=CF$$ .....(2)
$$AC=AF+GF+CF$$
$$AC=3AF$$ (From $$(1)$$ and $$(2)$$)
Hence proved
In the figure, given below, $$2AD=AB, P$$ is mid-point of $$AB, Q$$ is mid-point of $$DR$$ and $$PR\parallel BS$$. Prove that:
$$AQ\parallel BS$$.
The side $$AC$$ of a triangle $$ABC$$ is produced to point $$E$$ so that $$CE=\dfrac{1}{2}AC.\ D$$ is the mid-point of $$BC$$ and $$ED$$ produced meets $$AB$$ at $$F$$. Lines through $$D$$ and $$C$$ are drawn parallel to $$AB$$ which meet $$AC$$ at point $$R$$ respectively. Prove that:
$$3DF=EF$$
The side $$AC$$ of a triangle $$ABC$$ is produced to point $$E$$ so that $$CE=\dfrac{1}{2}AC.\ D$$ is the mid-point of $$BC$$ and $$ED$$ produced meets $$AB$$ at $$F$$. Lines through $$D$$ and $$C$$ are drawn parallel to $$AB$$ which meet $$AC$$ at point $$R$$ respectively. Prove that:
$$4CR=AB$$
In parallelogram $$ABCD, E$$ and $$F$$ are mid-points of the sides $$AB$$ and $$CD$$ respectively. The line segments $$AF$$ and $$BF$$ meet the line segments $$ED$$ and $$EC$$ at points $$G$$ and $$H$$ respectively.
Prove that:
Triangles $$HEB$$ and $$FHC$$ are congruent.
In triangle $$ABC, M$$ is mid-point of $$AB, N$$ is mid-point of $$AC$$ and $$D$$ is any point in base $$BC$$. Use Intercept Theorem to show that $$MN$$ bisects $$AD$$.
The figure is shown below
Since $$M$$ and $$N$$ are the midpoint of $$AB$$ and $$AC, MN\parallel BC$$
According to intercept theorem Since $$MN\parallel BC$$ and $$AM=BM$$,
Therefore $$AX=DX$$. Hence proved
Since $$M$$ and $$N$$ are the midpoint of $$AB$$ and $$AC, MN\parallel BC$$
According to intercept theorem Since $$MN\parallel BC$$ and $$AM=BM$$,
Therefore $$AX=DX$$. Hence proved
Can the angles $$110^\circ, 80^\circ, 70^\circ and\ 95^\circ$$ be the angles of a quadrilateral? Why or why not?
If the quadrilateral formed by joining the mid-points of the adjacent sides of quadrilateral $$ABCD$$ is a rectangle, show that the diagonal $$AC$$ and $$BD$$ intersect at right angle.
The figure is shown below
Let $$ABCD$$ be quadrilateral where $$P, Q, R, S$$ are the midpoint of $$AB, BC, CD, DA, PQRS$$ is a rectangle. Diagonal $$AC$$ and $$BD$$ intersect at point $$O$$. We need to show that $$AC$$ and $$BD$$ intersect at right angle.
Proof:
$$PQ\parallel AC$$, therefore $$\angle AOD=\angle PXO$$ [Corresponding angle] ...... (1)
Again $$BD\parallel RQ$$.
therefore $$\angle PXO=\angle RQX=90^{o}$$ [Corresponding angle and angle of rectangle].... (2)
From $$(1)$$ and $$(2)$$ we get,
$$\angle AOD=90^{o}$$
Similarly $$\angle AOB=\angle BOC=\angle DOC=90^{o}$$
Therefore diagonals $$AC$$ and $$BD$$ intersect at right angle
Hence proved
Let $$ABCD$$ be quadrilateral where $$P, Q, R, S$$ are the midpoint of $$AB, BC, CD, DA, PQRS$$ is a rectangle. Diagonal $$AC$$ and $$BD$$ intersect at point $$O$$. We need to show that $$AC$$ and $$BD$$ intersect at right angle.
Proof:
$$PQ\parallel AC$$, therefore $$\angle AOD=\angle PXO$$ [Corresponding angle] ...... (1)
Again $$BD\parallel RQ$$.
therefore $$\angle PXO=\angle RQX=90^{o}$$ [Corresponding angle and angle of rectangle].... (2)
From $$(1)$$ and $$(2)$$ we get,
$$\angle AOD=90^{o}$$
Similarly $$\angle AOB=\angle BOC=\angle DOC=90^{o}$$
Therefore diagonals $$AC$$ and $$BD$$ intersect at right angle
Hence proved
In triangle $$ABC; D$$ and $$E$$ are mid-points of sides $$AB$$ and $$AC$$ respectively. Through $$E$$, a straight line is drawn parallel to $$AB$$ to meet $$BC$$ at $$F$$. Prove that $$BDEF$$ is a parallelogram
If $$AB=16\ cm, AC=12\ cm$$ and $$BC=18\ cm$$, find the perimeter of the parallelogram $$BDEF$$.
In quadrilateral $$ABCD , AD = BC$$ and $$BD = CA$$ prove that
$$ \angle ADB = \angle BCA $$
In parallelogram $$ABCD, E$$ is the mid-point of $$AB$$ and $$AP$$ is parallel to $$EC$$ which meets $$DC$$ at point $$O$$ and $$BC$$ produced at $$P$$. Prove that:
$$O$$ is mid-point of $$AP$$.
In quadrilateral ABCD , AD = BC and BD = CA prove that
$$ \angle DAB = \angle CBA $$
ABCD is a rectangle in which diagonal AC bisects A as well as C. Show that ABCD is a square.
Given ABCD is a rectangle
with $$\angle 1 = \angle 2$$ and $$\angle 3 = \angle 4$$
But $$\angle 1 =\angle 4$$ (alternate angles)
Therefore, we have $$\angle 2 = \angle 4,$$ which means $$AB = BC,$$ similarly $$AD = CD$$
Hence, ABCD is a square.
with $$\angle 1 = \angle 2$$ and $$\angle 3 = \angle 4$$
But $$\angle 1 =\angle 4$$ (alternate angles)
Therefore, we have $$\angle 2 = \angle 4,$$ which means $$AB = BC,$$ similarly $$AD = CD$$
Hence, ABCD is a square.
A quadrilateral ABCD will be a parallelogram for the following conditions.
$$\angle D + \angle B = 180^{0}$$
If true then enter $$1$$ and if false then enter $$0$$
$$ Let\quad us\quad consider\quad fig(i)\quad \& \quad fig(ii).\\ \angle B+\angle D={ 180 }^{ O }\quad (given)\Longrightarrow \angle A+\angle C={ 180 }^{ O }\\ It\quad does\quad not\quad mean\quad that\quad \angle B=\angle D\quad as\quad in\quad fig(i).\\ eveven\quad if\quad \angle B=\angle D\quad it\quad does\quad not\quad mean\quad that\quad \angle A=\angle C\\ as\quad in\quad fig(i).\\ For\quad a\quad quadrilateral\quad to\quad be\quad a\quad parallelogram\quad it\quad is\quad necessary\\ that\quad both\quad the\quad opposite\quad pairs\quad of\quad angles\quad are\quad equal\quad as\quad in\\ fig(ii).\\ Here\quad the\quad sum\quad of\quad the\quad opposite\quad angles\quad are\quad { 180 }^{ O }.\\ i.e\quad ABCD\quad to\quad be\quad a\quad parallelogram\quad it\quad is\quad necessary\quad that\\ each\quad angle\quad should\quad be={ 90 }^{ o }.\quad It\quad may\quad or\quad may\quad not\quad occur\quad in\quad \\ each\quad quadrilateral\quad in\quad general.\\ Ans-\quad Can\quad be\quad but\quad need\quad not\quad be. $$
$$ABCD$$ is a convex quadrilateral with
$$AB + BC \leq AC + CD$$. Prove that $$AB < AC$$.
Let us suppose that $$AB \ge AC.$$
Then $$A$$ lies on or to the left of the line $$L$$. ( the perpendicular bisector of $$BC$$ )
From the question; $$AB+BC \le AC+CD$$
$$\Rightarrow AB=CD $$ and $$AD=BC$$
[ Opposite sides are equal ]
Let us consider that the sides are of $$'a'\ cm$$. Then the diagonal length is $$a \sqrt {2}\ cm.$$
$$\Rightarrow AB< AC$$ i.e, $$a<a \sqrt {2}.$$
Hence, the answer is $$AB< AC.$$
If a quadrilateral ABCD is such that $$\displaystyle \vec{AB}= b, \vec{AD}= d$$ and $$\displaystyle \vec{AC}= pb+qd(p+q\geq 1),$$ then show that the area of the quadrilateral is $$\displaystyle \frac{1}{2}\left ( p+q \right )\left | b\times d \right |.$$
Area $$\displaystyle = \Delta _{1}+\Delta_{2}$$
$$\displaystyle = \frac{1}{2}\left | b\times \left ( pb+qd \right ) \right |+\frac{1}{2}\left | \left ( pb+qd \right )\times d \right |$$
$$\displaystyle = \frac{1}{2}\left | q\left ( b\times d \right ) \right |+\frac{1}{2}\left | p\left ( b\times d \right ) \right |$$
$$\displaystyle = \frac{1}{2}\left | \left ( p+q \right ) \right |\left | \left ( b\times d \right ) \right |$$
$$\displaystyle = \frac{1}{2}\left ( p+q \right )\left | b\times d \right |$$ ($$\because \displaystyle p+q \geq 1$$ so positive only).
$$\displaystyle = \frac{1}{2}\left | b\times \left ( pb+qd \right ) \right |+\frac{1}{2}\left | \left ( pb+qd \right )\times d \right |$$
$$\displaystyle = \frac{1}{2}\left | q\left ( b\times d \right ) \right |+\frac{1}{2}\left | p\left ( b\times d \right ) \right |$$
$$\displaystyle = \frac{1}{2}\left | \left ( p+q \right ) \right |\left | \left ( b\times d \right ) \right |$$
$$\displaystyle = \frac{1}{2}\left ( p+q \right )\left | b\times d \right |$$ ($$\because \displaystyle p+q \geq 1$$ so positive only).
If all angles of a quadrilateral are congruent, then find their measures.
Use the information given in the following figure to find :$$x$$ (in degrees).
The bisectors of angles $$B$$ and $$C$$ of a parallelogram $$ABCD$$ meet at point $$O.$$ If the triangle $$OBC$$ formed is an isosceles right triangle, then $$ABCD$$ is a rectangle.
If the above statement is true then mention the answer as $$1,$$ else mention $$0$$ if false.
Given:
$$ABCD$$ is a parallelogram. $$OB$$ and $$OC$$ bisect $$\angle B$$ and $$\angle C$$, $$OBC$$ is a right isosceles triangle.
To prove:
$$ABCD$$ is a rectangle.
Proof:
If we can prove that adjacent angles of parallelogram $$ABCD$$ are of $$90^\circ$$ then that will implies that $$ABCD$$ is a rectangle.
In $$\triangle OBC$$,
$$\angle BOC = 90^{\circ}$$
And,
$$OB =OC$$
By using the theorem that, angles opposite to equal sides are equal we can assume that,
$$\angle OBC = \angle OCB = x$$
Now, by applying angle sum property,
$$\angle OBC + \angle OCB + \angle BOC = 180^{\circ}$$
$$\Rightarrow x + x + 90^{\circ} = 180^{\circ}$$
$$\Rightarrow 2x = 90^{\circ}$$
$$\Rightarrow x = 45^{\circ}$$
Hence,
$$\angle OBC = \angle OCB = 45^{\circ}$$
But it is given that, $$OB$$ and $$OC$$ bisect $$\angle B$$ and $$\angle C$$ that implies $$\angle B = \angle C = 90^{\circ}$$
Since, adjacent angles are right angles. $$ABCD$$ is a rectangle.
State True or False: The figure obtained by joining the mid-points of the adjacent sides of a rectangle is a Trapezium
Construct a rhombus whose diagonals are $$47\ cm$$ and $$54\ cm$$.
A rhombus has diagonals perpendicular to each other.1. Draw $$ AC = 47 cm $$
2. Draw perpendicular bisector XY of AC intersecting AC at O.
3. Taking O as centre and $$ \frac {1}{2} \times 54 = 27 cm $$ as radius
draw two arcs intersecting OX and OY at B and D respectively.
4. Join AB, BC, CD and AD.
ABCD is the required rhombus.
In rectangle ABCD, the bisectors of angles B and C meet at point O then triangle OBC is an isosceles right triangles.
If the above statement is true then mention answer as 1, else mention 0 if false
In the fig. given measure of angle $$\angle PLN=110^o$$. Then the measure of $$\angle LMN$$ in degrees is:
$$ABCD$$ is a trapezium with $$AB\: // \:DC$$ . A line parallel to $$AC$$ intersects $$AB$$ at point $$M$$ and $$BC$$ at point $$N$$ . Prove that : $$area \: of \: \bigtriangleup ADM \: = \: area \: of \: \bigtriangleup \: ACN$$.
The diagonal BD of a parallelogram ABCD bisects angles B and D. Prove that ABCD is a rhombus.
Given parallelogram $$ABCD$$ in which diaginal $$BD$$ bisects $$\angle B$$ and $$\angle D$$
$$BD$$ bisect $$\angle B,$$
$$\Rightarrow \angle ABD=\angle CBD=\dfrac { 1 }{ 2 } \angle ABC$$
$$BD$$ bisect $$\angle D,$$
$$\Rightarrow \angle ADB=\angle CDB=\dfrac { 1 }{ 2 } \angle ADC$$
$$\therefore \angle ABC=\angle ADC$$ [ Opposite angles of parallelogram are equal ]
$$\therefore \dfrac { 1 }{ 2 } \angle ABC=\dfrac { 1 }{ 2 } \angle ADC$$
$$\Rightarrow \angle ABD=\angle ADB$$ and $$\angle CBD=\angle CDB$$
$$\therefore AD=AB$$ and $$CD=BC$$
As $$ABCD$$ is a parallelogram, opposite sides are equal
$$\Rightarrow AB=CD$$ and $$AD=BC$$
$$\Rightarrow AB=BC=CD=AD$$ i.e, $$ABCD$$ is a rhombus.
Hence, the answer is proved.
Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.
The angles of a quadrilateral $$ABCD$$ are in the ratio $$4 : 9 : 10: 13$$. Find the angles.
In the adjoining figure, $$ABCD$$ is a trapezium in which $$AB\;\parallel\;DC\;and\;AD=BC$$. If $$P,\,Q,\,R,\,S$$ be respectively the midpoints of $$BA,\,BD\;and\;CD,\;CA$$ then show that $$PQRS$$ is a rhombus.
In the given figure, find the values of $$x$$ and $$y$$.
$$\text{It is given that, ABCD is a cyclic quadrilateral.}$$
$$\Rightarrow\angle A+\angle C=180^\circ$$
$$\Rightarrow 2x+5^\circ+6x-15^\circ=180^\circ$$
$$\Rightarrow 8x=180^\circ+10^\circ$$
$$\Rightarrow 8x=190^\circ$$
$$\Rightarrow x=23.75^\circ$$
$$\textbf{Step -2: Find the value of y.}$$
$$\text{Again,}$$
$$\Rightarrow\text{}\angle B+\angle
D=180^\circ$$
$$\Rightarrow 3x+10^\circ+2y-30^\circ=180^\circ$$
$$\Rightarrow 3\times 23.75^\circ+2y=180^\circ+20^\circ$$
$$\Rightarrow 71.25^\circ+2y=200^\circ$$
$$\Rightarrow 2y=128.75^\circ$$
$$\Rightarrow y=64.375^\circ$$
$$\textbf{Hence, the values of }\mathbf{x\text{ and }y\text{ are }23.75^\circ\text{ and }64.375^\circ\text{ respectively.}}$$
Prove that all four angles of a quadrilateral cannot be obtuse angles.
Prove that all four angles of a quadrilateral cannot be acute angles
In Fig. 6.43, if $$PQ \perp PS, PQ\parallel SR,\angle SQR={ 28 }^{ 0}$$ and $$\angle QRT={ 65 }^{ 0}$$, then find the values of $$x$$ and $$y$$.
The following figures GUNS and RUNS are parallelograms. Find $$x$$ and $$y$$. (Lengths are in cm)
The above figure is a parallelogram. Find the angle measures $$x, y$$ and $$z$$. State the properties you use to find them.
The angles of quadrilateral are in the ratio $$3 : 5 : 9 : 13$$. Find all the angles of the quadrilateral.
Consider the following parallelograms. Find the values of the unknowns $$x, y, z$$
$${\textbf{Step - 1 :
State the properties}}{\text{.}}$$
$${\text{In a
parallelogram ABCD , The opposite angles are equal}}{\text{.}} \Rightarrow
\angle {\text{A = }}\angle {\text{C,}}\angle {\text{B = }}\angle
{\text{D}}{\text{.}}$$
$${\text{The consecutive angles of parallelogram is supplementary}}{\text{.}} \Rightarrow \angle {\text{A + }}\angle {\text{B = 180}}^\circ {\text{,}}\angle {\text{C + }}\angle {\text{D = 180}}^\circ {\text{.}}$$
$${\textbf{Step - 2 : Solve for (i)}}{\text{.}}$$
$${\text{Here }}100^\circ {\text{ + z}} = 180^\circ [\because {\text{Adjacent angles of a parallelogram]}}{\text{.}}$$
$$ \Rightarrow {\text{z = 80}}^\circ .$$
$${\text{Opposite angles are same}}{\text{.}} \Rightarrow {\text{x = z = 80}}^\circ {\text{,y = 100}}^\circ {\text{.}}$$
$${\textbf{Step - 3 : Solve for (ii)}}{\text{.}}$$
$${\text{Here x}}^\circ {\text{ + 50}}^circ {\text{ = 180}}^\circ {\text{. [Adjacent angle]}}$$
$$ \Rightarrow {\text{x}}^\circ {\text{ = 130}}^\circ {\text{.}}$$
$${\text{x and y are opposite angles}}{\text{.So x = y = 130}}^\circ {\text{.}}$$
$${\text{x and z will corresponding angles, z = 130}}^\circ {\text{. }}$$
$${\textbf{Step - 4 : Solve for (iii)}}{\text{.}}$$
$${\text{Here as x and 90}}^\circ {\text{ vertical opposite angles,So x = 90}}^\circ {\text{.}}$$
$${\text{As 90}}^\circ {\text{,30}}^\circ {\text{,y are in a triangle}}{\text{.}} \Rightarrow 90^\circ + 30^\circ + {\text{y = 180}}^\circ {\text{.}}$$
$$ \Rightarrow {\text{y = 60}}^\circ {\text{.}}$$
$$y{\text{ and z are alternate angles }}{\text{.}}$$
$$ \Rightarrow {\text{y = z = 60}}^\circ {\text{.}}$$
$${\textbf{Step - 5 : Solve for (iv)}}{\text{.}}$$
$${\text{Here z and 80}}^\circ {\text{ are corresponding angles}}{\text{.So z = 80}}^\circ {\text{.}}$$
$$y = 80^\circ {\text{ also,because they are opposite angles}}{\text{.}}$$
$$80^\circ {\text{ and x are adjacent angles , so x + 80}}^\circ {\text{ = 180}}^\circ {\text{.}}$$
$$ \Rightarrow {\text{x}} = 100^\circ .$$
$${\textbf{Step - 6 : Solve for (v),}}$$
$${\text{Here y = 112}}^\circ {\text{.As they are opposite angles}}{\text{.}}$$
$${\text{Now in one triangle there are three angles, x,y and 40}}^\circ {\text{.}} \Rightarrow {\text{x + 112}}^\circ {\text{ + 40}}^\circ {\text{ = 180}}^\circ $$
$${\text{x = 180}}^\circ - 40^\circ - 112^\circ .$$
$$ \Rightarrow {\text{x = 28}}^\circ .$$
$${\text{Now x and z are alternate angles }}{\text{. So x = z = 28}}^\circ {\text{.}}$$
$${\textbf{Hence, the values will be - (i) x = z = 80}}^\circ {\textbf{, y = 100}}^\circ (ii){\textbf{ x = y = z = 130}}^\circ {\textbf{(iii) x = 90}}^\circ {\textbf{, y = z = 60}}^\circ {\text{,}}$$
$${\textbf{(iv) x = 100}}^\circ {\textbf{, y = z = 80}}^\circ {\textbf{,(v) x = z = 28}}^\circ {\textbf{, y = 112}}^\circ .$$
Given a parallelogram $$ABCD$$. Complete each statement along with the definition or property used.
(i) $$AD =$$(ii) $$\angle DCB =$$(iii) $$OC =$$(iv) $$m \angle DAB + m \angle CDA =$$
Can a quadrilateral $$ABCD$$ be a parallelogram if
(i) $$\angle D + \angle B = 180^{\circ}$$? (ii) $$AB = DC = 8$$ cm, $$AD = 4$$ cm and $$BC = 4.4$$ cm? (iii) $$\angle A = 70^{\circ}$$ and $$\angle C = 65^{\circ}$$?
In the above figure both RISK and CLUE are parallelograms. Find the value of $$x$$.
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other
In $$\triangle ABC$$, $$P$$ and $$Q$$ are mid- points of $$AB$$ and $$BC$$ respectively.
Thus, a pair of opposite sides of a quadrilateral $$PQRS$$ are parallel and equal.
Therefore, quadrilateral $$PQRS$$ is a parallelogram.
Since the diagonals of a parallelogram bisect each other.
$$OP = OR$$ and $$OQ = OS$$
Therefore, diagonals $$PR$$ and $$QS$$ of a parallelogram $$PQRS$$
Using mid point theorem
$$PQ \parallel AC$$ and $$PQ=\dfrac{1}{2}AC$$ .....(1)
Similarly,
$$RS \parallel AC$$ and $$RS=\dfrac{1}{2} AC$$ ....(2)
$$PQ \parallel AC$$ and $$PQ=\dfrac{1}{2}AC$$ .....(1)
Similarly,
$$RS \parallel AC$$ and $$RS=\dfrac{1}{2} AC$$ ....(2)
From (1) and (2)
Therefore, $$PQ \parallel SR$$ and $$PQ = SR$$
Therefore, $$PQ \parallel SR$$ and $$PQ = SR$$
Thus, a pair of opposite sides of a quadrilateral $$PQRS$$ are parallel and equal.
Therefore, quadrilateral $$PQRS$$ is a parallelogram.
Since the diagonals of a parallelogram bisect each other.
$$OP = OR$$ and $$OQ = OS$$
Therefore, diagonals $$PR$$ and $$QS$$ of a parallelogram $$PQRS$$
i.e., the line segments joining the mid- points of opposite sides of quadrilateral $$ABCD$$ bisect each other.
In parallelogram $$ABCD$$, two point $$P$$ and $$Q$$ are taken on diagonal $$BD$$ such that $$DP = BQ$$. Show that
(i) $$\triangle APD\cong \triangle CQB$$
(ii) $$AP = CQ$$
(iii) $$\triangle AQB \cong \triangle CPD$$
(iv) $$AQ = CP$$
(v) $$APCQ$$ is a parallelogram
Construction: Join $$AC$$ to meet $$BD$$ in $$O$$.
Therefore, $$OB = OD$$ and $$OA=OC$$ ...(1)
(Diagonals of a parallelogram bisect each other)
But $$ BQ = DP$$ ...given
$$\therefore OB – BQ = OD – DP$$
$$\therefore OQ = OP$$ ....(2)
Now, in $$\Box APCQ$$,
$$OA = OC$$ ....from (1)
$$OQ = OP$$ ....from (2)
$$\therefore \Box APCQ$$ is a parallelogram.
In $$\triangle APD$$ and $$\triangle CQB$$,
$$AD = CB $$ ....opposite sides of a parallelogram
$$AP = CQ$$ ....opposite sides of a parallelogram
$$DP = BQ$$ ...given
$$\triangle APD \cong \triangle CQB$$ ...By SSS test of congruence
$$\therefore AP = CQ$$ ...c.s.c.t.
$$ AQ = CP$$ ...c.s.c.t. ...(3)
In $$\triangle AQB$$ and $$\triangle CPD$$,
$$AB = CD$$ ....opposite sides of a parallelogram
$$AQ = CP $$ ...from (3)
$$BQ = DP$$ ...given
$$\therefore \triangle AQB \cong \triangle CPD$$ ....By SSS test of congruence
$$ABC$$ is a triangle right angled at $$C$$. A line through the mid-point $$M$$ of hypotenuse $$AB$$ and parallel to $$BC$$ intersects $$AC$$ and $$D$$. Show that
(i) $$D$$ is the mid-point of $$AC$$
(ii) $$MB\perp AC$$
(iii) $$CM = MA = \dfrac {1}{2}AB$$
(i) In $$\triangle ABC$$,
$$M$$ is the mid point of $$AB$$ and $$MD \parallel BC$$.
Therefore, $$D$$ is the mid- point of $$AC$$.
i.e., $$AD = DC $$ ... (1)
(ii) Since $$MD \parallel BC$$, corresponding angles has to be equal
$$\angle ADM= \angle ACB$$
$$\angle ADM =90^{\circ}$$
As $$ \angle ACB = 90^{\circ}$$ is given
Since $$\angle ADM$$ and $$ \angle CDM$$ are angles of a linear pair
But $$\angle ADM + \angle CDM = 180^{\circ}$$
Therfore $$ 90^{\circ} + \angle CDM = 180^{\circ} \Rightarrow \angle CDM = 90^{\circ}$$
Thus, $$\angle ADM= \angle CMD=90^o$$ ... (2)
$$\Rightarrow MD \bot AC$$
$$AD = CD$$ (from(1))
$$\angle ADM= \angle CDM$$
and $$MD = MD$$ (Common)
Using SAS criterion of congruence
$$\triangle AMD \cong \triangle CMD$$
$$\Rightarrow MA=MC$$ (Since corresponding parts of congruent triangles are equal)
Also,
Also,
$$MA= \dfrac{1}{2} AB$$, since $$M$$ is the mid-point of $$AB$$.
Hence, $$CM=MA=\dfrac{1}{2} AB$$
$$ABCD$$ is a trapezium in which $$AB \parallel DC, BD$$ is a diagonal and $$E$$ is the mid-point of $$AD$$. A line is drawn through $$E$$ parallel to $$AB$$, intersecting $$BC$$ at $$F$$ (see Fig). Show that $$F$$ is the mid-point of $$BC$$.
In $$\triangle DAB$$,
Let line $$EF$$ intersect side $$DB$$ at $$G$$, such that $$E-G-F$$ is a line.
$$E$$ is the midpoint of side $$AD$$ and $$EG \parallel AB$$.
By converse of midpoint theorem,
$$G$$ is the midpoint of side $$BD$$ .....(1)
$$AB \parallel DC$$ ....given
$$AB \parallel EF$$ ....given
$$\therefore AB \parallel DC \parallel EF$$ ....(2)
In $$\triangle BDC$$,
$$G$$ is the midpoint of side $$BD$$ ...from (1)
and $$DC \parallel GF$$ ....from (2)
By converse of midpoint theorem,
$$F$$ is the midpoint of side $$BC$$
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
$$O$$ is the mid point of $$AC$$ and $$BD$$. (diagonals of bisect each other)
In $$\triangle ABC, BO$$ is the median.
$$∴ ar(AOB) = ar(BOC) $$--- (i)
In $$\triangle ABC, BO$$ is the median.
$$∴ ar(AOB) = ar(BOC) $$--- (i)
Also,
In $$\triangle BCD, CO$$ is the median.
$$∴ ar(BOC) = ar(COD)$$ --- (ii)
In $$\triangle BCD, CO$$ is the median.
$$∴ ar(BOC) = ar(COD)$$ --- (ii)
In $$\triangle ACD, OD$$ is the median.
$$∴ ar(AOD) = ar(COD) $$--- (iii)
$$∴ ar(AOD) = ar(COD) $$--- (iii)
In $$\triangle ABD, AO$$ is the median.
$$∴ ar(AOD) = ar(AOB) $$--- (iv)
$$∴ ar(AOD) = ar(AOB) $$--- (iv)
From equations (i), (ii), (iii) and (iv),
$$ar(BOC) = ar(COD) = ar(AOD) = ar(AOB)$$
$$ar(BOC) = ar(COD) = ar(AOD) = ar(AOB)$$
So, the diagonals of a parallelogram divide it into four triangles of equal area.
$$ABCD$$ is a parallelogram and $$AP$$ and $$CQ$$ are perpendiculars from vertices $$A$$ and $$C$$ on diagonal $$BD$$ Show that
i) $$\triangle\ APB \cong \triangle\ CQD$$
ii) $$AP = CQ$$
(i) Since $$ABCD$$ is a parallelogram. Therefore, $$DC \parallel AB$$.
Now $$DC \parallel AB$$ and transversal $$BD$$ intersects them at $$B$$ and $$D$$.
Therefore, $$\angle ABD = \angle BDC$$ $$\mid$$ Alternate interior angles
in $$\triangle APB$$ and $$\triangle CQD$$, we have
$$\angle ABP = \angle QDC$$ ....(alternate interior angles of parallelogram $$ABCD$$ and $$DC \parallel AB$$)
$$\angle APB = \angle CQD $$ ....[each angle $$90^\circ$$]
and $$AB = CD$$ $$\mid$$ Opposite. sides of a || gm
Therefore, by AAS criterion of congruence
$$\triangle APB \cong \triangle CQD$$
(ii) Since $$\triangle APB \cong \triangle CQD$$
Therefore, $$AP = CQ $$ $$\mid$$ Since corresponding parts of congruent triangles are equal
In a parallelogram $$ABCD,\ E$$ and $$F$$ are the mid-points of sides $$AB$$ and $$CD $$ respectively (see Fig). Show that the line segments $$AF$$ and $$EC$$ trisect the diagonal $$BD$$.
If $$E, F, G$$ and $$H$$ are respectively the mid-points of the sides of a parallelogram $$ABCD$$, show that $$ar(EFGH) = \dfrac {1}{2} ar (ABCD)$$
Given:
$$E,F,G$$ and $$H$$ are respectively the mid-points of the sides of a parallelogram $$ABCD$$.
$$E,F,G$$ and $$H$$ are respectively the mid-points of the sides of a parallelogram $$ABCD$$.
To Prove:
$$ar (EFGH) = \dfrac {1}{2} ar(ABCD)$$
$$ar (EFGH) = \dfrac {1}{2} ar(ABCD)$$
Construction:
$$H$$ and $$F$$ are joined.
$$H$$ and $$F$$ are joined.
Proof:
$$AD \parallel BC$$ and $$AD = BC$$ (Opposite sides of a parallelogram)
$$\Rightarrow \dfrac {1}{2} AD = \dfrac {1}{2} BC$$
$$AD \parallel BC$$ and $$AD = BC$$ (Opposite sides of a parallelogram)
$$\Rightarrow \dfrac {1}{2} AD = \dfrac {1}{2} BC$$
Also,
$$AH \parallel BF$$ and and $$DH \parallel CF$$
$$AH \parallel BF$$ and and $$DH \parallel CF$$
$$\Rightarrow AH=BF$$ and $$DH=CF$$ $$\mid$$ $$H$$ and $$F$$ are mid points
Thus,
Thus,
$$ABFH$$ and $$HFCD$$ are parallelograms.
Now,
$$\triangle EFH$$ and ||gm $$ABFH$$ lie on the same base $$FH$$ and between the same parallel lines $$AB$$ and $$HF$$.
$$\therefore $$ Area of $$EFH =\dfrac{ 1}{2}ar(ABFH)$$ --- (i)
Also,
$$\therefore $$ Area of $$EFH =\dfrac{ 1}{2}ar(ABFH)$$ --- (i)
Also,
Area of $$GHF=\dfrac {1}{2}ar(HFCD)$$ --- (ii)
Adding (i) and (ii),
Area of $$\triangle EFH+$$ area of $$\triangle GHF$$ $$= \dfrac {1}{2}ar(ABFH) + \dfrac {1}{2}ar(HFCD)$$
$$\Rightarrow $$ Area of $$EFGH =$$ Area of $$ABFH$$
$$\Rightarrow ar(EFGH)= \dfrac {1}{2} ar (ABCD)$$
Area of $$\triangle EFH+$$ area of $$\triangle GHF$$ $$= \dfrac {1}{2}ar(ABFH) + \dfrac {1}{2}ar(HFCD)$$
$$\Rightarrow $$ Area of $$EFGH =$$ Area of $$ABFH$$
$$\Rightarrow ar(EFGH)= \dfrac {1}{2} ar (ABCD)$$
In the given Figure, $$ABCD$$ is a parallelogram, $$AE\perp DC$$ and $$CF \perp AD$$. If $$AB = 16\ cm, AE = 8\ cm$$ and $$CF = 10\ cm$$, find $$AD$$.
Prove that the rhombus with equal diagonals is a square.
$$ABCD$$ is a rhombus, in which diagonals AC and BD are equal.
We know that diagonals of rhombus bisect each other.
As $$AC=BD$$
$$\therefore AO=BO=CO=DO$$
In $$\triangle AOB,$$
$$\Rightarrow AO=OB$$ and $$\angle AOD=90°$$
$$\therefore \angle OAB=\angle OBA=\dfrac{90^o}{2}=45°$$
Similarly in $$\triangle AOD,$$
$$\Rightarrow \angle OAD=\angle ODA=45°$$
$$\therefore \angle A=\angle OAB+\angle OAD=45^o+45^o=90°$$
Similarly,
$$\Rightarrow \angle B=\angle C=\angle D=90°$$
here, $$AB=BC=CD=DA$$
$$\therefore$$ Quadrilateral ABCD is a square.
In the figure, suppose $$\angle P$$ and $$\angle Q$$ are supplementary angles and $$\angle R = 125^{\circ}$$. Find the measures of $$\angle S$$.
All rectangles are parallelograms, but all parallelograms are not rectangles. Justify this statements
Is parallelogram a rectangle? Can you call a rectangle a parallelogram?
Two angles of a quadrilateral are $$70^{\circ}$$ and $$130^{\circ}$$ and the other two angles are equal. Find the measure of these two angles.
Complete the following table by writing (YES) if the property holds for the particular Quadrilateral and (NO) if property does not holds.
| Properties | Trapezium | Parallelogram | Rhombus | Rectangle | Square |
| a. Consecutive angles are supplementary |
Complete the following table by writing (YES) if the property holds for the particular Quadrilateral and (NO) if the property does not hold.
| Properties | Trapezium | Parallelogram | Rhombus | Rectangle | Square |
| Diagonals are equal |
Suppose in a rectangle, the diagonals are perpendicular to each other. Prove that it is a square.
Consider two $$\triangle AED$$ $$\xi$$ $$\triangle AEB$$
Now, $$AE=AE$$ [ Because they are the same line ]
$$\Rightarrow ED=EB$$ [ Diagonals of rectangle bisect each other ]
and $$\angle AED= \angle AEB.$$ (Given)
$$\therefore \triangle AED\cong \triangle AEB$$ [ by SAS ]
Then we have $$AD=AB$$........(1)
Again in this rectangle $$AD=BC$$.........(2) and $$CD=AB$$......(3).
From (1) , (2) and (3) we have $$AB=BC=CD=DA$$ with the angles right-angle.
So, ABCD is a square.
Complete the following table by writing (YES) if the property holds for the particular Quadrilateral and (NO) if property does not holds.
| Properties | Trapezium | Parallelogram | Rhombus | Rectangle | Square |
| a. Opposite angles are equal |
Complete the following table by writing (YES) if the property holds for the particular Quadrilateral and (NO) if property does not holds.
| Properties | Trapezium | Parallelogram | Rhombus | Rectangle | Square |
| a. All sides are equal |
Complete the following table by writing (YES) if the property holds for the particular Quadrilateral and (NO) if property does not holds.
| Properties | Trapezium | Parallelogram | Rhombus | Rectangle | Square |
| a. One pair of opposite sides are parallel |
Suppose in a quadrilateral, the diagonals bisect the angles at the vertices. Prove that it is a rhombus.
Let the quadrilateral be ABCD with diagonals AC and BD intersecting at P.
Since they bisect each other and are perpendicular,
$$\triangle APB,\triangle BPC,\triangle CPD$$ and $$\triangle DPA$$ are right triangles.
They all are congruent by SAS.
$$\triangle APB$$ is congruent to the $$\triangle CPB$$ because they share common side BD,
Side AP and CP are congruent ( since P is the mid point of AC ) and the included angles are both right angles since they are all congruent, their third side ( the hypotenuse of each ) are congruent ( CPCTC ).
A rhombus is a quadrilateral in which all four sides are congruent, so ABCD is a rhombus.
Hence, the answer is proved.
Complete the following table by writing (YES) if the property holds for the particular Quadrilateral and (NO) if property does not holds.
| Properties | Trapezium | Parallelogram | Rhombus | Rectangle | Square |
| a. Two pairs of opposite sides are parallel |
The midpoint of the side AB of a triangle ABC is D and P is any point on BC. Suppose Q is a point on AC such that ADPQ is a parallelogram. Prove that DQ is parallel to BC.
Given that, in $$\triangle ABC,\ D$$ the midpoint of $$AB, \ P \ and \ Q$$ are points on $$BC \ and \ AC $$ respectively such that, $$ADPQ$$ is a parallelogram.
To prove: $$DQ \parallel BC$$
Construction: Join $$DQ.$$
Proof:
Given that $$ADPQ$$ is a parallelogram
$$AD=PQ$$ [Opposite sides of a parallelogram]
$$AD=DB$$ [$$D$$ is the mid-point of $$AB$$]
$$\Rightarrow DB=PQ$$ $$[\because \ AD=PQ]$$
$$\Rightarrow DB \parallel PQ$$ $$[\because \ PQ \parallel AD]$$
$$\angle BDP=\angle DPQ$$ [Alternate angles]
$$DP=DP$$ [Common side of $$\triangle BDP$$ and $$\triangle PDQ$$]
$$\therefore \ \triangle BDP \cong \triangle QPD$$ [By SAS rule]
$$\angle BPD=\angle QDP$$ [Corresponding angles of congruent triangles]
$$\therefore \ BP \parallel DQ$$ [Since alternate angles are equal]
Hence, $$DQ \parallel BC$$.
Complete the following table by writing (YES) if the property holds for the particular Quadrilateral and (NO) if property does not holds.
| Properties | Trapezium | Parallelogram | Rhombus | Rectangle | Square |
| a. Diagonals bisect each other |
$$ABCD$$ is quadrilateral in which $$AB\parallel CD$$. If $$AD = BC$$, show that $$\angle A = \angle B$$ and $$\angle C = \angle D$$
Here, AB$$\parallel$$CD [Given]
$$\Rightarrow$$ and AD$$\parallel$$EC [By construction]
$$\therefore$$ AECD is a parallelogram.
$$\Rightarrow$$ AD = EC [Opposite sides of parallelogram are equal]
$$\Rightarrow$$ But AD = EC [Given]
$$\therefore$$ EC = BC
$$\therefore$$ $$\angle$$CBE = $$\angle$$CEB ---- ( 1 )
$$\Rightarrow$$ $$\angle$$B + $$\angle$$CBE = 180$$^\circ$$ [Linear pair] ---- ( 2 )
$$\Rightarrow$$ AD$$\parallel$$EC [By construction]
$$\Rightarrow$$ and transeversal AE intersects them
$$\therefore$$ $$\angle$$A + $$\angle$$CEB = 180$$^\circ$$ ---- ( 3 )
[Sum of adjacent angles of parallelogram is supplementary ]
$$\Rightarrow$$ $$\angle$$B + $$\angle$$CEB = 180$$^\circ$$ [From ( 2 ) and ( 3 )]
$$\Rightarrow$$ But $$\angle$$CBE = $$\angle$$CEB [From ( 1 )]
$$\therefore$$ $$\angle A=\angle B$$ [Proved] -- ( 4 )
$$\Rightarrow$$ $$\because$$ AB$$\parallel$$CD
$$\Rightarrow$$ $$\angle$$A + $$\angle$$D = 180$$^\circ$$ [Sum
Supplementary angles of parallelogram is 180$$^\circ$$]
$$\Rightarrow$$ and $$\angle$$B + $$\angle$$C = 180$$^\circ$$
$$\therefore$$ $$\angle$$A + $$\angle$$D = $$\angle$$B + $$\angle$$C
$$\Rightarrow$$ But $$\angle$$A = $$\angle$$B [From ( 4 )]
$$\therefore$$ $$\angle C=\angle D$$
Three vertices of a parallelogram $$\text{ABCD}$$ taken in order are $$\text{A}\left( 3,6 \right)$$, $$\text{B}\left( 5,10 \right)$$ and $$\text{C}\left( 3,2 \right) $$ find :
(i) the coordinates of the fourth vertex $$\text{D}$$.
(ii) length of diagonal $$\text{BD}$$.
Let the coordinate of $$D$$ be $$\left( x,y \right) $$. In a parallelogram, mid point of diagonal $$AC$$ coincides with the mid-point of diagonal $$BD$$.
Mid point of $$AC=\left( \dfrac { 3+3 }{ 2 } ,\dfrac { 6+2 }{ 2 } \right) =\left( 3,4 \right) $$
Mid point of $$AC=\left( \dfrac { 3+3 }{ 2 } ,\dfrac { 6+2 }{ 2 } \right) =\left( 3,4 \right) $$
Mid point of $$BD=\left( \dfrac { x+5 }{ 2 } ,\dfrac { y+10 }{ 2 } \right) $$
Equating,
(i) $$3=\dfrac { x+5 }{ 2 } $$ and $$4=\dfrac { y+10 }{ 2 } $$
$$\implies x=1$$ and $$y=-2$$
Coordinates of $$D\left( 1,-2 \right) $$
Equating,
(i) $$3=\dfrac { x+5 }{ 2 } $$ and $$4=\dfrac { y+10 }{ 2 } $$
$$\implies x=1$$ and $$y=-2$$
Coordinates of $$D\left( 1,-2 \right) $$
(ii) $$BD=\sqrt { { \left( 5-1 \right) }^{ 2 }+{ \left( 10+2 \right) }^{ 2 } }$$
$$=\sqrt { 16+144 } $$
$$=\sqrt { 160 } $$
$$=4\sqrt { 10 } $$ units.
$$=\sqrt { 16+144 } $$
$$=\sqrt { 160 } $$
$$=4\sqrt { 10 } $$ units.
Show that the line segments joining the midpoints of the opposite sides of a quadrilateral and bisect each other
Given $$ABCD$$ is quadrilateral , $$S$$ and $$R$$ are the midpoints of $$AD$$ and $$DC$$ respectively.
That the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and half of it.
Hence $$SR\parallel AC$$ $$SR=\dfrac{1}{2}AC$$........ (1)
Similarly, in $$\Delta ABC$$, $$P$$ and $$Q$$ are midpoints of $$AB$$ and $$BC$$ respectively.
$$PQ\parallel AC$$ $$PQ=\dfrac{1}{2}AC$$....... (2) [By midpoint theorem]
From equations (1) and (2), we get
$$PQ\parallel SR$$ $$SR=PQ$$................(3)
Clearly, one pair of opposite sides of quadrilateral $$PQRS$$ is equal and parallel.
Hence $$PQRS$$ is a parallelogram
we know that the diagonals of parallelogram $$PQRS$$ bisect each other.
Thus $$PR$$ and $$QS$$ bisect each other.
$$ABCD$$ is a parallelogram $$AP$$ and $$CQ$$ are perpendicular drawn from vertices $$A$$ and $$C$$ on diagonal $$BD$$ (see figure) show that
$$AP = CQ$$
Given ABCD is a parallelogram AP and CQ are perpendicular drawn from verities A and C on BD
Then area of triangle $$BDC$$ =$$\dfrac{1}{2}\times BD\times CQ$$
And area of triangle $$ABD$$ =$$\dfrac{1}{2}\times BD\times AP$$
We know that diagonal BD divided ABCD is a parallelogram in two equal area
$$\dfrac{1}{2}\times BD\times CQ= \dfrac{1}{2}\times BD\times AP$$
$$\Rightarrow CQ=AP$$
The four angles of a quadrilateral are in the ratio $$1 : 2 : 3 : 4$$. Find the measure of each angle of the quadrilateral.
Complete the following table by writing (YES) if the property holds for the particular Quadrilateral and (NO) if property does not holds.
| Quadrilateral name | Trapezium | Parallelogram | Rhombus | Rectangle | Square |
| Each angle is a right angle |
Complete the following table by writing (YES) if the property holds for the particular Quadrilateral and (NO) if property does not holds.
| Properties | Trapezium | Parallelogram | Rhombus | Rectangle | Square |
| Diagonals are perpendicular to each other |
The opposite angles of a parallelogram are $$(3x - 2)^{\circ}$$ and $$(x + 48)^{\circ}$$. Find the measure of each angle of the parallelogram.
$$ABCD$$ is a rectangle $$AC$$ is diagonal. Find the angles of $$\triangle ACD$$. Give reasons
Given, $$ABCD$$ is a rectangle $$AC$$ is diagonal
Then $$AD=BC$$ and $$AB=CD$$
And $$ \angle ABC=\angle BCA=\angle CDA=\angle DAB={90}^{0} $$
In $$\Delta ABC$$ and $$\Delta ADC$$
$$\angle ABC=\angle ADC$$ (Angle of rectangle)
$$AB=DC$$ (Opposite side of rectangle )
$$AC=AC$$ (Common side)
$$\therefore \Delta ABC\cong \Delta ADC$$
$$\therefore \angle ACD=\angle ACB$$
$$\because \angle ACD+\angle ACB=90^{0}$$ ..........[Angle ACB is the angle of rectangle]
$$\therefore 2\angle ACD=90^{0}$$
$$\Rightarrow \angle ACD=45^{0}$$
Prove that a line drawn through the mid-point of one side of a triangle parallel to second side bisects the third side.
Given : In $$\triangle ABC$$ ,$$D$$ is the mid point of $$AB$$ and $$DE$$ is drawn parallel to $$BC$$
To prove $$AE=EC$$ :
Draw $$CF$$ parallel to $$BA$$ to meet $$DE$$ produced to $$F$$
$$DE||BC$$ (given)
$$CF||BA$$ (by construction)
Now $$BCFD$$ is a parallelogram
$$\therefore BD=CF$$
$$BD=AD$$ (as $$D$$ is the mid point of $$AB$$)
$$\Rightarrow AD=CF$$
In $$\triangle ADE$$ and $$\triangle CFE$$
$$AD=CF$$
$$\angle ADE= \angle CFE$$ (alternate angles)
$$\angle ADE=\angle CEF$$ (vertically opposite angle)
$$\therefore \triangle ADE\cong \triangle CFE$$ (by AAS criterion)
$$\Rightarrow AE=EC$$ (Corresponding sides of congruent triangles are equal.)
So $$E$$ is the mid point of $$AC$$
Hence proved.
In the given figure, $$AED$$ is an isosceles triangle with $$AE = AD. ABCD$$ is a parallelogram and $$EGF$$ is a line segment. If $$\angle DCF = 65^{\circ}$$ and $$\angle EFB = 100^{\circ}$$, then the number of diagonals of a regular polygon having its each exterior angle equal to the measure of $$\angle AEG$$, is
If the measures of three angles of a quadrilateral are $$100^{\circ}, 84^{\circ}$$ and $$76^{\circ}$$ then, find the measure of fourth angle.
In $$\Delta ABC$$, $$AD$$, is a median and $$E$$ is the mid point of $$AD$$. If $$BE$$ is produced to meet $$AC$$ in $$F$$, show that $$AF=\dfrac { 1 }{ 3 } AC$$.
$$AD$$ is median $$\Rightarrow BD=CD$$
$$E$$ is mid point of $$AD$$
Draw a line $$DG$$, such that $$DG||BF$$
In $$\triangle ADG$$
$$EF||DG$$ and $$E$$ is mid point of $$AD$$
$$\therefore F$$ is midpoint of $$AG$$ [converse of midpoint theorem]
$$\Rightarrow AF=FG ---1$$
In $$\triangle BCF$$
$$D$$ is midpoint of $$BC$$ and $$DG||BF$$
$$\therefore FG=GC ---2$$
From $$1$$ and $$2$$
$$\therefore AF=FG=GC$$
$$AC=AF+FG+GC$$
$$\Rightarrow \cfrac { 1 }{ 3 } AC=AF$$
Find the four angles of a cyclic quadrilateral ABCD in which $$\angle A=(2x-1)^o$$, $$\angle B=(y+5)^o \angle C=(2y+15)^o$$ and $$\angle D =(4x-7)^o$$.
Find the angles of a rhombus whose diagonal is equal to a side.
The perimeter of a rhombus is equal to $$48$$, and the sum of the lengths of the diagonals is equal to $$26$$. Find the area of the rhombus.
In trapezium ABCD, side AB $$\parallel$$ side DC, diagonals AC and BD intersect in point O. If AB=$$20$$, DC=$$6$$, OB=$$15$$, then find OD.
If the area of a rectangle is $$200$$ sq.m and the difference of its length and breadth is $$10$$ cm, find its length, breadth and perimeter.
In the adjoining figure, $$ABCD$$ is a square. A line segment $$CX$$ cuts $$AB$$ at $$X$$ and the diagonal $$BD$$ at $$O$$ such that $$\angle$$ $$COD = $$ $$80^0$$ and $$\angle$$ $$OXA =$$ $$x^0$$ . Find the value of $$x$$.
State and prove mid point theorem.
Mid point Theorem :
The line segment joining the mid points of any two sides of a triangle is parallel to the third side.
Given :
A $$\triangle ABC$$ in which D and E are the mid points of AB and AC, respectively.
To prove :
$$DE \parallel BC$$.
Proof :
Since D and E are the mid points of AB and AC, respectively, we have $$AD=DB$$ and $$AE=EC$$.
Therefore,
$$\dfrac{AD}{DB}=\dfrac{AE}{EC}$$ ( each equal to 1 )
Therefore, by the converse of thales theorem, $$DE \parallel BC$$.
In the given figure, AB $$\parallel$$ DC Prove that(i) $$\triangle DMU \sim \triangle BMV$$(ii) $$DM \times BV = BM \times DU$$
The diagonals of a quadrilateral intersect at right angles. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is a rectangle.
Here, $$ABCD$$ is a quadrilateral. $$AC$$ and $$BD$$ are diagonals, which are perpendicular to each other.
$$P,Q,R$$ and $$S$$ are the mid-point of $$AB,BC,CD$$ and $$AD$$ respectively.
In $$\triangle ABC,$$
$$P$$ and $$Q$$ are mid points of $$AB$$ and $$BC$$ respectively.
$$\therefore$$ $$PQ\parallel AC$$ and $$PQ=\dfrac{1}{2}AC$$ ----- ( 1 ) [ By mid-point theorem ]
Similarly, in $$\triangle ACD,$$
$$R$$ and $$S$$ are mid-points of sides $$CD$$ and $$AD$$ respectively.
$$\therefore$$ $$SR\parallel AC$$ and $$SR=\dfrac{1}{2}AC$$ ----- ( 2 ) [ By mid-point theorem ]
From ( 1 ) and ( 2 ), we get
$$PQ\parallel SR$$ and $$PQ=SR$$
$$\therefore$$ $$PQRS$$ is a parallelogram.
Now, $$RS\parallel AC$$ and $$QR\parallel BD.$$
$$\Rightarrow$$ Also, $$AC\perp BD$$ [ Given ]
$$\therefore$$ $$RS\perp QR$$
$$\therefore$$ $$PQRS$$ is a rectangle.
If $$OC$$ and $$OD$$ are the bisectors of $$\angle C$$ and $$\angle D$$ respectively and $$\angle A =115^{\circ}$$, find $$\angle B$$.
The following figure shows a trapezium $$ABCD$$ in which $$AB$$ is parallel to $$DC$$ and $$AD=BC$$.Prove that :$$i)\; \;\;\angle DAB = \angle CBA$$$$ii)\;\; \angle ADC = \angle BCD$$$$iii) \;AC=BD$$$$iv)\;\; OA=OB$$ and $$OC=OD$$
$$(i)$$ $$\angle DAB = \angle CBA$$
Construction :- Produce $$AB$$ to $$E$$ and draw $$CE \parallel DA,\;MO \parallel AB$$
Proof: So $$AECD$$ is a parallelogram.
In $$\triangle BEC$$
$$BC=CE$$ ($$AD=BC=CE$$,opposite sides of parallelogram)
$$\angle CBE=\angle CEB$$ (opposite to equal sides) ---- $$(1)$$
$$\angle ADC =\angle BEC$$ (opposite angle of a parallelogram) ----- $$(2)$$
$$\angle DAB+\angle ADC=180^{\circ}$$ (sum of co-interior angles) ------ $$(3)$$
$$\angle ABC+\angle EBC=180^{\circ}$$ (L.P.P.)
$$\angle ABC+\angle ADC=180^{\circ}$$ ---- $$(4)$$ from $$(2)$$
From equation $$(3)$$ and $$(4)$$
$$\angle DAB = \angle CBA$$
$$(ii)$$ $$\angle BCD = \angle EBC$$ ---- $$(5)$$ (alternate interior angles)
From equations $$(1),(2),$$ and $$(5),$$
$$\angle ADC = \angle BCD$$ proved
$$(iii)$$ In $$\triangle ABD$$ and $$\Delta BAC$$
$$\angle BAD=\angle ABC$$
$$AB=AB$$ (common)
$$AD=BC$$ (Given)
$$\triangle ABD\cong \triangle BAC$$ (by SAS rule)
$$BD=AC$$ (By C.P.C.T)
$$(iv)$$ In $$\triangle ABD$$
$$MO \parallel AB$$
$$\dfrac{AM}{MD}=\dfrac{BO}{OD}$$ --- $$A$$, (by lemma of B.P.T)
In $$\triangle ADC$$
$$MO \parallel DC$$
$$\dfrac{AM}{MD}=\dfrac{AO}{OC}$$ ---- $$B$$ (by B.P.T.)
From equations $$A$$ and $$B,$$
$$\dfrac{BO}{OD}=\dfrac{AO}{OC}$$ ---- $$C$$
$$\dfrac{{BO}}{{OD}} + 1 = \dfrac{{AO}}{{OC}} + 1$$
$$\dfrac{{BO + DO}}{{OD}} = \dfrac{{AO + OC}}{{OC}}$$
$$\dfrac{{BD}}{{OD}} = \dfrac{{AC}}{{OC}}$$
$$OD=OC$$ ($$AC=BD$$)
From equation $$C,$$
$$OB=OA$$ $$(OD=OC)$$
$$ABCD$$ is a rectangle in which diagonal AC bisects $$\angle A $$ as well as$$\angle C$$. Show that,
(i) $$ABCD$$ is square
(ii) diagonal $$BD$$ bisects $$\angle B$$ as well as $$\angle D$$.
(i) $$ABCD$$ is square
(ii) diagonal $$BD$$ bisects $$\angle B$$ as well as $$\angle D$$.
$$(i)$$ABCD is a rectangle , in which diagonal AC bisect $$\angle A$$ as well as $$\angle C$$. Therefore,
$$\angle DAC=\angle CAB\rightarrow (1)$$
$$\angle DCA=\angle BCA\rightarrow (2)$$
A square is a rectangle when all sides are equal. Now,
$$AD\parallel BC$$ & $$AC$$ is transversal, therefore
$$\angle DAC=\angle BCA$$ [Alternate angles]
From (1), $$\angle CAB=\angle BCA\rightarrow (3)$$
In $$\bigtriangleup ABC$$,
$$\angle CAB=\angle BCA $$ , therefore
$$BC=AB$$ $$\rightarrow (4)$$[sides opposite to equal angles]
But $$BC=AD$$ & $$AB=DC$$$$\rightarrow (5)$$ [Opposite sides of rectangle]
Therefore from (4)& (5),
$$AB=BC=CD=AD$$
Hence, ABCD is a square.
$$(ii)$$ ABCD is a square and we know that diagonals of a square bisect its
angles.
Hence, BD bisects $$\angle B$$ as well as $$\angle D$$.
Three angles of a quadrilateral are equal to $${100^ \circ },{60^ \circ }$$ and $${80^ \circ }$$, respectively. Find its fourth angle.
Three angles of a quadrilateral $$ABCD$$ are $$\angle A=70^o,\angle B=50^o$$ and $$\angle C=82^o$$. Find $$\angle D$$.
In a quadrilateral ABCD, show that $$AB+BC+CD+DA>AC+BD$$.
In the adjacent figure the $$AB$$ and $$AC$$ of $$\triangle ABC$$ are produced to point $$E$$ and $$D$$ respectively If bisectors $$BO$$ and $$CO$$ of $$\angle CBD$$ and $$\angle BCD$$ respectively meet at point $$O$$, then prove that $$\angle BOC=90^{o}-\dfrac {1}{2} \angle BAC$$.
Draw two angles of $${90^ \circ }$$ at the ends of a line segment. Extend these rays. Do you get a triangle. Join these 2 rays by another line segment which shape can yo get? What will be the measure of all the angles?
Let $$AB$$ be another line segment
Draw two angles of $$90^o$$ at the ends of line segment
we get two parallel lines $$AD$$ and $$BC$$ ad they dont meet each other $$\Delta ^{le}$$ cannot be formed
Join $$C$$ and $$D$$ with a straight line
$$ABCD$$ is a quadrilateral
$$\therefore$$ measure of all the angle of a quadrilateral is $$360^o$$
A quadrilateral $$ABCD$$ is drawn to circumscribe a circle (see Fig). prove that $$AB+CD=AD+BC$$.
Angles of a quadrilateral are in the ratio $$1: 2: 3: 4$$. Find all angles. What special name can you give to this quadrilateral and why?
ABCD is a quadrilateral in which AD$$=$$BC and $$\angle$$ DAB$$=\angle$$ CBA. Prove that $$\angle$$ABD $$=\angle$$BAC.
$$ABCD$$ is a quadrilateral, where $$AD=BC$$ and $$\angle DAB=\angle CBA$$
In $$\triangle ABD$$ and $$\triangle BAC,$$
$$\Rightarrow$$ $$AD=BC$$ [ Given ]
$$\Rightarrow$$ $$\angle DAB=\angle CBA$$ [ Given ]
$$\Rightarrow$$ $$AB=BA$$ [ Common side ]
$$\therefore$$ $$\triangle ABD\cong \triangle BAC$$ [ SAS Congruence rule ]
$$\therefore$$ $$\angle ABD=\angle BAC$$ [ CPCT ]
The perimeter of a square is $$(4x+20)cm$$. What will be the length of its diagonal?
Find the measurements of the angles of quadrilateral if they are in ratio of $$2:3:4:6$$
In $$\square$$PQRS, $$m\angle Q$$ is $$10^o$$ less, $$m\angle R$$ is $$20^o$$ less and $$m\angle S$$ is $$30^o$$ less, than $$m\angle P$$, then find measure of each angle.
In $$\Box ABCD$$, $$\bar{AB}\parallel\bar{CD}$$. Prove that $$AD=BC$$ and $$AB=CD$$.
Join $$AC$$ and $$BD$$ which intersect at point $$O$$.
In $$\triangle AOB$$ and $$\triangle COD$$
$$\Rightarrow$$ $$\angle BAO=\angle DCO$$ [ Alternate angles ]
$$\Rightarrow$$ $$\angle ABO=\angle CDO$$ [ Alternate angles ]
$$\Rightarrow$$ $$\angle AOB=\angle COD$$ [ Vertically opposite angles ]
$$\therefore$$ $$AB=CD$$ [ CPCT ]
$$\Rightarrow$$ $$AD=BC$$ [ The line segment between the parallel line are equal ]
The diagonals of a rhombus $$ABCD$$ are $$6\ cm$$ and $$8\ cm$$. Find the length of a side of the rhombus.
Diagonals of a rhombus bisect each other at right angles.
Let point of intersectio of diagonals be $$O$$
So, $$OD=3cm, OA=4cm$$
From Right angled triangle $$\triangle AOD$$,
$$\Rightarrow AD^2=OA^2+OD^2$$ (pythoagorus theorem)
$$\Rightarrow AD^2=4^2+3^2$$
$$\Rightarrow AD^2=16+9$$
$$\Rightarrow AD^2=25$$
$$\Rightarrow AD=5cm $$
$$AD$$ is side of the rhombus .
So, length of side of the rhombus is $$5cm$$
The measure of three angles of a quadrilateral are equal. If measure of each angle is $$95^\circ$$, then find measure of its fourth angle.
In the adjoining figure, $$D$$ and $$E$$ are respectively the midpoints of sides $$AB$$ and $$AC$$ of $$\triangle{ABC}$$. If $$PQ\parallel BC$$ and $$CDP$$ and $$BEQ$$ are straight lines, then prove that $$ar(\triangle{ABQ})=ar(\triangle{ACP})$$.
In $$\triangle PAC$$
$$PA\parallel DE$$ and $$E$$ is the midpoint of $$AC$$
So, $$D$$ is the midpoint of $$PC$$
$$PA\parallel DE$$ and $$E$$ is the midpoint of $$AC$$
So, $$D$$ is the midpoint of $$PC$$
By Converse of midpoint theorem
Also, $$DE=\cfrac { 1 }{ 2 } PA$$ ----- $$(1)$$
Similarly, $$DE=\cfrac { 1 }{ 2 } AQ$$ -------$$(2)$$
From $$(1)$$ and $$(2)$$ we have $$PA=AQ$$
$$\triangle ABQ$$ & $$\triangle ACP$$ are on same base PQ and between same parallels PQ and BC
$$\therefore ar\left( \triangle ABQ \right) =ar\left( \triangle ACP \right) $$ (proved)
Also, $$DE=\cfrac { 1 }{ 2 } PA$$ ----- $$(1)$$
Similarly, $$DE=\cfrac { 1 }{ 2 } AQ$$ -------$$(2)$$
From $$(1)$$ and $$(2)$$ we have $$PA=AQ$$
$$\triangle ABQ$$ & $$\triangle ACP$$ are on same base PQ and between same parallels PQ and BC
$$\therefore ar\left( \triangle ABQ \right) =ar\left( \triangle ACP \right) $$ (proved)
In a quadrilateral ABCD. AO and BO are bisectors of angle A and angle B respectively. Prove that $$\angle$$AOB$$=\dfrac{1}{2}\{\angle C+\angle D\}$$.
ABCD is a quadrilateral
To prove : $$ \angle AOB = \dfrac{1}{2}(\angle C+\angle D)$$
$$AO$$ and $$BO$$ is bisector of $$A \ and \ B$$
$$ \angle 1 = \angle 2\, \angle 3 = \angle 4 ...(1)$$
$$ \angle A+\angle B+\angle C+\angle D = 360 $$
(Angle sum property)
$$ \dfrac{1}{2}(\angle A+\angle B+\angle C+\angle D) = 180...(2)$$
In $$ \triangle AOB $$
$$ \angle 1+\angle 3+\angle 5 = \dfrac{1}{2}(\angle A +\angle B +\angle C+\angle D)$$
$$ \angle 1 + \angle 3 + \angle 5 = \angle 1 + \angle 3 +\dfrac{1}{2}(\angle C+\angle D)$$
$$ \boxed {\angle AOB = \dfrac{1}{2}(\angle C+ \angle D)}$$
Hence,Proved.
In triangle $$ABC, XY || AC$$ and divide the triangle into two parts of equal areas. Find the ratio $$\dfrac{AX}{AB}$$.
In $$\Delta ABC$$
$$XY$$ is parallel to $$AC$$
To find : $$\dfrac{AX}{AB}$$
In $$\Delta ABC$$ & $$\Delta XBY$$
$$\angle ABC = \angle XBY$$
$$\angle ACB = \angle XYB$$
$$\Delta ABC \sim \Delta XBY$$ [AA similarity]
Now,
We know that in similar triangles,
Ratio of area of triangle is equal to ratio of square of corresponding sides
$$\dfrac{\text{Area of} \, \Delta ABC}{\text{Area of}\, \Delta XBY } = \left(\dfrac{AB}{XB}\right)^2$$
$$\dfrac{\text{Area of}\, \Delta XBY + \text{Area of} \, XYCA}{\text {Area of} \,\Delta XBY} = \left(\dfrac{AB}{XB}\right)^2$$
It is given that $$XY$$ divides triangle in two equal parts.
$$\dfrac{\text{Area of}\, \Delta XBY + \text{Area of} \,\Delta XBY}{\text {Area of} \,\Delta XBY} = \left(\dfrac{AB}{XB}\right)^2$$
$$\left(\dfrac{AB}{XB}\right) = \sqrt{2}$$
$$\dfrac{AB}{\sqrt{2}} = XB$$
Now, we need to find $$\dfrac{AX}{AB}$$
Since $$AX + XB = AB$$
$$AX + \dfrac{AB}{\sqrt{2}} = AB$$
$$AX = AB - \dfrac{AB}{\sqrt{2}}$$
$$\dfrac{AX}{AB} =\left (1 - \dfrac{1}{\sqrt{2}}\right)$$
$$\dfrac{AX}{AB} = \dfrac{\sqrt{2} - 1}{\sqrt{2}}$$
In a $$\Delta ABC$$, the internal bisectors of $$\angle B$$ and $$\angle C$$ meet at $$P$$ and the external bisectors of $$\angle B$$ and $$\angle C$$ meets at Q. Prove that $$\angle BPC + \angle BQC = {180^0}$$.
$$\Rightarrow 2\angle PBC+2\angle QBC=180^o$$
$$\Rightarrow \angle PBC+\angle QBC=90^o$$
$$\Rightarrow \angle PBQ=90^o$$
$$\angle ACB+\angle YCB=180°\quad \because ACY $$ is a straight line
$$\Rightarrow 2\angle PCB+2\angle QCB=180^o$$
$$\Rightarrow \angle PCB+\angle QCB=90^o$$
$$\Rightarrow \angle PCQ=90^o$$
Now, in quadrilateral BPCQ, sum of all angles$$ = 360^o$$
So,$$\angle BPC+\angle BQC+\angle PBQ+\angle PCQ=360°$$
$$\Rightarrow \angle BPC+\angle BQC+90°+90°=360°\Rightarrow \angle BPC+\angle BQC=180°$$
The interior angles of a quadrilateral are in the ratio $$3:5:6:4.$$ Find the angles.
If BC is a diameter of a circle of circle of centre O and OD is perpendicular to the chord $$\overline { AB } $$ of a circle then $$CA\equiv$$________
$$\because OB=OC=radius =r$$
Join $$OA$$ which is radius i.e $$OA=r$$
$$OD\bot AB$$ in isosceles triangle $$AO$$ and so divide $$\overline {AB}$$ into two square sides $$\overline{AD}$$ and $$\overline{BD}$$
By applying mid point theorem in $$\triangle ABC$$,
$$\Rightarrow OD\parallel AC$$
$$\Rightarrow \overline{OD}=\dfrac{1}{2}\overline{AC}$$
$$\Rightarrow \overline{AC}=2\overline{OD}$$
Two opposite angles $$A$$ and $$C$$ of a parallelogram $$ABCD$$ are $$(6x-17)^{o}$$ and $$(x+63)^{o}$$ respectively as shown the figure. Find the measure of angle $$A$$.
In figure given, we have $$X$$ and $$Y$$ are the midpoints of $$AB$$ and $$BC$$ and $$AX=CY$$. Show that $$AB=BC$$.
Three angles of a quadrilateral are equal. The fourth angle measure $${120^ \circ }$$. What is the measure of each of the equal angles?
In a quadrilateral ABCD, the bisector of $$\angle$$C and $$\angle$$D intersect at O. Prove that $$\angle$$COD$$=\dfrac{1}{2}(\angle A+\angle B)$$.
In $$ABCD$$, by angle sum property
$$\angle A+\angle B+\angle C+\angle D=360^{o}$$
$$\therefore \angle A+\angle B=360^{o}-(\angle C+\angle D)$$
$$\dfrac{1}{2}(\angle A+\angle B)=180^{o}-\dfrac{1}{2}(\angle C+\angle D)$$ ...... $$(1)$$
Now, In $$\triangle COD$$
$$\dfrac{1}{2}\angle C+\dfrac{1}{2}\angle D+\angle COD=180^{o}$$
$$\angle COD=180^{o}-\dfrac{1}{2}(\angle C+\angle D)$$ ......... $$(2)$$
From $$(1)$$ & $$(2)$$
$$\angle COD=\dfrac{1}{2}(\angle A+\angle B)$$
Hence proved.
Show that the point A (3,5) B (6,0) C (1,-3) D (-2,2) are the vertex of the square ABCD.
From the given groups of angles which are possible group angles of a quadrilateral:i) $$60^{\circ}, 70^{\circ}, 80^{\circ}$$, and $$145^{\circ}$$ii) $$90^{\circ}, 90^{\circ}, 90^{\circ},$$ and $$90^{\circ}$$
Find the area of a rhombus if its vertices are $$(3, 0), (4, 5), (-1, 4)$$ and $$(-2, -1)$$ taken in order.
The sum of two angles of a quadrilateral is $$150^{o}$$ and the other angles are in the ratio $$2:3$$. Find the measure of two angles.
In Fig $$ABCD$$ is a rhombus in which $$\alpha ABC=56^{o}$$. Which is the measure of $$\angle ACD$$?
In the quadrilateral $$PQRS$$, the angles are in the ratio of $$3: 4: 5: 6$$. Find the angles of $$PQRS$$. What is the special name given to $$PQRS$$?
Prove that each of the four sides of a rhombus is of the same length.
Given :
A rhombus ABCD such that AB = BC.
To prove :
$$AB=BC=CD=DA$$.
Proof :
Since ABCD is a rhombus.
Therefore, ABCD is a parallelogram.
$$\implies AB=CD$$ and $$BC=AD$$
But, $$AB=BC$$
Therefore, $$AB=BC=CD=AD$$
Hence, all the four sides of a rhombus are equal.
What is parallelogram ?
$$\textbf{Step 1: Definition and its properties.}$$
$$\text{A parallelogram is a quadrilateral whose opposite sides are equal and parallel.}$$
$$\text{Number of sides = 4}$$
$$\text{Number of vertices = 4}$$
$$\text{Mutually Parallel sides = 2 (in pair)}$$
$$\text{Area = Base x Height}$$
$$\text{Perimeter = 2 (Sum of adjacent sides length)}$$
$$\text{Type of polygon = Quadrilateral}$$
$$\textbf{Step-2: Shape of a parallelogram}$$
$$\textbf{Hence, A parallelogram is a quadrilateral whose opposite sides are equal and parallel.}$$
In the adjoining figure, $$A C \perp O C \text { and } A D \perp O D$$. Find $$\angle C A D$$
Show that the line segment joining the mid points of the opposite sides of a quadrilateral bisect each other.
PR & SQ bisect each other i.e. OP=OR & OQ=OS.
In $$\Delta ADC$$,
$$\therefore SR || AC$$
$$SR =\frac {1}{2}AC$$...(1)
In $$\Delta ABC$$,
$$\therefore PQ || AC$$
$$PQ =\frac {1}{2}AC$$...(1)
from eq (1) and (2)
PQ= SR & PQ || SR
So, in PQRS, one pair of opposite sides is parallel and equal.
$$\therefore PQRS$$ is parallelogram
PR & SQ are its diagonals and diagonals of parallelogram bisect each other
$$\therefore OP= OR $$ & $$ OQ = OS$$
Show that the points $$A(1,2),B(5,4),C(3,8)$$ and $$D(-1,6)$$ are the vertices of a square.
We are going to find distance between sides
$$AB=\sqrt{(5-1)^2+(4-2)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$$$$BC=\sqrt{(3-5)^2+(8-4)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$$
$$CD=\sqrt{(-1-3)^2+(6-8)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$$
$$DA=\sqrt{(1+1)^2+(2-6)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$$
$$\therefore$$ $$AB=BC=CD=DA$$
Now, we are going to find distance between diagonals.
$$AC=\sqrt{(3-1)^2+(6-2)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$$
$$BD=\sqrt{(-1+5)^2+(6-4)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$$
$$\therefore$$ $$AC=BD$$
So, here we can see distance between all sides and diagonals are equal.
$$\therefore$$ Points $$A,B,C$$ and $$D$$ are the vertices of a square.
ABCD is a rectangle in which diagonal AC bisects $$\angle \mathrm { A }$$ as well as $$\angle \mathrm { C }$$ . Show that
(i) ABCD is a square
(ii) diagonal BD bisects $$\angle \mathrm { B }$$ as well as $$\angle \mathrm { D }$$.
$$AC$$ bisects $$\angle A$$.
$$\therefore$$ $$\angle DAC=\angle CAB$$ ----- ( 1 )
$$AC$$ bisects $$\angle C$$.
$$\therefore$$ $$\angle DCA=\angle ACB$$
Now, $$AD\parallel BC$$ and $$AC$$ as transversal.
$$\therefore$$ $$\angle DAC=\angle ACB$$ [ Alternate angles ] ---- ( 2 )
From ( 1 ) and ( 2 ) we get,
$$\Rightarrow$$ $$\angle CAB=\angle ACB$$ ----- ( 3 )
In $$\triangle ABC,$$
$$\Rightarrow$$ $$\angle CAB=\angle ACB$$ [ From ( 3 ) ]
$$\Rightarrow$$ $$BC=AB$$ ----- ( 4 ) [ Sides opposite to equal angles are equal ]
But $$BC=AD$$ and $$ AB=DC$$ ------ ( 5 ) [ Opposite sides of rectangle are equal ]
From ( 4 ) and ( 5 )
$$\Rightarrow$$ $$AB=BC=CD=DA$$
So, $$ABCD$$ is a rectangle with all sides equal.
$$\therefore$$ $$ABCD$$ is a square. ------ Hence proved.
Since, $$ABCD$$ is a square and diagonals of a square bisects its angles.
$$\therefore$$ Diagonal $$BD$$ bisects $$\angle B$$ and $$\angle D.$$ --- Hence proved
If the diagonals of a quadrilateral bisect each other, then prove that it is a parallelogram.
$${\textbf{Step - 1: Drawing a quadrilateral and its diagonals}}.$$
$${\text{ABCD is a quadrilateral with AC and BD are diagonals intersecting at O}}{\text{.}}$$
$${\text{It is given that diagonals bisect each other}}{\text{.}}$$
$$\therefore \;\;OA = OC\;{\text{and}}\;OB = OD$$
$${\textbf{Step - 2: Proving that the quadrilateral is a parallelogram}}.$$
$${\text{In}}\;{\text{ }}\Delta AOD{\text{ }}\;{\text{and }}\Delta COB,$$
$$OA = OC\;{\text{ }}\;{\text{ }}$$
$$\;\angle AOD = \angle COB\;...({\text{Vertically opposite angles}})$$
$$OD = OB$$
$${\text{Therefore }}\Delta AOD \cong \Delta COB{\text{ by SAS Congruence rule}}{\text{.}}$$
$$\therefore \;\angle OAD = \angle OCB\;$$
$${\text{Similarly, we can prove that}}\;\;\Delta AOB \cong \Delta COD$$
$$\therefore \angle ABO = \angle CDO\;$$
$${\text{So, for lines AB and CD with transversal BD, we can write}}$$
$$\angle ABO{\text{ and }}\angle CDO{\text{ are alternate angles and are equal i}}{\text{.e}}{\text{. AB||CD}}{\text{.}}$$
$${\text{Similarly, for lines AD and BC, with transversal AC, }}\angle OAD{\text{ and }}\angle OCB{\text{}}$$
$${\text{are alternate angles and are equal}}{\text{. Therefore, BC||AD}}{\text{.}}$$
$${\text{Thus, in ABCD, both pairs of opposite sides are parallel}}{\text{.}}$$
$${\text{Therefore, ABCD is a parallelogram}}{\text{.}}$$
$${\textbf{Hence, Proved that quadrilateral whose diagonals bisect each other is a parallelogram}}{\text{.}}$$
$$E$$ and $$F$$ are points on diagonals $$AC$$ of a parallelogram $$ABCD$$ such that $$AE=CF$$. Show that $$BFDE$$ is a parallelogram.
The four angles of a quadrilateral are equal. Draw this quadrilateral in your notebook. Find each of them.
REF.Image.
Given: four angles of a quadrilateral are equal
To find : all the angles
Draw the quadrilateral.
Let $$ \angle A, \angle B, \angle C, \angle D $$ be the 4 angles of the
quadrilateral
Since, the sum of angles of quadrilateral $$ = 360^{\circ}$$
$$ \therefore \angle A+\angle B+\angle C+\angle D = 360^{\circ}$$
$$ \angle A+\angle A+\angle A\angle A= 360^{\circ}$$ (since all are equal we all $$ = \angle A$$ )
$$ 4\angle A = 360^{\circ}$$
$$ \angle A = 90^{\circ}$$
$$ \therefore $$ each bangle of quadrilateral $$= 90^{\circ}$$
It must look like this with all the angles $$ = 90^{\circ}$$
In a quadrilateral, the angles are $$x^{o},(x+10)^{o},(x+20)^{o},(x+30)^{o}$$. Find the angles.
Construct Rhombus MATH with $$\mathrm { AT } = 4 \mathrm { cm } , \angle \mathrm { MAT } = 120 ^ { \circ }$$
Constructing steps.
$$i)$$ Draw a line segment $$AT=4cm$$.
$$ii)$$ Construct $$\angle MAT=120.$$
$$iii)$$ With $$A$$ as the centre and radius equal $$4cm.$$
$$iv)$$ Draw a arc on $$AX$$
$$v)$$ Mark the intersection as $$M$$
$$vi)$$ With $$M$$ as the centre and radius of $$4cm.$$Draw an arc.
$$vii)$$ With $$T$$ as the centre with same radius cut the previous arc as mark the point as $$H.$$
$$viii)$$ Join points $$MH$$ and $$HT.$$
$$\therefore$$ $$MATH$$ the required Rhombus is constructed.
If $$A , B . C$$ are $$( 3,3,3 ) , ( 0,6,3 )$$ and $$( 1,7,7 )$$ respectively. find $$D ( x , y , z )$$ such that ABCD is a square
ABCD is a trapezium is which $$AB\parallel DC$$, BD is a diagonal and E is the midpoint of AD. A lie is drawn through E parallel. AB IN intersecting BC at F. show that F is the midpoint of BC
Given ABCD is a trapezium.
We have to prove, F is the mid point of BC, i.e., BF$$=$$CF
Let EF intersect DB at G.
In $$\Delta$$ABD E is the mid point of AD and EG$$||$$AB.
$$\therefore$$ G will be the mid-point of DB.
Now EF$$||$$AB and AB$$||$$CD
$$\therefore$$ EF$$||$$CD
$$\therefore$$ In $$\Delta$$BCD, GF$$||$$CD
$$\Rightarrow$$ F is the mid point of BC.
If ABCD is a rohmbus with $$\angle ABC={ 56 }^{ 0 }$$ Find the measure of $$\angle ADC$$
$$As\>opposite\>angles\>are\>same\>in\>a\>rhombus,\>hence\>\angle\>ADC\>=\>\angle\>ABC\>=\>56^\circ$$
Prove that, the sum of either pair of opposite angles of a cyclic quadrilateral is 180.
Given : ABCD is a cyclic quadrilateral of a circle with centre at O.
To prove : $$\angle{BAD}+\angle{BCD}=180^{o}$$
$$\angle{ABC}+\angle{ADC}=180^{o}$$
Chord $$AB $$$$\angle{5}=\angle{8}.....(1)$$ [Angle in same segment are equal]
Chord $$BC$$
$$\angle{1}=\angle{6}.....(2)$$ [Angle in same segment are equal]
Chord $$CD$$
$$\angle{2}=\angle{4}.....(3)$$ [Angle in same segment are equal]
Chord $$AD$$
$$\angle{7}=\angle{3}.....(4)$$ [Angle in same segment are equal]
By angle sum property of qudrilateral
$$\angle{A}+\angle{B}+\angle{C}+\angle{D}=360^{o}$$
$$\angle{1}+\angle{2}+\angle{3}+\angle{4}+\angle{7}+\angle{8}+\angle{5}+\angle{6}=360^{o}$$
$$(\angle{1}+\angle{2}+\angle{7}+\angle{8})+(\angle{3}+\angle{4}+\angle{5}+\angle{6})=360^{o}$$
$$(\angle{1}+\angle{2}+\angle{7}+\angle{8})+(\angle{7}+\angle{2}+\angle{8}+\angle{1})=360^{o}$$ [From (1), (2) ,(3) and (4)]
$$2(\angle{1}+\angle{2}+\angle{7}+\angle{8})=360^{o}$$
$$\angle{1}+\angle{2}+\angle{7}+\angle{8}=180^{o}$$
$$(\angle{1}+\angle{2})+(\angle{7}+\angle{8})=180^{o}$$
$$\angle{BAD}+\angle{BCD}=180^{o}$$
Similarly,
$$\angle{ABC}+\angle{ADC}=180^{o}$$
Hence proved.
In $$\triangle PQR$$,point M is on side PQ and point S is on the side PR such that QRSM is what kind of figure?
$$QRSM$$ is a Trapezium
because opposite sides are parallel and sides intersect each other.
Find the lengths of diagonals $$AC$$ and $$BD$$. Given $$AB = 60\ cm$$ and $$\angle BAD = 60^{\circ}$$.
Diagonals of a rhombus bisect each other at right angles.
Diagonals of a rhombus bisect the angle of vertex.
$$OA = OC = \dfrac {1}{2}AC, OB = OD = \dfrac {1}{2} BD, \angle AOB = 90^{\circ}$$
$$\angle OAB = \dfrac {60^{\circ}}{2} = 30^{\circ}$$
$$AB = 60\ cm$$
In right triangle $$AOB$$
$$\sin 30^{\circ} = \dfrac {OB}{AB}$$
$$\dfrac {1}{2} = \dfrac {OB}{60}$$
$$OB = 30\ cm$$
$$\cos 30^{\circ} = \dfrac {OA}{AB}$$
$$\dfrac {\sqrt {3}}{2} = \dfrac {OA}{60}$$
$$OA = 51.96\ cm$$
Length of $$AC = 2\times OA = 2\times 51.96 = 103.92\ cm$$
Length of $$BD = 2\times OB = 2\times 30 = 60\ cm$$.
In the given figure, ABCD and PQRB are rectangles where Q is the mid point of BD. If QR=5cm, then find the length AB.
$$AB$$ and $$CD$$ bisect each other of $$K$$.
Prove that $$AC =BD$$
The angles of quadrilateral are in the ratio $$3:5:9:13$$. Find all the angles of the quadrilateral.
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
If one side of a square is represented by $$4x-7$$ and the adjacent side is represented by $$3x+5$$, find the value of $$x$$.
The four angles of a quadrilateral are in the ratio $$1 : 2 : 3 : 4$$.Find the measure of each angle of the quadrilateral.
Fill in the blanks
If three acute angles of a quadrilateral measure $$70^0$$ each, then the measure of the fourth angle is ____ .
Let the fourth angle be x.
So, $$70^o+70^o+70^o+x=360^o$$
$$210^o+x=360^o$$
$$x=360^o-210^o$$
$$x=150^o$$.
Prove that the sum of all the angle of a quadrilateral is $$360^\circ.$$
One single of quadrilateral it of $${ 108 }^{ \circ }$$ and the remaining three angles are equal. Find each of the three equal angles.
In the adjoining figure $$PQRS$$ is a rectangle. Identify the congruent triangles formed by the diagonals.
The congruent triangles formed by diagonals are $$\triangle POQ$$ & $$\triangle ROS$$ and $$\triangle POS$$ & $$\triangle ROQ$$.
Explanation :-
In $$\triangle POQ$$ & $$ \triangle $$ROS$$
$$PO=RO$$ ( diagonals of rectangle bisect other)
$$QO=SQ$$ ( diagonals of a rectangle bisect other).
$$ \angle POQ =\angle ROS$$ ( Vertically opposite angles).
$$\therefore \triangle POQ \cong \triangle ROS$$.
Similarly :-
$$1) PO=RO$$ (diagonals of rectangle bisect other).
$$2) SO=QO$$ ( diagonals of rectangle bisect other).
$$3) PS=QR$$ ( opposite sides of rectangle are equal)
$$\therefore \triangle POS\cong \triangle ROQ$$.
The angles of a quadrilateral are in the ratio $$2 : 4 : 5 : 7$$. Find the angles.
Show that the line segments joining the mid-points of the opposite side of a quadrilateral bisect each other.
Mid poinnt of
$$AB=\left (\dfrac {x_1+x_2}{2},\ \dfrac {y_1+y_2}{2}\right)$$
$$BC=\left (\dfrac {x_2+x_3}{2},\ \dfrac {y_2+y_3}{2}\right)$$
$$CD=\left (\dfrac {x_4+x_3}{2},\ \dfrac {y_4+y_3}{2}\right)$$
$$DA=\left (\dfrac {x_4+x_1}{2},\ \dfrac {y_4+y_1}{2}\right)$$
Mid point of bisectors is (both bisectors) $$\left (\dfrac {x_1+x_2+x_3+x_4}{2},\ \dfrac {y_1+y_2+y_3+y_4}{2}\right)$$
$$\therefore \ $$ Mid point of both side of quadrilateral bisect each other.
In the given figure, ABCD is a cyclic quadrilateral , whose diagonals intersect at $$E$$ such that $$\angle BAC= 30^{0}$$ , $$\angle DBC= 70^{0}$$ and AB=BC,Find $$\angle ACD$$.
Find the unknown angle.
Show that the four triangles as shown in the adjoining fig. formed by diagonals and sides of rhombus are congruent.
In the given figure, $$AC$$ is the bisector of $$\angle A$$. If $$AB=AC, AD=CD$$ and $$\angle ABC=75^{o}$$, find the value of $$x$$ and $$y$$.
In a trapezium $$ABCD$$ with $$AB$$ parallel to $$CD,$$ the diagonals intersect at $$P.$$ The area of $$\triangle A B P$$ is $$72\mathrm { cm } ^ { 2 }$$ and of $$\Delta C D P$$ is $$50\mathrm { cm } ^ { 2 } .$$ Find the area of the trapezium.
If $$A B C D$$ is a parallelogram, $$A E \perp D C$$ and $$C F \perp A D .$$ If $$A B = 16 \mathrm { cm }$$, $$ A E = 8 cm$$ and $$C F = 10 \mathrm { cm } ,$$ find $$A D .$$
Given, $$DC=AB=16cm$$
Area of parallelogram
Area of parallelogram
$$=\cfrac{1}{2}\times DC\times AE=\cfrac{1}{2}\times AD\times CF\\ \Rightarrow DC\times AE=AD\times CF\\ \Rightarrow 16\times 8=AD\times 10\\AD=\cfrac{144}{10}cm=14.4cm$$
In parallelogram $$ A B C D$$, $$\angle D = 115 ^ \circ ,$$ find the remaining angles .
$$\angle B=\angle D=115°$$ (Opposite angle)
$$\angle C+\angle D=180°$$ (Adjacent angle)
$$\ \angle C=180°-\angle D=180°-115°=65°$$
$$\angle C+\angle D=180°$$ (Adjacent angle)
$$\ \angle C=180°-\angle D=180°-115°=65°$$
$$\therefore \angle B=115°, 2C=65°,\angle B=\angle C=65°$$
$$ABCD$$ is a quadrilateral. $$AO$$ and $$BO$$ are the angle bisectors of angles $$A$$ and $$B$$ which meet at $$O$$. If $$\angle C=70^{o},\ \angle D=50^{o}$$, find $$\angle AOB$$.
We have,
ABCD is a quadrilateral.
Given that,
$$\angle C={{70}^{o}},\,\,\angle D={{50}^{o}}$$
So,
$$\angle A+\angle B+\angle C+\angle D={{360}^{o}}$$
$$ \left( A+B \right)={{360}^{o}}-\left( C+D \right) $$
$$ \Rightarrow \left( A+B \right)={{360}^{o}}-\left( {{70}^{o}}+{{50}^{o}} \right) $$
$$ \Rightarrow \left( A+B \right)={{360}^{o}}-{{120}^{o}} $$
$$ \Rightarrow \left( A+B \right)={{240}^{o}} $$
According to given question,
$$ \angle OAB=\dfrac{\angle A}{2} $$
$$ \angle OAB=\dfrac{\angle B}{2} $$
Now,
$$ \angle OAB+\angle OBA=\dfrac{\angle A+\angle
B}{2}={{120}^{o}} $$
$$ \Rightarrow \angle AOB+{{120}^{o}}={{180}^{o}} $$
$$ \Rightarrow \angle AOB={{180}^{o}}-{{120}^{o}} $$
$$ \Rightarrow \angle AOB={{60}^{o}} $$
Hence, this is the
answer.
Find the unknown angle.
A photo frame is in the shape of a quadrilateral. With one diagonal longer than the other. Is it a rectangle ? why or why not ?
If the angles of a quadrilateral are in the ratio $$1:2:3:4$$ find the angles
Two angles of a quadrilateral are each of measure $$75^o$$ and the other two angles are equal. What is the measure of these two angles ? Name the possible figure so formed.
Construct a parallelogram $$ABCD$$ such that $$BC=4.5\ cm,\ BD=4\ cm$$ and $$AC=5.6\ cm$$
We know diagonals in a parallelogram bisect each other.
so lets consider the intersecting point if the diagonals as $$O$$
Therefore,
$$OB = 2\ cm\,\,\, OC =2.8\ cm$$
and $$BC = 4.5\ cm$$
So, we can construct a triangle.
now extend $$BO$$ to $$D$$ and $$CO$$ to $$A$$ with $$2\ cm$$ and $$2.8\ cm$$ respectively.
join $$AD, CD$$ and $$AB$$.
11 gm MNPQ, MP=3.8 cm , NQ=4.5 cm and angle between MP and NQ is $$60^{\circ}$$.
We know that, the diagonals of a parallelogram bisect one another at common point O, and equal angles are formed. Sum of all $$\angle S$$ around $$=0=360^o$$
$$x+x+x+x=360$$
$$4x=360$$
$$x=90$$.
Prove that the points $$A(2, 3), B(-2, 2), C(-1, -2)$$ and $$D(3, -1)$$ are the vertices of a square $$ABCD$$.
Given a rectangle with a fixed perimeter of $$24$$ meters, if we increase the length by $$1\ \text{m}$$, then the width and area will vary accordingly. Use the following table of values to look at how the width and area vary as the length varies.
What do you observe? Write your observations in your notebooks.
| Len. (in cm) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Wid. (in cm) | 11 | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 |
| Area (in $$cm^{2}$$) | 11 | 20 | 27 | 32 | 35 | 36 | 35 | 32 | 36 |
The dimensions of a rectangular room (cubiodal) are l, b and h. What is the area of its four walls?
If $$D,\ E,\ F$$ are the mid-points of the sides $$BC,\ CA$$ and $$AB$$ respectively of $$\triangle ABC$$, prove that $$BDEF$$ is a parallelogram. And also show that $$ar(\triangle DEF)=\dfrac {1}{4}ar(\triangle ABC)$$.
We have,
$$Given\,that:\,\Delta ABC$$
Where $$D,E,F$$ are mid points of $$BC,CA$$ and $$AB$$ respectively.
To prove:-$$BDEF$$ is a parallelogram
Proof:- In $$\Delta ABC$$,
$$F$$ is mid point of $$AB$$,
$$D$$ is the midpoint of $$BC$$
$$E$$ is the midpoint of $$CA$$
$$ \therefore FE\parallel BC\Rightarrow \therefore
FE\parallel BD $$
$$ \therefore DE\parallel AB\Rightarrow \therefore DE\parallel FB $$
$$ \therefore FE\parallel BD\,\,and\,\,DE\parallel FB $$
So, in $$BDEF$$,
Both pairs of opposite sides are parallel,
$$\therefore BDEF$$ is a parallelogram.
Hence proved.
Now,
$$BDEF$$ is a parallelogram
$$ \therefore \Delta DBF\cong \Delta DEF $$
$$ ar\,\left( \Delta DBF \right)=ar\left( \Delta DEF \right)\,\,......\,\,\left( 1 \right) $$
Similarly, we can prove that,
$$FDCE$$ is a parallelogram
$$ \therefore \Delta DEC\cong \Delta DEF $$
$$ ar\left( \Delta DEC \right)=ar\left( \Delta DEF \right)\,\,......\,\,\left( 2 \right) $$
Similarly, we can prove that,
$$AFDE$$ is a parallelogram
$$ \therefore \Delta AFE\cong \Delta DEF $$
$$ ar\left( \Delta AFE \right)=ar\left( \Delta DEF \right)\,\,......\,\,\left( 3 \right) $$
From equation (1), (2), (3) to, and we get,
$$ar\left( \Delta DBF \right)=ar\left( \Delta DEC \right)=ar\left( \Delta AFE \right)=ar\left( \Delta DEF \right)$$
Now,
$$ ar\left( \Delta DBF \right)+ar\left( \Delta DEC
\right)+ar\left( \Delta AFE \right)+ar\left( \Delta DEF \right)=ar\left( \Delta
ABC \right) $$
$$ \Rightarrow ar\left( \Delta DEF \right)+ar\left( \Delta DEF \right)+ar\left( \Delta DEF \right)+ar\left( \Delta DEF \right)=ar\left( \Delta ABC \right) $$
$$ \Rightarrow 4ar\left( \Delta DEF \right)=ar\left( \Delta ABC \right) $$
$$ \Rightarrow ar\left( \Delta DEF \right)=\dfrac{1}{4}ar\left( \Delta ABC \right) $$
Hence proved.
i\\In the figure, $$Q$$ is the centre of the circle
and $$PM$$ and $$PM$$ are tangents to the circle.
If $$\angle MPN = {40^ \circ }.$$ find $$\angle MQN.$$
R.E.F. Image.
$$\sum $$ angles of quadrilateral = $$360^{\circ}$$
$$40+90+90+Q=360^{\circ}$$
$$Q= 140^{\circ}= \angle MQN$$
If three angles of quadrilateral are each equal to $$75 ^ { \circ } $$ , the fourth angle is
$$ABCD$$ is a trapezium $$\overline { A B } \| \overline { C D }$$ If $$A B = 20 \mathrm { cm }\mathrm { BC } = 8 \mathrm { cm } , \mathrm { CD } = 10 \mathrm { cm }$$ and $$\mathrm { AD } = 6 \mathrm { cm } .$$ Find the area of $$\mathrm { ABCD } .$$
R.E.F. Image.
ABCD is trapezium. Draw CM.|| AD.
In $$ \Delta $$ CMB ; 6, 8, 10 cm.
$$ 8 = \frac{a+b+c}{2} = \frac{6+8+10}{2} = 12 cm$$ a = 6 b = 8 c = 10
$$ \Delta = \frac{1}{2}\times b\times h = \sqrt{S(S-a)(S-b)(S-c)}$$
$$ = \frac{1}{2}\times 8\times h = \sqrt{12(12-6)(12-8)(12-10)}$$
$$ = 4h = \sqrt{12.6.4.2} = \sqrt{576} = 24$$
$$ x = h = 6 $$
Area of trapezium = Area of AMCD + Area of $$ \Delta $$ MBC
$$ = 10\times 6+24 = 84 cm^{2}$$
In the adjoining figure $$, ABCD$$ is a parallelogram. Any line through A cuts $$DC$$ at a point $$P$$ and $$B C$$ produced at $$Q .$$ Prove that : $$ar ( \Delta B P C ) = ar (\Delta DPQ).$$
R.E.F image
Prove:- Area $$\bigtriangleup BPC$$ = Area $$\bigtriangleup DPQ$$
form figure ---$$\bigtriangleup ACQ $$ & $$ \bigtriangleup DQC$$
$$CQ\left | \right |AD$$
CQ=CQ same base
Area $$\bigtriangleup ACQ= area \bigtriangleup DQC$$
(Between parallel lines & same base area of the triangle
should be equal)
Area $$\bigtriangleup ACQ$$ = Area $$\bigtriangleup DQC$$
subtract the area $$\bigtriangleup CPQ$$ form both side
Area $$\bigtriangleup ACQ$$ - Area $$\bigtriangleup CPQ$$ = Area $$\bigtriangleup DQC$$ - Area $$\bigtriangleup CPQ.$$
Area $$\bigtriangleup $$ APC=Area $$\bigtriangleup $$ DPQ----(1)
sin a $$AB\left | \right |PC$$ $$(\because$$ PC is a part of DC)
$$\bigtriangleup APC$$ & $$ \bigtriangleup BCP$$ are on the same base PC & between the same parallels $$PC\left | \right |AB$$ so they are equal in
Area ie Area $$ (\bigtriangleup APC)$$ = Aera $$(\bigtriangleup BCP)$$-----(ii)
form eq(i) & eq(ii)
Area $$(\bigtriangleup BCP)$$ = Aera $$\bigtriangleup DPQ$$
Diagonals $$AC$$ and $$BD$$ of a trapezium $$ABCD$$ with $$AB \parallel DC$$ intersect each other at $$O.$$ Prove that $$ar ( A O D ) = a r ( B O C )$$
REF.Image
A trapezium ABCD where AB||DC & Diagonals AC & BD intersect each other at O.
$$\Delta ADC$$ & $$\Delta BDC$$ Lie on the same base DC & between the same parallels AB & CD
$$\therefore Area (\Delta ADC)= Area (\Delta BDC)$$
Triangle with same base & between parallels are equal in Area.
subtracting ar $$(\Delta DOC)$$ both side
$$\Rightarrow Ar (\Delta ADC) - Area (\Delta DOC)= Ar (\Delta BDC)- Ar (\Delta DOC)$$
$$\Rightarrow Ar(AOD)= Ar (\Delta BOC)$$ proved.
The angles of quadrilateral are $$x,80^{\circ},120^{\circ},100^{\circ}$$ find $$x$$
Find the value of $$x$$ in given figure.
$${\textbf{Step-1: Finding the value of x.}}$$ 
$${\text{Produce CD to cut AB at E.}}$$
$${\text{Now, in}}$$ $$ \Delta BDE$$ $${\text{, we have, }}$$
$${\text{Exterior}}$$ $$\angle CDB=\angle CEB+\angle DBE $$
$$\Rightarrow x=∠CEB+45 …..(i)$$
$${\text{In}}$$ $$\Delta AEC,$$ $${\text{we have, }}$$
$${\text{Exterior}}$$ $$\angle CEB=\angle CAB+\angle ACE $$
$$=55+30=85$$
$${\text{Putting}}$$ $$ \angle CEB=85 $$ $${\text{in (i), we get, }}$$
$$x=85+45=130$$
$$∴x=130^\circ$$
$${\textbf{Hence , The value of x is }}$$ $$\mathbf{130^\circ.}$$
Prove that in a parallelogram the opposite angles are equal .
PQRS is a square. Determine $$\angle SRP$$.
In the adjoining fig. $$10.33 , A B C D$$ is a rhombus.
Find the measure of the following angles, if $$\angle A C B = 30 ^ { \circ }$$
(a) $$\quad \angle B O C$$
(b) $$\angle C B O$$
(c) $$\quad \angle O A D $$
(d) $$\quad \angle A B O$$
From the figure:-
$$\angle DAO=\angle BCO$$ (Alternate interior Angle)
$$=30^o$$
(Two parallel lines are crossed by a transversal)
We know that Diagonals in a rhombus intersect at right angle so $$\angle BOC=90$$ …………..$$(1)$$
through vertically opposite Angles $$AOD=90^o$$
Now takes $$\Delta BOC$$
$$\angle BOC+\angle OCB+\angle CBO=180^o$$
$$90^o+30^o+\angle CBO=180^o$$
$$\angle CBO=180-120^o$$
$$\angle CBO=60^o$$ ……………$$(2)$$ Ans
$$\angle OCB=\angle OAD$$ (Alternate interior Angle)
$$=30^o$$ …………..$$(3)$$ Ans
All the four triangle should be similar by this in $$\Delta AOB$$.
$$\angle AOB=90^o$$
$$\angle OAB=30^o$$
$$\angle ABO=60^o$$.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angle, then it is a square.
Two opposite angles of a parallelogram are $$(3x-2)^{0}$$ and $$(50-x)^{0}$$. Find $$x$$
Find the area of the quadrilateral ABCD in which AB= 3 cm, BC= 4 cm,CD= 4 cm , DA=5 cm and AC= 5 cm
Area of quad $$ABCD=$$ area ($$\triangle ABC$$)+ar ($$\triangle ACD$$)
$$ABC$$ is a right angled triangle and $$ACD$$ is an isoscles triangle.
ar($$\triangle ABC$$)$$=\cfrac { 1 }{ 2 } \times AB\times BC=\cfrac { 1 }{ 2 } \times 3\times 4\quad { cm }^{ 2 }=6\quad { cm }^{ 2 }$$
area of isoscles $$\triangle =\cfrac { 1 }{ 2 } \times 6\times \sqrt { { a }^{ 2 }-\cfrac { { b }^{ 2 } }{ 4 } } $$
Here $$a=5cm$$ and $$b=4cm$$
ar($$\triangle ACD$$) $$=\cfrac { 1 }{ 2 } \times 4\times \sqrt { { 5 }^{ 2 }-\cfrac { { 4 }^{ 2 } }{ 4 } } $$
$$=2\times \sqrt { { 5 }^{ 2 }-4 } =2\times \sqrt { 25-4 } { cm }^{ 2 }$$
$$=2\times \sqrt { 21 } { cm }^{ 2 }$$
$$ =$$ $$2\sqrt {21}$$ $${ cm }^{2}$$
$$\therefore$$ ar($$ABCD$$)$$= (6+2\sqrt { 21 } ){ cm }^{ 2 }$$
The bisectors of $$\angle B$$ and $$\angle C$$ of quadrilateral ABCD meet at a point $$R$$ inside the quadrilateral.If $$\angle A = 95 ^ { \circ }$$ and $$\angle D = 111 ^0,$$ find $$\angle B R C$$
The angles of a quadrilateral are in the ratio 2:3:4:Find the measure of the smallest and greatest angle.
If the mid points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral.
Given: A quadrilateral $$ABCD$$ in which the mid-point of the sides of it are joined in order of form parallelogram $$PQRS$$
To prove: $$ar(||gmPQRS)=\dfrac{1}{2}ar(ABCD)$$
Construction: Join $$BD$$ and draw perpendicular from $$A$$ and $$BD$$ which intersect $$SR$$ and $$BD$$ at $$X$$ and $$Y$$ respectively.
Proof: In $$\Delta ABD$$, $$S$$ and $$R$$ are the mid-points of sides $$AB$$ and $$AD$$ respectively.
$$\therefore SR||BD$$
$$\Rightarrow SX||BY$$
$$\Rightarrow X$$ is the mid-point oy $$AY$$ [Converse of mid-point theorem]
$$\Rightarrow AX||XY$$...........(1)
($$\because\, S$$ is the mid-point of $$AB$$ and $$SX||BY$$)
$$SR=\dfrac{1}{2}BD$$.......(2) ($$\because$$ Mid-point theorem)
Now, $$ar(\Delta ABD)=\dfrac { 1 }{ 2 } \times BD\times AY$$
$$ar(\Delta ASR)=\dfrac { 1 }{ 2 } \times SR\times AX$$
$$ar(\Delta ASR)=\dfrac { 1 }{ 2 } \times \left(\dfrac{1}{2}BD\right)\times \left(\dfrac{1}{2}AY\right)SR\times AX$$ (Using (1) and (2))
$$ar(\Delta ASR)=\dfrac { 1 }{ 4 } \times \left(\dfrac{1}{2}BD\times AY\right)$$ (Using (1) and (2))
$$\Rightarrow ar (\Delta ASR)=\dfrac{1}{4}\times(\Delta ABD)$$.......(3)
Similarly,
$$\Rightarrow ar (\Delta CPQ)=\dfrac{1}{4}\times(\Delta CBD)$$...........(4)
$$\Rightarrow ar (\Delta BPS)=\dfrac{1}{4}\times(\Delta BAC)$$...........(5)
$$\Rightarrow ar (\Delta DRQ)=\dfrac{1}{4}\times(\Delta DAC)$$...........(6)
Adding (3),(4),(5) and (6), we get,
$$ar(\Delta ASR)+ar(\Delta CPQ)+ar(\Delta BPS)+ar(\Delta DRQ)$$
$$=\dfrac { 1 }{ 4 } ar(\Delta ABD)+\dfrac { 1 }{ 4 } ar(\Delta CBD)+\dfrac { 1 }{ 4 } ar(\Delta ABC)+\dfrac { 1 }{ 4 } ar(\Delta DAC)$$
$$=\dfrac { 1 }{ 4 } [ar(\Delta ABD)+ ar(\Delta CBD)+ ar(\Delta BAC)+ar(\Delta DAC)]$$
$$=\dfrac { 1 }{ 4 } [ar(ABCD)+ ar(\Delta ABCD)[$$
$$=\dfrac { 1 }{ 4 } \times 2ar(ABCD)$$
$$=\dfrac { 1 }{ 2} \times 2ar(ABCD)$$
$$ar(\Delta ASR)+ar(\Delta CPQ)+ar(\Delta BPS)+ar(\Delta DRQ)$$
$$=\dfrac { 1 }{ 2} \times 2ar(ABCD)$$
$$\Rightarrow ar(ABCD)-ar(||gm PQRS)=\dfrac{1}{2}ar(ABCD)$$
$$\Rightarrow ar(||gm PQRS)=ar(ABCD)-\dfrac{1}{2}ar(ABCD)$$
$$\Rightarrow ar(||gm PQRS)=dfrac{1}{2}ar(ABCD)$$
Hence, proved.
If $$E,F,G$$ and $$H$$ are respectively the mid-points of he sides of a parallelogram $$ABCD,$$ show that
$$ ar ( \mathrm { EFGH } ) = \dfrac { 1 } { 2 } ar ( \mathrm { ABCD } )$$
Diagonal AC of a parallelogram ABCD bisects $$\angle A$$Show that:
(i) it bisects $$\angle C$$ also,
(ii) ABCD is a rhombus
(ii) ABCD is a rhombus
$$\therefore$$ $$\angle DAC=\angle BAC$$ ---- ( 1 )
Now,
$$AB\parallel DC$$ and $$AC$$ as traversal,
$$\therefore$$ $$\angle BAC=\angle DCA$$ [ Alternate angles ] --- ( 2 )
$$AD\parallel BC$$ and $$AAC$$ as traversal,
$$\therefore$$ $$\angle DAC=\angle BCA$$ [ Alternate angles ] --- ( 3 )
From ( 1 ), ( 2 ) and ( 3 )
$$\angle DAC=\angle BAC=\angle DCA=\angle BCA$$
$$\therefore$$ $$\angle DCA=\angle BCA$$
Hence, $$AC$$ bisects $$\angle C.$$
$$(ii)$$ In $$\triangle ABC,$$
$$\Rightarrow$$ $$\angle BAC=\angle BCA$$ [ Proved in above ]
$$\Rightarrow$$ $$BC=AB$$ [ Sides opposite to equal angles are equal ] --- ( 1 )
$$\Rightarrow$$ Also, $$AB=CD$$ and $$AD=BC$$ [ Opposite sides of parallelogram are equal ] ---- ( 2 )
From ( 1 ) and ( 2 ),
$$\Rightarrow$$ $$AB=BC=CD=DA$$
Hence, $$ABCD$$ is a rhombus.
Draw a quadrilateral $$A B C D.$$ Measure the four angles $$\angle A , \angle B , \angle C$$ and $$\angle D.$$ Verify whether their sum is $$360 ^ { \circ }.$$
In a quadrilateral $$ABCD$$, $$\angle A=60^{o},\angle B=105^{o}$$ and $$\angle D=50^{o}$$. Find $$\angle C$$.
In a quadrilateral $$ABCD,$$
$$\angle A=60^o,\,\angle B=105^o$$ and $$\angle D=50^o$$.We know, by angle sum property, in a quadrilateral sum of four angles is $$360^o.$$
$$\therefore$$ $$\angle A+\angle B+\angle C+\angle D=360^o$$
$$\Rightarrow$$ $$60^o+105^o+\angle C+50^o=360^o$$
$$\Rightarrow$$ $$215^o+\angle C=360^o$$
$$\Rightarrow$$ $$\angle C=145^o$$.
Hence, $$145^o$$ is the answer.
Can all the four angles of a quadrilateral be obtuse (or) acute angles? For reason for your answer.
In given figure, $$P$$ and $$Q$$ are points on the sides $$AB$$ and $$AC$$ respectively of $$\triangle ABC$$ such that $$AP=3.5\ cm,PB=7\ cm,AQ=3\ cm$$ and $$QC=6\ cm$$. If $$PQ=4.5\ cm$$, find $$BC$$.
We have
$$\left(\dfrac{AQ}{AC}\right)=\dfrac{3}{9}=\dfrac{1}{3}$$
$$\left(\dfrac{AP}{AB}\right)=\dfrac{3.5}{10.5}=\dfrac{1}{3}$$
In $$\Delta APQ$$ & $$\Delta ABC$$
$$\left(\dfrac{AQ}{AC}\right)=\left(\dfrac{AP}{AB}\right)$$
& $$\angle PAQ=\angle BAC$$
Thus $$\Delta APQ\sim \Delta ABC$$ by SAS criterion.
Hence $$\left(\dfrac{AQ}{AC}\right)=\left(\dfrac{PQ}{BC}\right)$$ (by cpst)
$$\Rightarrow \dfrac{1}{3}=\left(\dfrac{4.5}{BC}\right)$$
$$\Rightarrow BC=13.5$$cm.
A field in the form of parallelogram has one of its diagonals $$42\ m$$ and the perpendicular distance of this diagonals from either of the outlying vertices in $$10\ m\ 8\ dm$$. Find the area of the field.
R.E.F image
As 1dm=0.1m $$\Rightarrow 10m \, 8dm=10.8m$$
Area of ABCD=Aera of $$\bigtriangleup ADC+Aera(\bigtriangleup ABC)$$
Area of triangle = $$ \frac{1}{2} \times base \times height $$
$$ \Rightarrow $$Area of ABCD =$$\frac{1}{2}\times 10.8\times 42+\frac{1}{2}\times 10.8\times 42$$
=$$10.8\times 42=453.6m^{2}.$$
A quadrilateral has three actute angles each measuring $${ 70 }^{ 0 }$$, what is the measure of the fourth angle?
$$ABCD$$ is trapezium in which $$AB \parallel DC, AB=78\ cm, CD = 52\ cm, AD=28\ cm$$ and $$BC=30\ cm$$. Find the area of the trapezium.
R.E.F image
AS $$AB\left | \right |CD$$ and DM and
$$CN \perp AB \Rightarrow DM=CN=h$$
In $$\bigtriangleup DMA,DA^{2}=AM^{2}+DM^{2}$$
$$\Rightarrow 28^{2}=(26-X)^{2}+h^{2}...(1)$$
In $$\bigtriangleup CNB,CB^{2}=CN^{2}+NB^{2}$$
$$\Rightarrow 30^{2}=X^{2}+H^{2}...(2)$$
$$Eq^{n}(1)...Eq^{2}(2)-58\times 2=(26-x)^{2}-x^{2}$$
$$116=x^{2}-(x-26)^{2}=(2x-26)26$$
$$\Rightarrow x-13=\dfrac{116}{26\times 2}=\dfrac{29}{13}$$
By$$ eq^{n},h^{2}=30^{2}-\dfrac{29^{2}}{13^{2}}=900-\dfrac{841}{169}=895.02$$
$$\Rightarrow h=29.9cm$$
Area trapezium=$$\frac{1}{2}$$(sum of parallal sides)height=$$\frac{1}{2}(130)(29.2)$$
$$\Rightarrow Area = 1898cm^{2}$$
One side of a parallelogram is $$\dfrac{3}{4}$$ times its adjacent side. If the perimeter of the parallelogram is $$70\ cm$$, find the sides of the parallelogram .
In the given fig. $$AB||CD,\angle BDC=40^{o}$$, and $$\angle BAD=75^{o}$$. Find $$x,y,z$$.
Two angles of a quadrilateral measure $$80^{o}$$ each. The other two differ by $$20^{o}$$. What is the measure of each of the other angles.
In a parallelogram if the area, base & corresponding altitude are $$y^{2}, y-3, y+4$$. Then find the area of parallelogram in square unit.
In quadrilateral $$A B C D , \angle A + \angle D = 180 ^ { \circ } .$$ What special name can be given to this quadrilateral?
In quadrilateral $$ABCD,\angle A+\angle D=180^0$$ i.e, the sum of two consecutive angles is $$180^0$$, SO pair of opposite side $$AB$$ and $$CD$$ are parallel.
Therefore, quadrilateral $$ABCD$$ is trapezium. Hence, special name which can be given to this quadrilateral $$ABCD$$ is a trapezium
Prove that the line segments joining the mid points of the opposite sides of any quadrilateral bisect each other.
Now , since ABCD is a parallelogram thus it's diagonals bisect each other.
Hence proved.
In a Quadrilateral, the angles are $${x^0},{(x + 10)^0},\,{(x + 20)^0},\,{(x + 30)^{0.}}$$ Find the angles.
The ratio of two adjacent sides of a parallelogram is $$2:3$$. Its perimeter is $$50\ cm$$. Find its area if altitude corresponding to large side is $$10\ cm$$.
Prove that the points $$A(2, 3), B(-2, 2), C(-1, -2)$$ and $$D(3, -1)$$ are the vertices square $$ABCD$$.
$$ABCD$$ is a parallelogram, $$AE$$ is perpendicular to $$DC$$, If $$DC=20\ cm$$, $$AD=5\ cm$$ and the area of the parallelogram is $$40\ cm^{2}$$. Find $$DE$$.
The perimeter of a parallelogram is $$156\ cm$$. If one side measures $$36\ cm$$, find the measure of its adjacent side.
If the diagonals of a quadrilateral bisect each other then quadrilateral is a parallelograms. Prove it
R.E.F image In $$\bigtriangleup $$ AOB & $$\bigtriangleup $$ COD
AO=CO
$$\angle AOB=\angle COP$$
BO=DO
$$\therefore $$ $$\bigtriangleup AOB\cong \bigtriangleup COD by S.A.S.$$
Hence AB=CD (by cpct)
$$\angle 1=\angle 2$$
Hence $$,AB\left | \right |CD $$
Now in a quadrilateral
if me pair of opposite
side are equal & parallel
then it is a parallelograms
Thus,ABCD is a parallelograms proved.
Can the angles $$110 ^ { 0 } , 80 ^ { \circ } , 70 ^ { \circ }$$ and $$95 ^ { \circ }$$ be the angles of a quadrilateral? Give your reasons.
In the figure, $$\Box ABCD$$ is a parallelogram. $$AB=8, BC=12$$ and $$\angle B=30^{o}$$. Find $$A(\Box ABCD)$$.
Show that if the diagonal of a quadrilateral interest each other at right angle, then the sum of square of opposite sides are equal;.
$$ABCD$$ is a cyclic quadrilateral. If $$\angle{BAC}={50}^{\circ}$$ and $$\angle{DBC}={60}^{\circ}$$ then find $$\angle{BCD}$$
$$ABCD$$ is a parallelogram and $$AP$$ and $$CQ$$ are perpendiculars from vertices $$A$$ and $$C$$ on diagonal $$BD$$.Show that $$AP = CQ$$
What is the area of the square $$ABCD$$ shown in the diagram?
If the angles of quadrilateral are in the ratio $$3:5:9:13$$, find the value of smallest angle of quadrilateral.
Angles of a quadrilateral are $$(4x)^{\circ},\ 5(x + 2)^{\circ},\ (7x - 20)^{\circ}$$ and $$6(x + 3)^{\circ}$$.
(i) Find the value of $$x$$ (in degrees).
(ii) Each angle of the quadrilateral
$$ABCD$$ is a parallelogram and $$AP$$ and $$CQ$$ are perpendiculars from vertices $$A$$ and $$C$$ on diagonal $$BD$$.Show that $$\triangle{APB}\cong \triangle{CQD}$$
The parallel sides of a trapezium are 28 cm and 21 cm and each of the non-parallel sides is 17 cm. Find the area of the trapezium.
In triangle ACE-
By pythagoras theorem
$$AE^2=AC^2+CE^2$$
$$AE^2=289-12.25$$
$$AE=16.67 cm$$
Therefore height$$=16.67 cm$$
Hence area of the trapezium $$=\dfrac{1}{2}\times{sum \ of \ parallel \ sides}\times{height}$$
$$=\dfrac{1}{2}\times{49}\times{16.67}$$
$$=408.415 cm^2$$
The diagonal of a rhombus are $$15$$cm and $$18$$cm. Find its area and side.
In a quadrilateral $$ABCD$$, $$ AB \parallel CD $$ . If $$ \angle A : \angle D = 2:3 $$ and $$ \angle B : \angle C = 7:8 $$, find the measure of each angle.
Let $$\angle A$$ and $$\angle D$$ be $$2x$$ and $$3x.$$
$$\Rightarrow$$ $$\angle B:\angle C=7:8$$
Let $$\angle A$$ and $$\angle D$$ be $$7y$$ and $$8y.$$
Since, $$AB\parallel CD$$ and $$AD$$ is a transversal.
$$\therefore$$ $$\angle A+\angle D=180^o.$$ [ Interior angles on same side of transversal ]
$$\Rightarrow$$ $$2x+3x=180^o.$$
$$\Rightarrow$$ $$5x=180^o.$$
$$\Rightarrow$$ $$x=36^o$$
$$\Rightarrow$$ $$\angle A=2\times 36^o=72^o$$
$$\Rightarrow$$ $$\angle D=3\times 36^o=108^o.$$
Since, $$AB\parallel CD$$ and $$BC$$ is a transversal.
$$\therefore$$ $$\angle B+\angle C=180^o.$$ [ Interior angles on same side of transversal ]
$$\Rightarrow$$ $$7y+8y=180^o$$
$$\Rightarrow$$ $$15y=180^o$$
$$\therefore$$ $$y=12^o$$
$$\Rightarrow$$ $$\angle B=7\times 12^o=84^o$$
$$\Rightarrow$$ $$\angle C=8\times 12^o=96^o$$
Prove that any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.
Let $$ABCD$$ be a trapezium $$DC \parallel AB$$ and $$EF$$ is a line parallel to $$DC$$ and $$AB$$.
To prove :
$$\dfrac{AE}{ED}=\dfrac{BF}{FC}$$
Construction :
Join $$AC$$, meeting $$EF$$ in $$G$$.
Proof :
In $$\triangle ADC$$, we have
$$EG \parallel DC$$
$$\dfrac{AE}{ED}=\dfrac{AG}{GC}$$ [By Thale's theorem] .....$$(1)$$
In $$\triangle ABC$$, we have
$$GF \parallel AB$$
$$\dfrac{AG}{GC}=\dfrac{BF}{FC}$$ [By Thales's theorem] ......$$(2)$$
From $$(1)$$ & $$(2)$$, we get,
$$\dfrac{AE}{ED}=\dfrac{BF}{FC}$$
Hence, it is proved that any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.
In the figure $$ABCD$$ is a rectangle and $$DEC$$ is an equilateral. Find $$\angle DAE$$.
The given fig is drawn using more than one triangle:
$$\angle D A B + \angle A B C + \angle B C D + \angle C D A$$
Find $$x+y+z+w$$, in the given figure.
The angles of a quadrilateral are $$(x+10), (2x+5), (2x-20)$$ & $$(2x-5)$$, then find their measures.
Find the measure of the fourth angle of the quadrilateral if the three interior angles $$48^{o},\ 85^{o}$$ and $$140^{o}$$.
Find the four angles of equilateral, if the angles are in the ratios:
$$5:7:11:13$$
Find the angle measure $$x$$ in the following figures.
If $$D,E,F$$ are midpoint of the sides $$BC, CA$$ an $$AB$$ of triangle $$ABC$$, then show that $$BDEF$$ is a parallelogram
Find the four angles of quadrilateral, if the angles are in the ratios:
$$2:3:4:6$$
ABCD is equilateral $$\triangle BEC$$ is rectangle to BD cross ED at 'O' point $$\angle BOC=$$?
The angles in Quadrilateral are $$105,35,45,x$$ find $$x$$
In figure, $$ABCD$$ is a parallelogram
Prove that $$\dfrac{DQ}{DP}=\dfrac{AQ}{DC}$$
In the figure given above, $$ABCD$$ is a cyclic quadrilateral in which $$\angle BAD=75^{\circ},\; \angle ABD=58^{\circ}$$ and $$\angle ADC=77^{\circ}.$$ Find the measure of $$\angle CBD.$$
Do $$120,45,64,76$$ form Quadrilateral .give reasons.
Find the measure of angles x,y,z and a in the following parallelograms .
Is it possible to have a quadrilateral whose angles are of measures 105, 165, 55 and 45 . Give reason ?
From the following figure find ,
(i) x
(ii) $$\angle ABC$$
(iii) $$\angle ACD$$
$$A B C D$$ is a rhombus in which diagonals $$A C$$ and $$B D$$ intersect each other at $$O$$ . If $$\angle A C D = 55 ^ { \circ }$$ find $$\angle B D C$$
In the figure, given below, $$ABCD$$ is a cyclic quadrilateral in which $$\angle BAD=75^{o}; \angle ABD=58^{o}$$ and $$\angle ADC=77^{o}$$. Find: $$1)\angle CBD$$ $$2)\angle BCD$$
Two adjacent angles of a parallelogram are in the ratio $$1:5$$. Find all the angles of the parallelogram.
Two opposite angles of a parallelogram are $${( 3x-2)}^{\circ}$$ and $${(50-x)}^{\circ}$$.find the measure of each angle of the parallelogram.
$$ \ln a \Delta A B C, P $$ and $$ Q $$ are points on sides $$ A B $$ and $$ A C $$ respectively, such that PQ $$ \|\mathrm{BC} $$ . If $$ A P=2.4 \mathrm{cm}, A Q=2 \mathrm{cm}, Q C=3 cm$$ and $$ B C=10\mathrm{cm}, $$ find $$ A B $$ and $$ P Q $$
Find the area of a square whose perimeter is 160 $$ \mathrm{m} $$ . A floor 3.0 $$ \mathrm{m} $$ long and 2.0 $$ \mathrm{m} $$ wide is to be covered with square shaped tiles of length 10 $$ \mathrm{cm}. $$Find the cost of flooring it, if the cost of tiles is $$Rs.$$ $$150$$ per $$100$$ tiles.
In the figure, $$PQRS$$ is a parallelogram with $$PQ = 16$$ cm and $$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$$.Find the lengths of $$PN$$ and $$RM$$
The angles of a quadrilateral are in the ratio $$3:5:9:13$$. Find all the angles of the quadrilateral.
A square $$PQRS$$ is of side $$10\, \text{cm}$$ each. Write the length of its diagonal $$PR$$.
All the angles of a quadrilateral are equal. What special name is given to this quadrilateral?
All the angle of a quadrilateral are equal. Also ths sum of angles of a quadrilateral is $$360^0$$. Therefore, each angle of quadrilateral is $$90^0$$. So, given the quadrilateral is a rectangle. Hence, special name given quadrilateral is rectangle.
In the given fig. $$XY\parallel BC,\,\,BE\parallel CA\,\,and\,\,FC\parallel AB.$$
Prove that $$ar\left( {\Delta ABE} \right) = ar\left( {\Delta ACF} \right).$$
In given figure $$ABCD$$ is a rhombus. If $$\angle BAC=38^{o}$$, find:
$$\angle DAC$$
In the adjoining figure, $$ABC$$ is triangle. Through $$A,$$ $$B$$ and $$C$$ lines are drawn parallel to $$BC,$$ $$CA$$ and $$AB$$ respectively, which forms a $$\Delta PQR.$$ Show that $$2\left( {AB + BC + CA} \right) = PQ + QR + RP.$$
If two sides of a cyclic quadrilateral are parallel, prove that the remaining two sides equal and the diagonals are also equal.
Let $$ABCD$$ be the cyclic quadrilateral with $$AD$$ parallel to $$BC.$$ The center of the circle is $$O.$$
Join $$AC$$
Now, since $$AD$$ is parallel to $$BC$$ and $$AC$$ is the transverse
$$\angle DAC = \angle ACB$$ (interior opposite angles)
$$\angle DAC$$ is subtended by chord $$DC$$ on the circumference and $$\angle ACB$$ is subtended by chord AB.
If the angles subtended by $$2$$ chords on the circumference of the circle are equal, then the lengths of the chords are also equal.
Hence, $$DC = AB$$ i.e. the $$2$$ remaining sides of the cyclic quadrilateral are equal.
Now, $$BD$$ and $$AC$$ are the two diagonals.
In $$\triangle ABD$$ and $$\triangle DCA,$$
$$AB = CD$$ $$($$proved above$$)$$
$$\angle DCA = \angle DAB$$ $$($$angles subtended by the same chord $$AD)$$
$$\angle ACB = \angle DAC$$ $$($$proved above$$)$$
Hence, $$\triangle ABD$$ is congruent to $$\triangle DCA\ ($$by $$ASA)$$
Hence, $$BD = AC$$ (by CPCT)
In given figure $$ABCD$$ is a rhombus. If $$\angle BAC=38^{o}$$, find:
$$\angle ADC$$
In the quadrilateral (1) given below, $$AB||DC$$, $$E$$ and $$F$$ are mid point of $$AD$$ and $$BD$$ respectively. Prove that $$G$$ is mid point of $$BC$$.
Given:- $$AB \parallel DC, \; E$$ and $$F$$ are mid-points of $$AD$$ and $$BD$$ respectively.
To prove:- $$G$$ is the mid-point of $$BC$$
Proof:-
In $$\triangle{ABD}$$,
$$DF = BF \quad \left( \because \text{F is the mid-point of BD} \right)$$
Also, $$E$$ is the mid-point of $$AD \quad \left( \text{Given} \right)$$
Therefore,
$$EF \parallel AB$$ and $$EF = \cfrac{1}{2} AB ..... \left( 1 \right)$$
$$\Rightarrow EG \parallel CD \quad \left( \because AB \parallel CD \right)$$
Now,
$$F$$ is the mid-point of $$BD$$ and $$FG \parallel DC$$
$$\therefore G$$ is the mid-point of $$BC$$
Hence proved.
Show that opposite angles of parallelogram are equal.
We know, in parallelogram opposites sides are parallel.
So, $$AB\parallel DC$$ and $$AD\parallel BC$$
Since, $$AB\parallel DC$$ and $$AC$$ is the transversal
$$\Rightarrow$$ $$\angle BAC=\angle DCA$$ ---- ( 1 ) [ Alternate angles ]
Similarly, $$AD\parallel BC$$ and $$AC$$ is the transversal.
$$\Rightarrow$$ $$\angle DAC=\angle BCA$$ ---- ( 2 ) [ Alternate angles ]
Adding ( 1 ) and ( 2 ),
$$\Rightarrow$$ $$\angle BAC+\angle DAC=\angle DCA+\angle BCA$$
$$\Rightarrow$$ $$\angle BAD=\angle DCB$$
Similarly, we can prove, $$\angle ADC=\angle ABC$$
$$\therefore$$ We have proved that, opposite angles of a parallelogram are equal.
$$ABCD$$ is a quadrilateral in which $$\overline {AB} = \overline {CD} $$ and $$\overline {AD} = \overline {BC} $$. Show that it is a parallelogram.
REF.image.
Given:In $$\square ABCD$$
$$AB= DC \,and \,AD=BC$$
To prove: ABCD is a parallelogram
Construction: Draw diagonal AC and DB.
Proof: in $$\bigtriangleup ABC$$ and $$\bigtriangleup ADC$$
$$AD=BC$$ [Given]
$$AB=DC$$ [Given]
$$AC=AC$$ [Common side]
$$\therefore$$ By $$SSS$$ property
$$\bigtriangleup ADC \cong \bigtriangleup ACB$$
$$\therefore \angle DAC=\angle DCA$$
$$\therefore AB||DC$$ [By theorem]
In$$\bigtriangleup ABD\, and \,\bigtriangleup DCB$$
DB=DB [Common side]
AD=BC [Given]
AB=DC [Given]
$$\therefore \bigtriangleup ABD \cong \bigtriangleup DCB$$
$$\therefore AD||BC$$
Since $$AB||DC$$ and $$AD||BC$$.
$$\bigtriangleup ABCD$$ is parallelogram
In given figure $$ABCD$$ is a rhombus. If $$\angle BAC=38^{o}$$, find:
$$\angle ACB$$
Prove that opposite sides of a quadrilatral circumscribing a ciecle subtend supplementary angle at the centre of the circle.
Given: Let ABCD be the quadrilateral circumscribing the circle with centre O.
ABCD touches the circle at points P, Q, R and S.
To prove: Opposite sides subtends supplementary angles at centre.
i.e., $$\angle AOB+\angle COD=180^o$$
& $$\angle AOD+\angle BOC=180^o$$
Construction: Join OP, OQ, OR and OS.
Proof: Let us rename the angles
In $$\Delta AOP$$ and $$\Delta AOS$$,
AP$$=$$AS
(Length of tangents drawn from external point to a circle are equal)
AO$$=$$AO (common)
OP$$=$$OS (both radius)
$$\Delta AOP\cong \Delta AOS$$(SSS congruence rule)
$$\angle AOP=\angle AOS$$(CPCT)
i.e., $$\angle 1=\angle 8$$ …………….$$(1)$$
Similarly, we can prove
$$\angle 2=\angle 3$$ ……………….$$(2)$$
$$\angle 5=\angle 4$$ ………………..$$(3)$$
$$\angle 6=\angle 7$$ ……………….$$(4)$$
Now,
$$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^o$$
(sum of angles round a point is $$360^o$$)
$$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 1=360^o$$
$$\angle 1+\angle 2+\angle 5+\angle 6=\dfrac{360^o}{2}$$
$$(\angle 1+\angle 2)+(\angle 5+\angle 6)=180^o$$
$$\angle AOB+\angle COD=180^o$$
Hence both angles are supplementary.
similarly, we can prove
$$\angle BOC+\angle AOD=180^o$$
Hence proved.
In the quadrilateral given below, $$AB||DC$$, $$E$$ and $$F$$ are mid point of of non-parallel sides $$AD$$ and $$BC$$ respectively. Calculate $$EF$$ if $$AB=6 \text{ cm}$$ and $$DC=4 \text{ cm}$$
In the quadrilateral (1) given below, $$AB||DC$$, $$E$$ and $$F$$ are mid point of $$AD$$ and $$BD$$ respectively. Prove that
$$EG=\dfrac{1}{2}(AB+DC)$$
In the quadrilateral given below, $$AB||DC$$, $$E$$ and $$F$$ are mid point of of non-parallel sides $$AD$$ and $$BC$$ respectively. Calculate $$AB$$, if $$DC=8 cm$$ and $$EF=9 cm$$.
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
To prove:- $$ar{\left( \triangle{AOB} \right)} = ar{\left( \triangle{BOC} \right)} = ar{\left( \triangle{COD} \right)} = ar{\left( \triangle{AOD} \right)}$$
Proof:-
Let $$ABCD$$ be a parallelogram with diagonals $$AC$$ and $$BD$$
intersecting at $$O$$. Since the diagonals of a parallelogram bisect each other at the point of intersection.
Therefore,
$$AO = OC$$ and $$BO = OD$$
We know that the median of a triangle divides it into two equal parts.
Now,
In $$\triangle{ABC}$$,
$$\because BO$$ is median.
$$ar{\left( \triangle{AOB} \right)} = ar{\left( \triangle{BOC} \right)} ..... \left( 1 \right)$$
In $$\triangle{BCD}$$,
$$\because CO$$ is median.
$$ar{\left( \triangle{BOC} \right)} = ar{\left( \triangle{COD} \right)} ..... \left( 2 \right)$$
In $$\triangle{ACD}$$,
$$\because DO$$ is median.
$$ar{\left( \triangle{AOD} \right)} = ar{\left( \triangle{COD} \right)} ..... \left( 3 \right)$$
From equation $$\left( 1 \right), \left( 2 \right) \& \left( 3 \right)$$, we get
$$ar{\left( \triangle{AOB} \right)} = ar{\left( \triangle{BOC} \right)} = ar{\left( \triangle{COD} \right)} = ar{\left( \triangle{AOD} \right)}$$
Hence proved.
$$ A B C D $$ is a quadrilateral.
Prove that $$ ( A B + B C + C D + D A ) > ( A C + B D ) $$
In parallelogram $$ABCD,E$$ is the mid-point of $$AD$$ and $$F$$ is the mid-point of $$BC$$. Prove that $$BFDE$$ is a parallelogram.
Given: Parallelogram $$ABCD$$, in which $$E$$ and $$F$$ are mid-points of $$AD$$ and $$BC$$
To Prove: $$BFDE$$ is a parallelogram.
Proof: $$E$$ is mid-point of $$AD.$$ (Given)
$$DE=\dfrac{1}{2}AD$$
Also $$F$$ is midpoint of $$BC$$ (Given)
$$BF=\dfrac{1}{2}BC$$
But $$AD=BC$$ (opp. sides of ||gm)
$$\therefore BF=DE$$
Again $$AD\parallel BC$$ (opp. sides of ||gm)
$$DE\parallel BF$$
Now, in quadrilateral $$BFDE$$, $$DE\parallel BF$$ and $$DE=BF$$
Hence $$BFDE$$ is a parallelogram.
If the points $$A(1,-2),B(2,3),C(a,2)$$ and $$D(-4,-3)$$ from a parallelogram. Find the value of $$'a'$$ and height of the parallelogram taking $$AB$$ as base.
Consider the given points $$A(1, -2), B(2, 3), C(0, 2)$$ and $$D(-4, -3)$$
Since ABCD form a parallelogram, the midpoint of the diagonal AC should coincide with the midpoint of BD.
Mid point of AC$$=$$ Mid point of BD
$$\left[\dfrac{1+a}{2}, \dfrac{-2+2}{2}\right]=\left[\dfrac{2-4}{2}, \dfrac{3-3}{2}\right]$$
$$\left[\dfrac{a+1}{2}, 0\right]=\left[\dfrac{-2}{2}, 0\right]$$
Since the mid points coincide, we have
$$\dfrac{1+a}{2}=a$$
$$\Rightarrow a+1=-2$$
$$\Rightarrow a=-2-1$$
$$\Rightarrow a=-3$$
Now, area of $$\Delta ABC$$
$$=\dfrac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$$
$$=\dfrac{1}{2}\left|1(3-2)+2(2-(-2))+(-3)(-2-3)\right|$$
$$=\dfrac{1}{2}\left|1(1)+2(4)+(-3)(-5)\right|$$
$$=\dfrac{1}{2}\left|1+8+15\right|$$
$$=\dfrac{24}{2}=12$$ sq. units
$$ar(ABCD)$$ parallelogram $$=2\times $$ Area of triangle
$$=2\times 12$$
$$=24$$ sq. units
Area of parallelogram $$=$$Base $$\times$$ Height
$$\dfrac{Area}{Base} =height$$
So by the distance formula
$$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
$$=\sqrt{(-3+4)^2+(2+3)^2}$$
$$=\sqrt{1+25}$$
$$=\sqrt{26}$$
Thus height $$=\dfrac{24}{\sqrt{26}}=\dfrac{24}{\sqrt{26}}\times \dfrac{\sqrt{26}}{\sqrt{26}}=\dfrac{24\sqrt{26}}{26}=\dfrac{12\sqrt{26}}{13}$$.
Let $$ABCD$$ be a rectangle and let $$P,Q,R,S$$ be the mid-points of $$AB, BC, CD,DA$$ respectively. Prove that $$PQRS$$ is a rhombus.
In $$\triangle APS$$ and $$\triangle CRQ$$,
$$AP=CR$$ ($$P$$ and $$R$$ are the midpoints of $$AB$$ and $$CD$$)
$$AS=CQ$$ ($$S$$ and $$Q$$ are the midpoints of $$AD$$ and $$BC$$)
$$AP=CR$$ ($$P$$ and $$R$$ are the midpoints of $$AB$$ and $$CD$$)
$$AS=CQ$$ ($$S$$ and $$Q$$ are the midpoints of $$AD$$ and $$BC$$)
$$\angle SAP=\angle QCR $$ ($$90^o$$ angles of a rectangle)
$$\therefore \triangle APS \cong \triangle CRQ$$ (SAS postulate)
$$\therefore \triangle APS \cong \triangle CRQ$$ (SAS postulate)
$$\therefore SP=QR$$ (Corresponding sides of congruent triangles)
Similarly, we can show $$PQ=RS$$
$$\therefore PQ=QR=RS=SP$$
$$\therefore \Box PQRS$$ is a rhombus
$$\therefore \Box PQRS$$ is a rhombus
The length of a diagonals of square is $$\sqrt {2}(6+2\sqrt {5})cm$$. Then finds the length of sides of the square.
The three angles of a quadrilateral are $${ 80 }^{ \circ }$$, $${ 70 }^{ \circ }$$and $${ 120 }^{ \circ }$$. The fourth angle is.
ABCD is a quadrilateral in which P, Q, R, and S are mid-points of the sides AB, BC, CD, and DA. AC is a diagonal. Show that: PQ=SR
In parallelogram $$PQRS$$, $$\angle Q=\left( 4x-5 \right) ^{ \circ }$$ and $$\angle S=\left( 3x-10 \right) ^{ \circ }$$. Calculate : $$\angle Q$$ and $$\angle R$$ .
ABCD is a quadrilateral in which P, Q, R, and S are mid-points of the sides AB, BC, CD, and DA. AC is a diagonal. Show that: PQRS is a parallelogram.
Given,
$$ABCD$$ is quadrilateral in which $$P,Q,R,S$$ are midpoints or
$$AB,BC,CD$$ and $$DA$$
In $$\Delta ADC$$,
$$SR||AC$$ and $$SR=\dfrac{1}{2}AC$$.............(i)
In $$\Delta ABC$$,
$$PQ||AC$$ and $$PQ=\dfrac{1}{2}AC$$.............(ii)
From (i) and (ii)
$$SR=PQ=\dfrac{1}{2}AC$$ and $$PQ||SR||AC$$
So, $$SR=PQ$$ and $$SR||PQ$$
So, $$PQRS$$ is a parallelogram.
Check whether the points $$A \left(2,-1\right), B\left(3,4\right), C\left(-2,3\right)$$ and $$D\left(-3,2\right)$$ represents a Rhombus
AB$$=\sqrt{(2-3)^2+(-1-4)^2}$$
$$=\sqrt{26}$$
BC$$=\sqrt{(3+2)^2+(4-3)^2}$$
$$=\sqrt{26}$$
CD$$=\sqrt{(-2+3)^2+(3+2)^2}$$
$$=\sqrt{26}$$
AD$$=\sqrt{(2+3)^2+(-1+2)^2}$$
$$=\sqrt{26}$$
All length of sides are equal
$$\therefore$$ It can be a square or rhombus
$$\therefore$$ Length of diagonals.
AC$$=\sqrt{(2+2)^2+(-1-3)^2}$$
$$=\sqrt{32}$$
BD$$=\sqrt{(3+3)^2+(4+2)^2}$$
$$=\sqrt{72}$$
$$\therefore$$ length of diagonals are not equal
$$\therefore$$ ABCD is a rhombus not a square.
Hence proved.
$$E$$ is the mid-point of the median $$AD$$ of $$\Delta ABC$$ and $$BE$$ is produced to meet $$AC$$ at $$F$$ .Show that $$AF\frac{1}{3}AC.$$
Given: A $$\Delta ABC$$ in which $$E$$ is the mid point of median $$AB$$ and $$BE$$ is produced to meet $$AC$$ at $$F$$.
Refer image,
To prove: $$AF=\dfrac{1}{3}AC$$
Construction: Draw $$DG||BF$$ intersecting $$AC$$ at $$G$$.
Proof: In $$\Delta ADG,E$$ is the mid-point of $$AD$$ and $$EF||DG$$
$$\therefore AF=FG$$........(1) [Converse of mid-point theorem]
In $$\Delta FBC,D$$ is the midpoint if $$BC$$ and $$DG||BF$$.
$$\therefore FG=GC$$...........(2) [Converse of mid-point theorem]
From equation (1) and (2), we get,
$$AF=FG=GC$$.........(3)
But, $$AC=AF+FG+GC$$
$$\Rightarrow AC=AF+AF+AF$$ (Using (3))
$$\Rightarrow AC=3AF$$
$$\Rightarrow AF=\dfrac{1}{3}AC$$
Hence proved.
In a right angled $$\Delta BAC,\angle BAC={ 90 }^{ 0 }$$ segment AD, BE and CF are the medians
Prove that $$2\left( { AD }^{ 2 }+{ BE }^{ 2 }+{ CF }^{ 2 } \right) ={ 3BC }^{ 2 }$$
D and F are midpoints of sides $$ BC$$ and $$AB$$ .
By midpoint theorm, $$DF$$ will be parallel to $$AC$$ and is equal to $$\dfrac{1}{2} AC$$
$$DF =\dfrac{1}{2} AC$$
Therefore, $$\triangle ADF, \triangle ABE, \triangle AFC$$ are all right angle triangles.
By pythagoras theorm, we have:
In $$ \triangle ADF, AD^2=AF^2+DF^2$$
In $$ \triangle ABE, BE^2=AB^2+AE^2$$
In $$\triangle AFC, CF^2=AF^2+AC^2$$
In $$\triangle ABC, BC^2= AB^2+AC^2$$
$$LHS=2(AD^{2}+BE^{2}+CF^{2})$$
$$\displaystyle =2[(AF^{2}+DF^{2})+(AB^{2}+AE^{2})+(AF^{2}+AC^{2})]$$
$$\displaystyle =2[(\frac{AB^{2}}{4}+\frac{AC^{2}}{4})+(AB^{2}+\frac{AC^{2}}{4})+(\frac{AB^{2}}{4}+AC^{2})]$$
$$\displaystyle=2[(\frac{AB^{2}}{4}+\frac{AC^{2}}{4})+(AB^{2}+AC^{2})+(\frac{AC^{2}}{4}+\frac{AB^{2}}{4})]$$
$$\displaystyle=2[\frac{BC^{2}}{2}+BC^{2}]$$
$$=3 BC^{2}=R.H.S.$$
$$ABCD$$ is a quadrilateral with $$\angle{A}={80}^{\circ}, \angle{B}={40}^{\circ}, \angle{C}={140}^{\circ},\angle{D}={100}^{\circ}$$.Is $$ABCD$$ is a trapezium?Justify your answer.
In a quadrilateral , the angle are $$x^{0},(x+10)^{o},(x+20)^o,(x+30)^o$$.Find the angles
D,E and F are respectively the mid points of the sides $$AB,BC$$ and $$CA,$$ respectively of a $$\Delta ABC$$. Prove that by joining these mid-points $$D,E$$ and $$F$$ the $$\Delta ABC$$ is divided into four congruent triangle
Given:
A $$\Delta ABC$$ and $$\Delta DEF$$ which is formed by joining the mid-point $$D,E$$ and $$F$$ of the sides $$AB,BC$$ and $$CA$$ of $$\Delta ABC$$
Refer image,
To prove: $$\Delta DEF\cong \Delta EDB\cong \Delta CFE\cong \Delta FAD$$
Proof: $$DF$$ joins mid-points of sides $$AB$$ and $$AC$$ respectively of $$\Delta ABC$$
$$\therefore DF||BC$$ [Mid-point theorem]
$$\Rightarrow DF||BE$$
SImilarly, $$EF||BD$$
So, quadrilateral $$BEFD$$ is a parallelogram.
$$\Rightarrow \Delta DEF\cong \Delta EDB$$.........(1)
($$\because$$ DIagonal of parallelogram divides it into two congruent triangles)
Similarly, $$\Delta DEF\cong \Delta CFE$$......(2)
and $$\Delta DEF\cong \Delta FAD$$..........(3)
From equation (1),(2) and (3), we get
$$\Delta DEF\cong \Delta EDB\cong$$$$\Delta CFE\cong \Delta FAD$$
Hence, proved.
The angles of the quadrilateral are in the ratio 3:5:9:Find all the angles of the quadrilateral.
If two angles of a quadrilateral are $$40^{\circ}$$ and $$110^{\circ}$$ and the other two are in the ratio $$3:4$$, find these angles.
The angles of a quadrilateral are $$2x,\; 2x+15,\; 4x-12$$ and $$3x+5.$$ Find the value of $$x$$ and all angles of the quadrilateral.
P,Q and R are the mid points of sides BC,CA, and AB of ABC, and AD. is . the perpendicular from A to BC. prove that P,Q,R and D are concyclic.
We have to prove that $$R,D,P$$ and $$Q$$ are concylic.
Refer image,
Join $$RD,QD,PR$$ and $$PQ$$
$$\because RP$$ joints $$R$$ and $$P$$, the mid-point of $$AB$$ and $$BC$$
$$\therefore RP||AC$$ [Mid-point theorem]
$$\because RP$$ joints $$R$$ and $$P$$, the mid-point of $$AB$$ and $$BC$$
$$\therefore RP||AC$$ [Mid-point theorem]
Similarly, $$PQ||AB$$
$$\therefore ARPQ$$ is a $$||gm$$
So $$\angle RAQ=\angle RPQ$$ [Opposite angles of a $$||gm$$]
$$\because ABD$$ is a rt. $$\angle D\Delta$$ and $$DR$$ is a median,
$$\therefore RA=DR$$ and $$\angle 1=\angle 2$$........(2)
Similarly $$\angle 3=\angle 4$$........(3)
Adding (2) and (3), we get,
$$\angle 1+\angle 3=\angle 2+\angle 4$$
$$\angle RDQ=\angle RAQ$$
$$\angle RPQ$$ [Proved above]
Here, $$R,D,P$$ and $$Q$$ are conyclic
[$$\because \angle D$$ abd $$\angle P$$ are substended by $$RQ$$ on the same soide of it]
In figure, ABCD is a cyclic quadrilateral. If $$\angle$$BAC$$=50^o$$ and $$\angle$$DBC$$=60^o$$, then find $$\angle$$BCD.
The ratio between three angles of a quadrilateral is $$7:11:13$$ respectively, the value of the fourth angle of the quadrilateral is $$112^o$$. what is the difference between the largest and smallest angles of the quadrilateral ?
Show that line segment joining the midpoints of the opposite sides of a quadrilateral bisects each other.
$$ABCD$$ is a quadrilateral. Is $$AB + BC + CD + DA > AC + BD?$$
The diagram shows a quadrilateral on a $$1\ cm^{2}$$ grid.
Write down the mathematical name of this quadrilateral.
If following statement is true enter $$1$$ else enter $$0$$.The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is $$ 60^{\circ},$$ then the other angles are $$ 120^{\circ},60^{\circ},120^{\circ} $$.
Given, $$AB || CD$$.
Also, $$\angle A = 60^{\circ}$$.
We know that
$$\angle A + \angle B = 180^{\circ}$$ ___(i) [Co-interior angle are supplementary]
$$\angle B + \angle C = 180^{\circ}$$ __(ii) [Co-interior angle are supplementary]
Then, from $$(i)$$ & $$(ii)$$,
$$\angle A = \angle C$$.
Similarly, $$\angle B = \angle D$$.
From equation $$(i)$$,
$$60^{\circ} + \angle B = 180^{\circ}$$
Then, $$\angle B = 120^{\circ}$$.
Other angles are $$120^{\circ} , 60^{\circ} , 120^{\circ}$$.
Therefore, the given statement is true.
Hence, $$1$$ is the answer.
Give reasons for the following.
A rectangle can be thought of as a special parallelogram.
Find the missing values:
| S.No. | Base | Height | Area of the parallelogram |
| a. | $$20\ cm$$ | $$246$$ $$cm^2$$ | |
| b. | $$15\ cm$$ | $$154.5$$ $$cm^2$$ | |
| c. | $$8.4\ cm$$ | $$48.72$$ $$cm^2$$ | |
| d. | $$15.6\ cm$$ | $$16.38$$ $$cm^2$$ |
in a given figure a circle passes through points A,B,C and D.$$ \angle $$BAD=70$$ ^{\circ} $$,find x
$$ABCD$$ is quadrilateral. Is $$AB+BC+CD+DA<2(AC+BD)$$?
Since, the sum of lengths of any two sides in a triangle should be greater than the length of the third side.
Therefore, In $$\Delta AOB, AB<OA+OB$$ ….(i)In $$\Delta BOC, BC<OB+OC$$ …..(ii)
In $$\Delta COD, CD<OC+OD$$ ….(iii)
In $$\Delta AOD, DA<OD+OA$$ ……(iv)
Adding equation (i), (ii), (iii), (iv),
$$AB+BC+CD+DA<OA+OB+OB+OC+OC+OD+OD+OA$$
$$\Rightarrow AB+BC+CD+DA<2OA+2OB+2OC+2OD$$
$$\Rightarrow AB+BC+CD+DA<2[(AO+OC)+(DO+OB)]$$
$$\Rightarrow AB+BC+CD+DA<2(AC+BD)$$
Hence, it is proved.
Give reasons for the following.
A square can be thought of as a special rectangle.
A quadrilateral has three acute angles each measure $${80}^{o}$$. What is the measure of the fourth angle?
A quadrilateral has all its four angles of the same measure. What is the measure of each?
In a quadrilateral define the following:
Adjacent sides
Two sides of a quadrilateral which have a common end point are called its adjacent sides.
In the given figure, (AB, BC), (BC, CD), (CD, DA) and (DA, AB) are four pairs of adjacent sides of quadrilateral ABCD.
The angles of a quadrilateral are $${110}^{o},{72}^{o},{55}^{o}$$ and $${x}^{o}$$. Find the value of $$x$$.
Two angles of a quadrilateral are of the measure $${65}^{o}$$ and the other two angles are equal. What is the measure of each of these two angles?
The four angles of a quadrilateral are as $$3:5:7:9$$. Find the angles.
In a quadrailateral define the following:
Diagonals
The line segment joining the opposite vertices of quadrilateral is the diagonal of the quadrilateral.
In the given figure, the two diagonals are AC and BD.
Three angles of a quadrilateral are equal. The fourth angle is of measure $${150}^{o}$$. What is the measure(in degree) of equal angles?
In a quadrilateral define the following:
Adjacent angles
Two angles of a quadrilateral having a common arm are called its adjacent angles.
In the given figure,$$ (∠A, ∠B), (∠B, ∠C), (∠C, ∠D)$$ and $$(∠D, ∠A)$$ are four pairs of adjacent angles of quadrilateral $$ABCD.$$
The three angles of a quadrilateral are respectively equal to $${110}^{o},{50}^{o}$$ and $${40}^{o}$$. Find its fourth angle. (in degrees)
The angle between the altitudes of a parallelogram, through the same vertex of an obtuse angle of the parallelogram is $$60^{\circ}$$. Find the angles of the parallelogram.
$${\textbf{Step - 1:
Draw the parallelogram and name it}}{\text{.}}$$
$${\text{Suppose the parallelogram is ABCD}}{\text{.Here two obtuse angles are }}\angle {\text{ADC and }}\angle {\text{ABC}}{\text{.}}$$
$${\text{So from }}\angle {\text{ADC specifically from D point,to opposite arms DE and DF are two }}$$
$${\text{altitudes}}{\text{.}}$$
$${\text{Given that the angle between them is 60}}^\circ {\text{.}} \Rightarrow \angle {\text{EDF = 60}}^\circ {\text{.}}$$
$${\text{Obviously , }}\angle BED = \angle BFD = 90^\circ .$$
$${\textbf{Step - 2 :}}{\textbf{ Find }}\angle {\textbf{B }}{\text{.}}$$
$${\text{So here we have quadrilateral BEDF }}$$
$${\text{From here }}\angle {\text{FBE = 360}}^\circ - \left( {\angle {\text{BED + }}\angle {\text{BFD + }}\angle {\text{FDE}}} \right).$$
$$ \Rightarrow \angle {\text{FBE = 360}}^\circ - \left( {90^\circ + 90^\circ + 60^\circ } \right).$$
$$ \Rightarrow \angle {\text{FBE = 120}}^\circ {\text{.}}$$
$${\textbf{Step - 3 : Find the required angles using properties}}{\text{.}}$$
$${\text{We already have showed }}\angle {\text{FBE = 120}}^\circ {\text{.}}$$
$${\text{So, }}\angle {\text{B = }}\angle {\text{ ABC = 120}}^\circ {\text{.}}$$
$${\text{So opposite angle , }}\angle {\text{D = 120}}^\circ {\text{.}}$$
$${\text{We know that, }}\angle {\text{A,}}\angle {\text{B two are cointerior angles of parallelogram , so }}$$
$$\angle {\text{A + }}\angle {\text{B = 180}}^\circ {\text{.}}$$
$$ \Rightarrow \angle {\text{A = 180}}^\circ - {\text{ 120}}^\circ {\text{.}}$$
$$ \Rightarrow \angle {\text{A = 60}}^\circ {\text{.}}$$
$${\text{Also , }}\angle {\text{C is opposite to }}\angle {\text{A }}{\text{.So }}\angle {\text{C = 60}}^\circ {\text{. }}$$
$${\textbf{Therefore , the angles of parallelogram are 60}}^\circ {\textbf{,120}}^\circ {\textbf{,60}}^\circ {\textbf{,120}}^\circ {\textbf{.}}$$
In the figure, find the measure of $$\angle MPN$$.
In a quadrilateral $$ABCD$$, the angles $$A,B,C,D$$ are in the ratio $$1:2:4:5$$. Find the measure of each angle of the quadrilateral.
State the midpoint theorem.
In a quadrilateral, $$ABCD$$, $$CO$$ and $$DO$$ are the bisectors of $$\angle C$$ and $$\angle D$$ respectively. Prove that $$\angle COD=\cfrac{1}{2}(\angle A+\angle B)$$
In $$\Delta COD$$
$$\angle1+$$$$\angle$$$$COD$$ + $$\angle 2$$$$=180^0$$ ..... [Angle sum property of a triangle]
$$=\angle$$$$COD=$$$$180$$ $$-\angle$$ $$1-\angle2$$
$$=\angle$$$$COD=$$$$180$$ - $$\angle$$ $$1+\angle2$$
$$=\angle$$$$COD=$$$$180-$$ [$$\dfrac{1}{2}\angle$$$$c+\dfrac{1}{2}\angle d]$$
[$$OC$$ and $$OD$$ are bisectors of $$\angle C$$ and $$\angle D$$ represents]
$$=\angle$$$$COD$$ $$= 180 - \dfrac{1}{2}$$ ($$\angle C$$ +$$\angle D$$) ] $$ ............(1)$$
In quadrilateral $$ABCD$$,
$$\angle A+$$ $$\angle B+$$ $$\angle C+$$ $$\angle D=$$ $$ 360^0 $$
$$\angle C+$$ $$\angle D=360^0-$$ $$\angle A+$$ $$\angle B$$ $$........(2)$$
Substituting $$(2)$$ in $$(1)$$
$$=\angle COD$$ $$=180 -\dfrac{1}{2} (360-$$$$\angle A+$$ $$\angle B)$$
$$=\angle COD=180-180+\dfrac{1}{2}(\angle A+\angle B)$$
$$=\angle COD=180-180+\dfrac{1}{2}(\angle A+\angle B)$$
Hence, proved.
In the figure, $$ABCD$$ is a trapezium in which $$AB\parallel DC$$. If $$\angle A={55}^{o}$$ and $$\angle B={70}^{o}$$, find $$\angle C$$ and $$\angle D$$.
In the adjoining figure $$OD$$ is perpendicular to the chord $$AB$$ of a circle with centre $$O$$. If $$BC$$ is a diameter, show that $$AC\parallel DO$$ and $$AC=2\times OD$$
Three angles of a quadrilateral are $${75}^{o},{90}^{o}$$ and$${75}^{o}$$. Find the measure of the fourth angle.
In figure, AB $$||$$ CD. Find the values of x, y, z.
If the sum of the two angles of a quadrilateral is $${180}^{o}$$. What is the sum of the remaining two angles in degree?
In a trapezium $$ABCD, AB || DC, AB=a\ cm$$, and $$DC=b\ cm$$. If $$M$$ and $$N$$ are the midpoint of the nonparallel sides, $$AD$$ and $$BC$$ respectively then find the ratio of $$ar(DCNM)$$ and $$ar(MNBA)$$.
Join the diagonal $$DB$$ which cuts the line $$MN$$ at the point $$Y$$.
We know that $$M$$ and $$N$$ are the midpoints of $$AD$$ and $$BC$$
So we get
$$MN\parallel AB \parallel CD$$
Consider $$\triangle ADB$$
We know that $$m$$ is the midpoint of $$AD$$ and $$MY \parallel AB$$
So we get $$Y$$ as the midpoint of $$DB$$
It can be written as
$$MY=\dfrac{1}{2}AB$$
Consider $$\triangle BDC$$
We know that
$$NY=\dfrac{1}{2}CD$$
We get
$$MN=MY+YN$$
By substituting the values
$$MN=\dfrac{1}{2} AB+\dfrac{1}{2} CD$$
By taking out $$\dfrac{1}{2}$$ as common
$$MN=\dfrac{1}{2} (AB+CD)$$
It is given that $$Ab=a$$ and $$CD=b$$
So we get
$$MN=\dfrac{(a+b)}{2}$$
Construct $$DQ \bot AB$$ where $$DQ$$ cuts the line $$MN$$ at the point $$P$$
So we know that $$P$$ is the midpoint of $$DQ$$
It can be written as
$$DP=PQ=h$$
We get
Area of trapezium $$DCNM=\dfrac{1}{2} \times (MN+CD)\times DP$$
By substituting
Area of trapezium $$DCNM=\dfrac{1}{2}\left( \dfrac{a+b}{2}+b\right)h$$
So we get
Area of trapezium $$DCNM=\dfrac{h}{4}(a+3b)$$.....(1)
Area of trapezium $$MNBA =\dfrac{1}{2}\times (MN+AB)\times PQ$$
By substituting
Area of trapezium $$MNBA=\dfrac{1}{2}\left( \dfrac{a+b}{2}+a\right)h$$
So we get
Area of trapezium $$MNBA=\dfrac{h}{4(3a+b)}.....(2)$$
By dividing both the equations
Area of trapezium $$DCNM$$ $$\div$$ Area of trapezium $$MNBA=\left\{ h/4 (a+3b)\right\} \div \left\{ h/4(3a+b)\right\}$$
By cancelling the similar terms
Area of trapezium $$DCNM$$ \div Area of trapezium $$MNBA=(a+3b)\div(3a+b)$$
It can be written as
Area of trapezium $$DCNM$$: Area of trapezium $$MNBA=(a+3b):(3a+b)$$
Therefore, ratio of $$ar(DCNM)$$ and $$ar(MNBA)$$ is $$(a+3b):(3a+b)$$
$$P, Q, R, S$$ are respectively the midpoints of the sides $$AB, BC, CD$$ and $$DA$$ of parallelogram $$ABCD$$. Show that $$PQRS$$ is a parallelogram and also show that $$ar (||gm PQRS)=\dfrac{1}{2} \times ar(||gm ABCD)$$.
In the adjoining figure, $$D$$ and $$E$$ are respectively the mid point of sides $$AB$$ and $$AC$$ of $$\triangle ABC$$. If $$PQ ||BC$$ and $$CDP$$ and $$BEQ$$ are straight lines then prove that $$ar(\triangle ABQ)=ar (\triangle ACP)$$.
In the figure, $$ABCD$$ is a parallelogram in which $$\angle A={72}^{o}$$. Calculate $$\angle B,\angle C$$ and $$\angle D$$.
Prove that the sum of all the angles of a quadrilateral is $${360}^{o}$$.
In the figure, $$ABCD$$ is a parallelogram in which $$\angle DAB={80}^{o}$$ and $$\angle DBC={60}^{o}$$
$$P,Q,R$$ and $$S$$ are respectively the midpoints of the sides $$AB,BC,CD$$ and $$DA$$ of a quadrilateral $$ABCD$$. Show that $$PQ\parallel SR$$
Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.
In the figure, $$\text{D, E}$$ and $$\text{F}$$ are the midpoints of the sides $$\text{BC, CA}$$ and $$\text{AB}$$ respectively, of $$\triangle \text{ABC}$$. Show that $$\angle \text{EDF}=\angle \text{A}$$, $$\angle \text{DEF}=\angle \text{B}$$ and $$\angle \text{DFE}=\angle \text{C}$$
$$D$$ is the midpoint of $$BC$$ of $$\triangle ABC$$ and $$E$$ is the midpoint of $$BD$$. If $$O$$ is the midpoint of $$AE$$, prove that $$ar(\triangle BOE)=\dfrac{1}{8} ar(\triangle ABC)$$.
In the figure given below, $$ABCD$$ is a parallelogram. $$E$$ is a point on $$AB, CE$$ intersects the diagonal $$BD$$ at $$O$$ and $$EF \parallel BC$$. If $$AE : EB = 2 : 3$$, find
(i) $$EF : AD$$
(ii) area of $$\triangle BEF$$ : area of $$\triangle ABD$$
(iii) area of $$\triangle ABD$$ : area of trapezium $$AFED$$
(iv) area of $$\triangle FEO$$ : area of $$\triangle OBC$$.
Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square.
Consider $$\triangle ABC$$
We know that $$E$$ and $$F$$ are the midpoints
Based on the midpoint theorem
We know that $$EF\parallel AC$$ and $$EF=1/2AC$$
Consider $$\triangle ADC$$
we know that $$H$$ and $$G$$ are the midpoints
Based on the midpoint theorem
We know that $$HG\parallel AC$$ and $$HG=1/2AC$$
so we get $$EF\parallel HG$$ and
$$EF=HG=1/2AC$$.....(1)
consider $$\triangle BAD$$
We know that $$H$$ and $$E$$ are the midpoints
Based on the midpoint theorem
We know that $$HE\parallel BD$$ and $$HE=1/2BD$$
Consider $$\triangle BCD$$
we know that $$G$$ and $$F$$ are midpoints
Based on the midpoint theorem
We know that $$GF\parallel BD$$ and $$GF=1/2BD$$
So we get $$HE\parallel GF$$ and
$$HE=GF=1/2BD$$.......(2)
We know that the diagonals of a square are equal
so we get
$$AC=BD$$....(3)
By using equations (1), (2), (3)
So we get $$GF\parallel BD$$ and $$HE\parallel GF$$
we have
$$EF=GH=GH=HE$$
we know that $$EFGH$$ is a rhombus
from the figure we know that the diagonals of a square are equal and intersect ar right angles
$$\angle DOC={90}^{o}$$
we know that the sum of adjacent angles of parallelogram is $${180}^{o}$$
it can be written as
$$\angle DOC+\angle GKO={180}^{o}$$
by substituting values
$${90}^{o}+\angle GKO={180}^{o}$$
$$\angle GKO={90}^{o}$$
from the figure we know that $$\angle GKO$$ and $$\angle EFG$$ are
corresponding angles
we get $$\angle GKO=\angle EFG={90}^{o}$$
Hence, $$EFGH$$ is a square.
therefore, it is proved that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is square.
Prove that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other.
Construct lines $$KH,BD$$ and $$GL$$
We know that $$K$$ and $$H$$ are the midpoints of $$AD$$ and $$AB$$
Consider $$\triangle ABD$$
Using the midpoint theorem
We get
$$KH=1/2BD$$
consider $$\triangle CBD$$
Based on the midpoint theorem
$$GL=1/2BD$$
so we get
$$KH=GL$$
Consider $$\triangle KOH$$ and $$\triangle GOL$$
We know that $$KH=GL$$
From the figure we know that alternate angles are equal
So we get
$$\angle OKH=\angle GLO$$ and $$\angle OHK=\angle OGL$$
By SAS congruence criterion
$$\triangle KOH \cong \triangle GOL$$
$$OG=OH$$ and $$OK=OL$$ (c.p.c.t)
So we know that $$GH$$ and $$GL$$ bisect each other
Therefore, it is proved that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other.
The diagonals of a quadrilateral $$ABCE$$ are equal. Prove that the quadrilateral formed by joining the midpoints of its sides is a rhombus.
consider $$\triangle ABC$$
we know that $$P$$ And $$Q$$ are the midpoints of $$AB$$ and $$BC$$
so we get $$PQ\parallel AC$$ and $$
PQ=\dfrac 12AC$$....(1)
Consider $$\triangle BCD$$
we know that $$Q$$ and $$R$$ are the midpoints of $$BC$$ and$$CD$$
so we get $$QR\parallel BD$$
$$QR=\dfrac 12BD$$....(2)
Consider $$\triangle ADC$$
we know that $$S$$ and $$R$$ are the midpoints of $$AD$$ and $$CD$$
so we get $$RS\parallel AC$$
$$RS=\dfrac 12AC$$
Consider $$\triangle ABD$$
we know that $$P$$ and $$S$$ are the midpoints of $$AB$$ and $$AD$$
so we get $$SP\parallel BD$$
$$SP=\dfrac 12BD$$....(4)
Using all equations
$$PQ\parallel RS$$ and $$QR\parallel SP$$
thus, $$PQRS$$ is a parallelogram
it is given that $$AC=BD$$
we can write it as
$$\dfrac 12AC=\dfrac 12BD$$
from equations we get
$$PQ=QR=RS=SP$$
$$PQRS$$ is a rhombus
therefore it is proved that the quadrilateral formed by joining the midpoints of its sides is a rhombus
Prove that the lines joining the middle points of opposite sides of a quadrilateral and the line joining the middle points of its diagonals meet in a point and bisect one another.
The lines joining the middle points of opposite sides of a quadrilateral and the line joining the middle points of its diagonals meet in a point and bisect one another, Hence proved.
In Figure $$,\dfrac{A O}{O C}=\dfrac{B O}{O D}=\dfrac{1}{2}$$ and $$A B=5 \ {cm} .$$ Find the value of $$DC$$.
The diagonals of a quadrilateral $$ABCD$$ are perpendicular to each other. Prove that the quadrilateral formed by joining the midpoints of its sides is a rectangle.
Consider $$\triangle ABC$$
We know that $$P$$ and $$Q$$ are the midpoints of $$AB$$ and $$BC$$
So we get $$PQ\parallel AC$$ and
$$PQ=\dfrac{1}{2}AC$$....(1)
Consider $$\triangle ADC$$
We know that $$R$$ and $$S$$ are the midpoints of $$CD$$ and $$AD$$
So we get $$RS\parallel AC$$ and
$$RS=\dfrac{1}{2}AC$$....(2)
Using the equations
We get $$PQ\parallel RS$$ and $$PQ=RS$$
We know that a pair of opposite sides are equal in a parallelogram
We know that $$AC$$ and $$BD$$ are the diagonals intersecting at point $$O$$
Consider $$\triangle ABD$$
We know that $$P$$ and $$S$$ are the midpoints of $$AD$$ and $$AB$$
So we get
$$PS\parallel BD$$
It can be written as
$$PN\parallel MO$$
Based on equation (1)
We get
$$PM\parallel NO$$
So we know that$$PM\parallel NO$$ and$$PN\parallel MO$$
We know that opposite angles are equal in parallelogram
$$\angle MPN=\angle MON$$
We know that
$$\angle BOA=\angle MON$$
So we get
$$\angle MPN=\angle BOA$$
We know that $$AC\bot BD$$ and $$\angle BOA={90}^{o}$$
It can be written as
$$\angle MPN={90}^{o}$$
So we get
$$\angle QPS={90}^{o}$$
So we know that$$PQRS$$ is a parallelogram with $$\angle QPS={90}^{o}$$
Therefore it is proved that the quadrilateral formed by joining the midpoints of its sides is a rectangle.
Fill in the blanks to make the statements true.
The sum of all _______ of a quadrilateral is $$ 360^{\circ} $$ .
Fill in the blanks to make the statements true.
In quadrilateral HOPE , the pairs of opposite sides are _________
Fill in the blanks to make the statements true.
A quadrilateral in which a pair of opposite sides is parallel is ________ .
Fill in the blanks to make the statements true.
In quadrilateral WXYZ , the pairs of opposite angles are _________.
Fill in the blanks to make the statements true.
If all sides of a quadrilateral are equal , it is a _______ .
Fill in the blanks to make the statements true.
________ is a regular quadrilateral.
Solve the following :
Two sticks each of length $$ 7 \,cm $$ are crossing each other such that they bisect each other at right angles. What shape is formed by joining their end points ? Give reason .
A line $$ l $$ is parallel to line $$m$$ and a transversal $$p$$ intersects them at $$X , Y$$ respectively . Bisectors of interior angles at $$X$$ and $$Y$$ intersect at $$P$$ and $$Q$$ . Is $$PXQY$$ a rectangle ? Give reason.
Given
$$l\parallel m $$ and
Bisectors of interior angles at $$X$$ and $$Y$$ intersect at $$P$$ and $$Q$$
Now,
$$\angle DXY=\angle XYA\\$$
$$\Rightarrow \dfrac{\angle DXY}{2} = \dfrac{\angle XYA}{2} \\$$
$$\Rightarrow \angle PXY=\angle QYX\\$$
$$\Rightarrow XP \parallel QY \qquad...(i)\\$$
Similarly,
$$\angle CXY=\angle BYX\\$$
$$\Rightarrow \dfrac{\angle CXY}{2} = \dfrac{\angle BYX}{2} \\$$
$$\Rightarrow \angle QXY=\angle PYX\\$$
$$\Rightarrow XQ \parallel PY \qquad...(ii)\\ $$
Now,
$$ \angle DXY + \angle XYB = 180^{\circ} \\$$
$$\Rightarrow \dfrac{\angle DXY}{2} + \dfrac{\angle XYB}{2} = \dfrac{180^{\circ}}{2} \\$$
$$\angle PXY+\angle PYX = 90^{\circ} \qquad...(iii)\\$$
In $$ \triangle XYP $$,
$$ \angle PXY+\angle PYX + \angle P = 180^{\circ} $$
$$ \Rightarrow 90^{\circ} + \angle P = 180^{\circ} $$ [From eq $$(iii)$$]
$$ \Rightarrow \angle P =90^{\circ} \qquad...(iv)]$$
$$\therefore$$ from $$(i),(ii)$$ and $$(iv)$$,
$$PXQY$$ is a parallelogram with one angle $$90^\circ$$
Hence, $$PXQY$$ is a rectangle.
Fill in the blanks to make the statement true.
If the diagonals of a quadrilateral bisect each other , it is a __________.
Fill in the blanks to make the statements true.
If only one diagonal of a quadrilateral bisects the other , then the quadrilateral is known as ________
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then
$$ar(\Delta BDE) = \dfrac{1}{4} ar (\Delta ABC).$$ Write True or False and justify your answer:
$$\Delta ABC$$ and $$\Delta BDE$$ are two equilateral triangles.
Let each sides of triangle $$ABC$$ be $$x.$$
Again, D is the mid-point of BC, so each side of triangle BDE is $$\dfrac{x}{2}.$$
Now, $$\dfrac{ar (\Delta BDE)}{ar (\Delta ABC)} = \dfrac{ \dfrac{\sqrt 3}{4} (\frac{x}{2})^2}{\dfrac{\sqrt 3}{4} x^2} = \dfrac{1x^2}{4x^2} = \dfrac{1}{4}$$
Hence, $$ar(\Delta BDE) = \dfrac{1}{4} ar(\Delta ABC)$$
$$\therefore$$ The given statement is true.
In the given figure, ABCD and EFGD are two parallelogram and G is the mid-point of CD. Then $$ar(\Delta DPC) = \dfrac{1}{2} ar (\parallel EFGD).$$ Write True or False and justify your answer:
Find the angle measure x in the following figures:
Find the angle measure x in the following figures:
In trapezium $$ABCD, AB \parallel DC$$ and L is the mid-point of BC. Through L, a line $$PQ \parallel AD$$ has been drawn which meets AB is P and DC produced in Q (Fig. 9.18). Prove that $$ar (ABCD) = ar(APQD)$$
Diagonals of a quadrilateral ABCD bisect each other. If $$\angle A = 35^\circ,$$ determine $$\angle B.$$
As the diagonals of quadrilateral ABCD bisect each other, so ABCD is a parallelogram.
Now, $$ABCD$$ is a parallelogram
$$\therefore \angle A + \angle B = 180^\circ$$
[$$\because$$ Adjacent angles of a parallelogram are supplementary]
$$\therefore 35^\circ + \angle B = 180^\circ$$
$$\Rightarrow \angle B = 180^\circ - 35^\circ = 145^\circ$$
Can all the angles of a quadrilateral be acute angles ? Give reason for your answer.
Can all the angles of a quadrilateral be right angles ? Give reason for your answer.
Find the angle measure x in the following figures:
We know that the sum of four angles of a quadrilateral is equal to $$ 360^{\circ}. $$
Consider Quadrilateral MNOP,
$$\begin{array}{c}\angle \mathrm{M}+\angle \mathrm{N}+\angle \mathrm{O}+\angle \mathrm{P}=360^{\circ} \\90^{\circ}+\mathrm{y}+83^{\circ}+110^{\circ}=360^{\circ} \\283^{\circ}+\mathrm{y}=360^{\circ} \\\mathrm{y}=360^{\circ}-283^{\circ} \\\mathrm{y}=77^{\circ}\end{array}$$
Therefore, the value of $$ y $$ is $$ 110^{\circ} $$
We know that, sum of angles linear pair is equal to $$ 180^{\circ} $$
$$\begin{array}{c}\text { So, } y+x=180^{\circ} \\77^{\circ}+x=180^{\circ}\end{array}$$
By transposing we get,
$$\begin{array}{l}x=180^{\circ}-77^{\circ} \\x=103^{\circ}\end{array}$$
Therefore, the value of $$ x $$
$$ 103^{\circ} $$.
Consider Quadrilateral MNOP,
$$\begin{array}{c}\angle \mathrm{M}+\angle \mathrm{N}+\angle \mathrm{O}+\angle \mathrm{P}=360^{\circ} \\90^{\circ}+\mathrm{y}+83^{\circ}+110^{\circ}=360^{\circ} \\283^{\circ}+\mathrm{y}=360^{\circ} \\\mathrm{y}=360^{\circ}-283^{\circ} \\\mathrm{y}=77^{\circ}\end{array}$$
Therefore, the value of $$ y $$ is $$ 110^{\circ} $$
We know that, sum of angles linear pair is equal to $$ 180^{\circ} $$
$$\begin{array}{c}\text { So, } y+x=180^{\circ} \\77^{\circ}+x=180^{\circ}\end{array}$$
By transposing we get,
$$\begin{array}{l}x=180^{\circ}-77^{\circ} \\x=103^{\circ}\end{array}$$
Therefore, the value of $$ x $$
$$ 103^{\circ} $$.
What type of quadrilateral do the points $$ A(2 , -2) , B (7 , 3) , C(11 , -1) $$ and $$ D(6 , -6) $$ taken in that order form?
Consider the following parallelograms. Find the values of $$ x, y, z $$ in each.
$${\textbf{Step - 1: Finding angle y}}{\text{.}}$$
$${\text{We know that the sum of angles on a straight line is 180}}^\circ {\text{,we can write,}}$$
$$\angle DAB + 120^\circ = 180^\circ $$
$$\angle DAB = 60^\circ $$
$${\text{Also ,the sum of adjacent interior angles of a parallelogram is 180}}^\circ .$$
$$\angle DAB + \angle CBA = 180^\circ $$
$$60^\circ + y = 180^\circ $$
$$y = 120^\circ $$
$${\textbf{Step - 2: Finding angle x}}{\text{.}}$$
$${\text{As opposite angles of a parallelogram are equal,}}$$
$$x = \angle DAB \Rightarrow x = 60^\circ $$
$${\textbf{Step - 3: Finding angle z}}{\text{.}}$$
$${\text{As the sum of adjacent interior angles of a parallelogram is 180}}^\circ .$$
$$\angle CDA + \angle BAD = 180^\circ $$
$$z + \angle BAD = 180^\circ $$
$$z + 60^\circ = 180^\circ $$
$$z = 120^\circ $$
$${\textbf{Thus , x = 60}}\mathbf{^\circ ,y = 120^\circ {\text{ and z = 120}}^\circ .}$$
Consider the following parallelograms. Find the values of $$ x, y, z $$ in each.
$${\textbf{Step - 1: Finding angle x}}{\text{.}}$$
$${\text{As we know that the sum of the interior angles of a parallelogram is 180}}^\circ ,$$
$$\angle CDA + \angle BAD = 180^\circ $$
$$70^\circ + \angle BAC + \angle OAD = 180^\circ $$
$$70^\circ + \angle OAD + x = 180^\circ ..(i)$$
$${\text{Also the sum of angles of a straight line is 180}}^\circ ,{\text{we can write}}$$
$$\angle BOA + \angle AOD = 180^\circ $$
$$\angle AOD = 180^\circ - 100^\circ = 80^\circ $$
$${\text{In }}\Delta AOD,{\text{sum of all the angle is 180}}^\circ ,{\text{we can write}}$$
$$\angle ODA + \angle DAO + \angle AOD = 180^\circ $$
$$\angle OAD = 180^\circ - 40^\circ - 80^\circ = 180^\circ - 120^\circ = 60^\circ $$
$${\text{Therefore, from equation (i)70}}^\circ + 60^\circ + x = 180^\circ $$
$$x = 50^\circ $$
$${\textbf{Step - 2: Finding angle z}}{\text{.}}$$
$${\text{As we know that the alternate angles are equal, and we can observe that }}$$
$$\angle CBD{\text{ and }}\angle BDA{\text{ are alternate angles, we can write}}$$
$$z = 40^\circ $$
$${\textbf{Step - 3: Finding angle y}}{\text{.}}$$
$${\text{Also we know that vertically opposite angles are equal and from the figure we can see that}}$$
$$\angle DOA = \angle COB \Rightarrow \angle COB = 80^\circ $$
$${\text{In }}\Delta OBC,{\text{ sum of all the angles is 180}}^\circ ,{\text{we can write}}$$
$$\angle COB + \angle OBC + \angle BCO = 180^\circ $$
$$80^\circ + 40^\circ + y = 180^\circ $$
$$y = 60^\circ $$
$${\textbf{Thus, x = 50}}\mathbf{^\circ ,y = 60^\circ ,z = 40^\circ .}$$
Name the quadrilateral whose diagonals are perpendicular bisectors of each other.
In the following figures, HOPE is parallelograms. Find the measures of angles $$ \mathrm{x}, \mathrm{y} $$ and $$ \mathrm{z} . $$ State the properties you use to find them.
Consider the parallelogram HOPE,
We know that, sum of interior and exterior angle is equal to $$ 180^{\circ} $$
$$\begin{array}{l}\angle \mathrm{HOP}+\angle \mathrm{POM}=180^{\circ} \\\angle \mathrm{HOP}+70^{\circ}=180^{\circ}-70^{\circ} \\\angle \mathrm{HOP}=110^{\circ}\end{array}$$
Then, $$ \angle \mathrm{HEP}=\angle \mathrm{HOP} $$
$$x=110^{\circ}$$... [because in parallelogram opposite angles are equal]
$$\angle \mathrm{OPH}=\angle \mathrm{PHE}$$
$$y=40^{\circ}$$... [because alternate angles are equal]
Now, consider the triangle HOP
We know that, sum of measures of interior angles of triangle is equal to $$ 180^{\circ} $$.
$$\begin{array}{l}\angle \mathrm{PHO}+\angle \mathrm{HOP}+\angle \mathrm{OPH}=180^{\circ} \\\mathrm{z}+110^{\circ}+40^{\circ}=180^{\circ} \\\mathrm{z}+150^{\circ}=180^{\circ} \\\mathrm{z}=180^{\circ}-150^{\circ} \\\mathrm{z}=30^{\circ}\end{array}$$
Therefore, value of $$ x=110^{\circ}, y=40^{\circ} $$ and $$ z=30^{\circ} $$
We know that, sum of interior and exterior angle is equal to $$ 180^{\circ} $$
$$\begin{array}{l}\angle \mathrm{HOP}+\angle \mathrm{POM}=180^{\circ} \\\angle \mathrm{HOP}+70^{\circ}=180^{\circ}-70^{\circ} \\\angle \mathrm{HOP}=110^{\circ}\end{array}$$
Then, $$ \angle \mathrm{HEP}=\angle \mathrm{HOP} $$
$$x=110^{\circ}$$... [because in parallelogram opposite angles are equal]
$$\angle \mathrm{OPH}=\angle \mathrm{PHE}$$
$$y=40^{\circ}$$... [because alternate angles are equal]
Now, consider the triangle HOP
We know that, sum of measures of interior angles of triangle is equal to $$ 180^{\circ} $$.
$$\begin{array}{l}\angle \mathrm{PHO}+\angle \mathrm{HOP}+\angle \mathrm{OPH}=180^{\circ} \\\mathrm{z}+110^{\circ}+40^{\circ}=180^{\circ} \\\mathrm{z}+150^{\circ}=180^{\circ} \\\mathrm{z}=180^{\circ}-150^{\circ} \\\mathrm{z}=30^{\circ}\end{array}$$
Therefore, value of $$ x=110^{\circ}, y=40^{\circ} $$ and $$ z=30^{\circ} $$
Consider the following parallelograms. Find the values of $$ x, y, z $$ in each.
$${\textbf{Step - 1: Finding angle y}}{\text{.}}$$
$${\text{As the opposite angles of a parallelogram are equal, we can write}}$$
$$y = 120^\circ $$
$${\textbf{Step - 2: Finding angle x}}{\text{.}}$$
$${\text{Considering }}\Delta ACD,{\text{we know that the sum of all angles of a triangle is 180}}^\circ $$
$$\angle DCA + \angle CAD + \angle ADC = 180^\circ $$
$$35^\circ + x + y = 180^\circ $$
$$x = 180^\circ - 35^\circ - y$$
$$x = 180^\circ - 35^\circ - 120^\circ $$
$$x = 25^\circ $$
$${\textbf{Step - 3: Finding angle z}}{\text{.}}$$
$${\text{As we know that alternate angles are equal, and from the figure we can see that}}$$
$$\angle DAC{\text{ and }}\angle ACB{\text{ are alternate angles, we can write}}$$
$$\angle DAC = \angle ACB$$
$$z = x = 25^\circ $$
$${\textbf{Thus, x = 25}}\mathbf{^\circ ,y = 120^\circ ,z = 25^\circ .}$$
$$ABCD$$ is a rhombus with $$P, Q$$ and $$R$$ as midpoints of $$AB, BC$$ and $$CD$$ respectively. Prove that $$PQ \bot QR$$.
It is given that ,
$$ABCD$$ is a rhombus with $$P, Q$$ and $$R$$ as mid-points of $$AB, BC$$ and $$CD$$
To prove:
$$PQ \bot QR$$
Construct: Join $$AC$$ and $$BD$$
Proof:
We know that,
Diagonals of a rhombus intersect at right angle.
$$\therefore \angle MON=90^o $$ ...(1)
In $$\triangle BCD$$
$$Q$$ and $$R$$ are mid-points of $$BC$$ and $$CD$$.
$$RQ \parallel DB$$ and $$RQ=1/2 DB$$ ...(2) [ By basic proportionality theorem ]
Now, $$RQ \parallel DB \Rightarrow MQ \parallel ON$$
Similarly, $$ QP\parallel CA\Rightarrow QN \parallel MO$$
Therefore, in quadrilateral $$MQNO$$,
$$ MQ \parallel ON\,, \ QN \parallel MO$$ and $$ \angle MON=90^o$$
$$\Rightarrow MQNO$$ is a square.
$$\Rightarrow NQ \bot MQ$$ or $$PQ \bot QR$$
Hence, it is proved.
In the given figure, find x+y+z+w.
Let MNOP is a quadrilateral,
We know that, sum of four angles of quadrilateral is equal to $$ 360^{\circ} $$.
$$\begin{array}{c}\angle \mathrm{M}+\angle \mathrm{N}+\angle \mathrm{O}+\angle \mathrm{P}=360^{\circ} \\130^{\circ}+80^{\circ}+70^{\circ}+\angle \mathrm{P}=360^{\circ} \\280^{\circ}+\angle \mathrm{P}=360^{\circ} \\\angle \mathrm{P}=360^{\circ}-280^{\circ} \\\angle \mathrm{P}=80^{\circ}\end{array}$$
We know that, sum of angles linear pair is equal to $$ 180^{\circ} $$
$$\text { So, } x+130^{\circ}=180^{\circ}$$
By transposing we get,
$$\begin{array}{l}x=180^{\circ}-130^{\circ} \\x=50^{\circ}\end{array}$$
Therefore, the value of $$ x $$ is $$ 50^{\circ} $$.
Then, $$ y+80^{\circ}=180^{\circ} $$
By transposing we get,
$$\begin{array}{l}y=180^{\circ}-80^{\circ} \\y=100^{\circ}\end{array}$$
Therefore, the value of $$ y $$ is $$ 100^{\circ} $$.
$$\text { Similarly, } z+70=180^{\circ}$$
By transposing we get,
$$\begin{array}{l}z=180^{\circ}-70^{\circ} \\z=110^{\circ}\end{array}$$
Therefore, the value of $$ z $$ is $$ 110^{\circ} $$.
$$\text { Similarly, } w+80=180^{\circ}$$
By transposing we get,
$$\begin{array}{l}z=180^{\circ}-80^{\circ} \\z=100^{\circ}\end{array}$$
Therefore, the value of $$ z $$ is $$ 110^{\circ} $$.
Hence, $$ x+y+z+w $$
$$\begin{array}{l}=50^{\circ}+100^{\circ}+110^{\circ}+100 \\=360^{\circ}\end{array}$$
We know that, sum of four angles of quadrilateral is equal to $$ 360^{\circ} $$.
$$\begin{array}{c}\angle \mathrm{M}+\angle \mathrm{N}+\angle \mathrm{O}+\angle \mathrm{P}=360^{\circ} \\130^{\circ}+80^{\circ}+70^{\circ}+\angle \mathrm{P}=360^{\circ} \\280^{\circ}+\angle \mathrm{P}=360^{\circ} \\\angle \mathrm{P}=360^{\circ}-280^{\circ} \\\angle \mathrm{P}=80^{\circ}\end{array}$$
We know that, sum of angles linear pair is equal to $$ 180^{\circ} $$
$$\text { So, } x+130^{\circ}=180^{\circ}$$
By transposing we get,
$$\begin{array}{l}x=180^{\circ}-130^{\circ} \\x=50^{\circ}\end{array}$$
Therefore, the value of $$ x $$ is $$ 50^{\circ} $$.
Then, $$ y+80^{\circ}=180^{\circ} $$
By transposing we get,
$$\begin{array}{l}y=180^{\circ}-80^{\circ} \\y=100^{\circ}\end{array}$$
Therefore, the value of $$ y $$ is $$ 100^{\circ} $$.
$$\text { Similarly, } z+70=180^{\circ}$$
By transposing we get,
$$\begin{array}{l}z=180^{\circ}-70^{\circ} \\z=110^{\circ}\end{array}$$
Therefore, the value of $$ z $$ is $$ 110^{\circ} $$.
$$\text { Similarly, } w+80=180^{\circ}$$
By transposing we get,
$$\begin{array}{l}z=180^{\circ}-80^{\circ} \\z=100^{\circ}\end{array}$$
Therefore, the value of $$ z $$ is $$ 110^{\circ} $$.
Hence, $$ x+y+z+w $$
$$\begin{array}{l}=50^{\circ}+100^{\circ}+110^{\circ}+100 \\=360^{\circ}\end{array}$$
Name the quadrilateral whose diagonals bisect each other.
Name the quadrilateral whose diagonals are equal
Consider the following parallelograms. Find the values of $$ x, y, z $$ in each.
$${\textbf{Step - 1: Finding angle x}}{\text{.}}$$
$${\text{As we know that the sum of interior angles of a parallelogram is 180}}^\circ ,{\text{we can write}}$$
$$\angle DCB + \angle ABC = 180^\circ $$
$$\angle DCA + \angle ACB + \angle ABC = 180^\circ $$
$$67^\circ + x + 70^\circ = 180^\circ $$
$$x = 180^\circ - 70^\circ - 67^\circ $$
$$x = 43^\circ $$
$${\textbf{Step - 2: Finding angle z}}{\text{.}}$$
$${\text{As we know that the sum of interior angles of a parallelogram is 180}}^\circ ,{\text{we can write}}$$
$$\angle ADC + \angle BCD = 180^\circ $$
$$\angle ADC + \angle BCA + \angle ACD = 180^\circ $$
$$z + 67^\circ + x = 180^\circ $$
$$z = 180^\circ - 67^\circ - 43^\circ $$
$$z = 70^\circ $$
$${\textbf{Step - 3: Finding angle y}}{\text{.}}$$
$${\text{As we can observe from the figure that }}$$
$$\angle DCA = \angle FAE$$
$$\angle DAF = \angle CBA$$
$${\text{And we can write,}}y = \angle DAF + \angle FAE$$
$$y = 70^\circ + 67^\circ = 137^\circ $$
$${\textbf{Thus, x = 43}}\mathbf{^\circ ,y = 137^\circ ,z = 70^\circ .}$$
In the figure given, $$ABCD$$ is a kite in which $$BC = CD, AB =AD$$. And $$E,\ F,\ G$$ are midpoints of $$CD,\ BC$$ and $$AB$$. Prove that the line drawn through $$G$$ and parallel to $$FE$$ bisects DA.
Given that,
$$ABCD$$ is a kite in which $$BC=CD,\ AB=AD$$
$$E,\ F,\ G$$ are midpoints of $$CD,\ BC$$ and $$AB$$
To prove :
The line drawn through $$G$$ and parallel to $$FE$$ bisect $$DA$$
i.e., $$DH=HA$$
Construction :
Join $$AC$$ and $$BD$$
Construct $$GH$$ through $$G$$ parallel to $$FE$$
Proof:
In $$\triangle BCD$$,
$$E$$ and $$F$$ are midpoints of $$CD$$ and $$BC$$
$$\Rightarrow EF\parallel DB\quad...(i)\qquad$$[ Midpoint theorem ]
Now, $$GH\parallel FE$$
$$\Rightarrow GH\parallel DB\quad$$[ Using $$(i)$$]
In $$\triangle ABD$$
$$G$$ is midpoints of $$AB$$ and $$GH \parallel DB$$
$$\Rightarrow H $$ is the midpoint of $$AD\quad$$[Converse of midpoint theorem]
$$\Rightarrow DH=HA$$
Hence, the line drawn through $$G$$ and parallel to $$FE$$ bisect $$DA$$
In the adjoining figure, the lines $$l,\ m,$$ and n are parallel to each other, and G is mid-point of CD. Calculate: $$CF$$ if $$GE=2.3cm$$
In the adjoining figure, the lines $$l,\ m,$$ and n are parallel to each other, and G is mid-point of CD. Calculate:$$BG$$ if $$AD=6cm $$
In the given figure, $$ABCD$$ is a parallelogram. If $$P$$ and $$Q$$ are mid-points of sides $$CD$$ and $$BC$$ respectively, show that $$CR=\dfrac14 AC$$.
In given figure $$ABCD$$ is a quadrilateral in which $$P, Q, R$$ and $$S$$ are mid-points of the sides. $$AB, BC, CD$$ and $$DA$$. $$AC$$ is a diagonal. Show that: $$PQ = SR$$
The diagonal $$AC$$ and $$BD$$ of a parallelogram $$ABCD$$ intersect at $$O$$. If $$P$$ is the midpoint of $$AD$$, prove that $$PO=\dfrac12\ CD$$
It is given that
$$ABCD$$ is a parallelogram in which diagonals $$AC$$ and $$BD$$ intersect each other at $$O$$
$$ P$$ is the midpoint of $$AD$$
Join $$OP$$
To prove: $$PQ =\dfrac12 CD$$
Proof:
We know that,
The diagonals of a parallelogram bisect each other
$$\Rightarrow BO=OD$$
$$\Rightarrow O$$ is the mid-point of $$BD$$
In $$\triangle ABD$$
$$P$$ and $$O$$ is the mid point of $$AB$$ and $$BD$$
$$\Rightarrow PO =\dfrac 12 AB \quad$$[ Midpoint theorem ]
Now , in parallelogram $$ABCD$$,
$$AB =CD$$
$$\therefore PO =\dfrac 12 CD$$
Hence proved
The diagonal $$AC$$ and $$BD$$ of a parallelogram $$ABCD$$ intersect at $$O$$. If $$P$$ is the midpoint of $$AD$$, prove that $$PQ \parallel AB$$
It is given that,
$$ABCD$$ is a parallelogram in which diagonals $$AC$$ and $$BD$$ intersect each other at int $$O, P$$ is the midpoint of $$AD$$
Join $$OP$$
To prove: $$PQ || AB$$
Proof:
We know that,
The diagonals of parallelogram bisect each other
$$\Rightarrow BO=OD$$
$$\Rightarrow O$$ is the midpoint of $$BD$$
In $$\triangle ABD$$
$$P$$ and $$O$$ is the mid point of $$AB$$ and $$BD$$
$$\Rightarrow PO \parallel AB \quad$$[ Converse of midpoint theorem
$$\Rightarrow PQ \parallel AB$$
Hence, proved.
In the given figure, $$ABCD$$ is a parallelogram and $$E$$ is mid-points of $$AD. DL \parallel EB$$ produced at $$F$$. Prove that $$B$$ is mid-points of $$AF$$ and $$EB=LF$$.
In the figure given $$ABCD$$ is a parallelogram. $$E$$ and $$F$$ are mid-point of the sides $$AB$$ and $$CO$$ respectively. The straight lines $$AF$$ and $$BF$$ meet the straight lines $$ED$$ and $$EC$$ in points $$G$$ and $$H$$ respectively. Prove that $$GEHF$$ is a parallelogram.
In given figure $$ABCD$$ is a quadrilateral in which $$P, Q, R$$ and $$S$$ are mid-points of the sides. $$AB, BC, CD$$ and $$DA$$. $$AC$$ is a diagonal. Show that: $$PQRS$$ is a parallelogram
In the given figure, D, E and F are mid points of the sides BC, CA and AB respectively of $$ \Delta \mathrm{ABC} $$. Prove that BCEF is a trapezium and area of trap. BCEF $$ =3 / 4 $$ area of $$ \Delta \mathrm{ABC} $$.
In the adjoining figure, the lines $$l,\ m,$$ and n are parallel to each other, and G is mid-point of CD. Calculate: $$AB$$ if $$BC=2.4cm$$
Find the size of each lettered angle in the above figures.
As $$DE \parallel AB$$ [Given]
$$\angle ECB = \angle CBA$$ [Alternate angles]
$$75^{o} = \angle CBA$$
$$\Rightarrow \angle CBA = 75^{o}$$
Since, $$AD \parallel BC$$ we have
$$(x + 66^{o}) + 75^{o} = 180^{o}$$
$$x + 141^{o} = 180^{o}$$
$$x = 180^{o} - 141^{o}$$
$$x = 39^{o}$$ ...(i)
Now, in $$\triangle AMB$$
$$x + 30^{o} + \angle AMB = 180^{o}$$ [Angles sum property of a triangle]
$$39^{o} + 30^{o} + \angle AMB = 180^{o}$$ [From (i)]
$$69^{o} + \angle AMB + 180^{o}$$
$$\angle AMB = 180^{o} - 69^{o} = 111^{o}$$ ...(ii)
Since, $$\angle AMB = y$$ [Vertically opposite angles]
$$\Rightarrow y = 111^{o}$$
Hence, $$x= 39^{o}$$ and $$y= 111^{o}$$
$$\angle ECB = \angle CBA$$ [Alternate angles]
$$75^{o} = \angle CBA$$
$$\Rightarrow \angle CBA = 75^{o}$$
Since, $$AD \parallel BC$$ we have
$$(x + 66^{o}) + 75^{o} = 180^{o}$$
$$x + 141^{o} = 180^{o}$$
$$x = 180^{o} - 141^{o}$$
$$x = 39^{o}$$ ...(i)
Now, in $$\triangle AMB$$
$$x + 30^{o} + \angle AMB = 180^{o}$$ [Angles sum property of a triangle]
$$39^{o} + 30^{o} + \angle AMB = 180^{o}$$ [From (i)]
$$69^{o} + \angle AMB + 180^{o}$$
$$\angle AMB = 180^{o} - 69^{o} = 111^{o}$$ ...(ii)
Since, $$\angle AMB = y$$ [Vertically opposite angles]
$$\Rightarrow y = 111^{o}$$
Hence, $$x= 39^{o}$$ and $$y= 111^{o}$$
In the given figure, $$ABC$$ is an isosceles triangle in which $$AB = AC$$. $$AD$$ bisects exterior angle $$PAC$$ and $$CD \, || \, BA.$$ Show that
(i) $$\angle DAC=\angle BCA $$
(ii) $$ABCD$$ is a parallelogram.
(i) $$\angle DAC=\angle BCA $$
(ii) $$ABCD$$ is a parallelogram.
Given: In isosceles triangle $$ABC$$, $$AB = AC. AD$$ is the bisector of ext. $$\angle PAC$$ and $$CD || BA$$
To prove: (i) $$\angle DAC = \angle BCA$$
(ii)$$ABCD$$ is a $$||$$ gm
Proof:
In $$\triangle ABC$$
$$AB=AC$$[Given]
$$\angle C = \angle B$$ [Angles opposite to equal sides]
Since, ext. $$\angle PAC = \angle B + \angle C$$
$$= \angle C + \angle C$$
$$= 2 \angle C$$
$$\angle PAC= 2 \angle BCA$$
So, $$2\angle DAC = 2 \angle BCA$$
$$\angle DAC = \angle BCA$$
But these are alternate angles
Thus, $$AD || BC$$
But, $$AB || AC$$
Hence, $$ABCD$$ is a $$||$$ gm.
To prove: (i) $$\angle DAC = \angle BCA$$
(ii)$$ABCD$$ is a $$||$$ gm
Proof:
In $$\triangle ABC$$
$$AB=AC$$[Given]
$$\angle C = \angle B$$ [Angles opposite to equal sides]
Since, ext. $$\angle PAC = \angle B + \angle C$$
$$= \angle C + \angle C$$
$$= 2 \angle C$$
$$\angle PAC= 2 \angle BCA$$
So, $$2\angle DAC = 2 \angle BCA$$
$$\angle DAC = \angle BCA$$
But these are alternate angles
Thus, $$AD || BC$$
But, $$AB || AC$$
Hence, $$ABCD$$ is a $$||$$ gm.
In the figure ( 2 ) given below, $$ A B C $$ is right-angled triangle at A. AGFB is a square on the side $$ A B $$ and $$ B C D E $$ is a square on the hypotenuse BC. If $$ A N \perp ED$$, prove that:
(i) $$ \Delta \mathbf{B C F} \cong \Delta \mathbf{A B E} $$
(ii) area of square ABFG = area of rectangle BENM.
Prove that the quadrilateral obtained by joining the mid-points of an isosceles trapezium is a rhombus.
Given: $$ABCD$$ is an isosceles trapezium in which $$AB || DC$$ and $$AD = BC$$
$$P, Q, R$$ and $$S$$ are the mid-points of the sides $$ AB, BC, CD$$ and $$DA$$ respectively and $$PQ, QR, RS$$ and $$SP$$ are joined.
To prove: $$PQRS$$ is a rhombus
Construction: Join $$AC$$ and $$BD$$
Proof:
Since, $$ABCD$$ is an isosceles trapezium
Its diagonals are equal
$$AC = BD$$
Now, in $$\triangle ABC$$
$$P$$ and $$Q$$ are the mid-points of $$AB$$ and $$BC$$
So, $$PQ || AC$$ and
$$PQ = \dfrac{1}{2} AC$$ … (I)
$$P, Q, R$$ and $$S$$ are the mid-points of the sides $$ AB, BC, CD$$ and $$DA$$ respectively and $$PQ, QR, RS$$ and $$SP$$ are joined.
To prove: $$PQRS$$ is a rhombus
Construction: Join $$AC$$ and $$BD$$
Proof:
Since, $$ABCD$$ is an isosceles trapezium
Its diagonals are equal
$$AC = BD$$
Now, in $$\triangle ABC$$
$$P$$ and $$Q$$ are the mid-points of $$AB$$ and $$BC$$
So, $$PQ || AC$$ and
$$PQ = \dfrac{1}{2} AC$$ … (I)
Similarly, in $$\triangle ADC$$
$$S$$ and $$R$$ are the mid-point of $$AD$$ and $$CD$$
So, $$SR || AC$$ and $$SR = \dfrac{1}{2} AC$$ … (ii)
From (i) and (ii), we have
$$PQ || SR$$ and $$PQ = SR$$
$$S$$ and $$R$$ are the mid-point of $$AD$$ and $$CD$$
So, $$SR || AC$$ and $$SR = \dfrac{1}{2} AC$$ … (ii)
From (i) and (ii), we have
$$PQ || SR$$ and $$PQ = SR$$
Similarly using $$\Delta ADB $$ and $$\Delta DCB$$ we can prove that $$SP\parallel QR$$ and $$SP=QR$$
Thus, $$PQRS$$ is a parallelogram.
Now, in $$\triangle APS$$ and $$\triangle BPQ$$
$$AP = BP$$ [P is the mid-point]
$$AS = BQ$$ [Half of equal sides]
$$\angle A = \angle B$$ [As $$ABCD$$ is an isosceles trapezium]
So, $$\triangle APS \cong \triangle BPQ$$ by SAS Axiom of congruency
Thus, by C.P.C.T we have
$$PS = PQ$$
But these are the adjacent sides of a parallelogram
So, sides of $$PQRS$$ are equal
Hence, $$PQRS$$ is a rhombus
Hence proved
$$AP = BP$$ [P is the mid-point]
$$AS = BQ$$ [Half of equal sides]
$$\angle A = \angle B$$ [As $$ABCD$$ is an isosceles trapezium]
So, $$\triangle APS \cong \triangle BPQ$$ by SAS Axiom of congruency
Thus, by C.P.C.T we have
$$PS = PQ$$
But these are the adjacent sides of a parallelogram
So, sides of $$PQRS$$ are equal
Hence, $$PQRS$$ is a rhombus
Hence proved
In the figure ( 2) given below, DE is drawn parallel to the diagonal AC of the quadrilateral ABCD to meet BC produced at the point E. Prove that area of quad. $$ \mathrm{ABCD}= $$ area of $$ \Delta \mathrm{ABE} $$
In the figure ( 2) given below, DE is drawn parallel to the diagonal AC of the quadrilateral ABCD to meet BC produced at the point E. Prove that area of quad. $$ \mathrm{ABCD}= $$ area of $$ \Delta \mathrm{ABE} $$
Given:from fig (1)
In $$ \| \mathrm{gm} \mathrm{ABCD}, $$ points $$ \mathrm{P} $$ and $$ \mathrm{Q} $$ trisect $$ \mathrm{BC} $$ into three equal parts.
To prove:
area of $$ \Delta \mathrm{APQ}= $$ area of $$ \Delta \mathrm{DPQ}=1 / 6 $$ (area of $$ \| \mathrm{gm} $$ ABCD)
Firstly, let us construct: through $$ \mathrm{P} $$ and $$ \mathrm{Q} $$, draw PR and $$ \mathrm{QS} $$ parallel to $$ \mathrm{AB} $$ and $$ \mathrm{CD} $$.
Proof:
ar $$ (\triangle \mathrm{APD})=\operatorname{ar}(\Delta \mathrm{AQD})[\text { since, } \Delta \mathrm{APD} \text { and } \Delta \mathrm{AQD} \text { lie on the same base } \mathrm{AD} $$ and between
the same parallel lines $$ \mathrm{AD} $$ and $$ \mathrm{BC} $$ ]
ar $$ (\triangle \mathrm{APD})-\operatorname{ar}(\Delta \mathrm{AOD})=\operatorname{ar}(\Delta \mathrm{AQD})-\operatorname{ar}(\Delta \mathrm{AOD})[\text { On subtracting ar } \Delta \mathrm{AOD} $$ on bothsides]
$$ \operatorname{ar}(\Delta \mathrm{APO})=\operatorname{ar}(\Delta \mathrm{OQD}) \ldots \ldots(1) $$
$$ \operatorname{ar}(\Delta \mathrm{APO})+\operatorname{ar}(\Delta \mathrm{OPQ})=\operatorname{ar}(\Delta \mathrm{OQD})+\operatorname{ar}(\Delta \mathrm{OPQ})[\text { On adding ar } \Delta \mathrm{OPQ} \text { on both sides }] $$
$$ \operatorname{ar}(\Delta \mathrm{APQ})=\operatorname{ar}(\Delta \mathrm{DPQ}) \ldots \ldots(2) $$
We know that, $$ \triangle \mathrm{APQ} $$ and $$ \| \mathrm{gm} \mathrm{PQSR} $$ are on the same base $$ \mathrm{PQ} $$ and between same parallel lines $$ \mathrm{PQ} $$ and $$ \mathrm{AD} $$.
$$ \operatorname{ar}(\Delta \mathrm{APQ})=1 / 2 \operatorname{ar}(\| \mathrm{gm} \mathrm{PQRS}) \ldots \ldots(3) $$
Now $$ [\text { ar }(\| g m A B C D) / \text { ar }(\| g m P Q R S)]=[(B C \times h e i g h t) /(P Q \times h e i g h t)]= $$
$$ [(3 P Q \times h e i g h t) /(1 P Q \times h i g h t)] $$
$$ \operatorname{ar}(\| g m P Q R S)=1 / 3 \operatorname{ar}(\| g m A B C D) \ldots .(4) $$
by using ( 2 ), ( 3 ), (4), we get $$ \operatorname{ar}(\Delta \mathrm{APQ})=\operatorname{ar}(\Delta \mathrm{DPQ}) $$
$$\begin{array}{l}=1 / 2 \text { ar }(\| g m P Q R S) \\=1 / 2 \times 1 / 3 \text { ar }(\| g m A B C D) \\=1 / 6 \text { ar }(\| g m A B C D)\end{array}$$
Hence proved.
In $$ \| \mathrm{gm} \mathrm{ABCD}, $$ points $$ \mathrm{P} $$ and $$ \mathrm{Q} $$ trisect $$ \mathrm{BC} $$ into three equal parts.
To prove:
area of $$ \Delta \mathrm{APQ}= $$ area of $$ \Delta \mathrm{DPQ}=1 / 6 $$ (area of $$ \| \mathrm{gm} $$ ABCD)
Firstly, let us construct: through $$ \mathrm{P} $$ and $$ \mathrm{Q} $$, draw PR and $$ \mathrm{QS} $$ parallel to $$ \mathrm{AB} $$ and $$ \mathrm{CD} $$.
Proof:
ar $$ (\triangle \mathrm{APD})=\operatorname{ar}(\Delta \mathrm{AQD})[\text { since, } \Delta \mathrm{APD} \text { and } \Delta \mathrm{AQD} \text { lie on the same base } \mathrm{AD} $$ and between
the same parallel lines $$ \mathrm{AD} $$ and $$ \mathrm{BC} $$ ]
ar $$ (\triangle \mathrm{APD})-\operatorname{ar}(\Delta \mathrm{AOD})=\operatorname{ar}(\Delta \mathrm{AQD})-\operatorname{ar}(\Delta \mathrm{AOD})[\text { On subtracting ar } \Delta \mathrm{AOD} $$ on bothsides]
$$ \operatorname{ar}(\Delta \mathrm{APO})=\operatorname{ar}(\Delta \mathrm{OQD}) \ldots \ldots(1) $$
$$ \operatorname{ar}(\Delta \mathrm{APO})+\operatorname{ar}(\Delta \mathrm{OPQ})=\operatorname{ar}(\Delta \mathrm{OQD})+\operatorname{ar}(\Delta \mathrm{OPQ})[\text { On adding ar } \Delta \mathrm{OPQ} \text { on both sides }] $$
$$ \operatorname{ar}(\Delta \mathrm{APQ})=\operatorname{ar}(\Delta \mathrm{DPQ}) \ldots \ldots(2) $$
We know that, $$ \triangle \mathrm{APQ} $$ and $$ \| \mathrm{gm} \mathrm{PQSR} $$ are on the same base $$ \mathrm{PQ} $$ and between same parallel lines $$ \mathrm{PQ} $$ and $$ \mathrm{AD} $$.
$$ \operatorname{ar}(\Delta \mathrm{APQ})=1 / 2 \operatorname{ar}(\| \mathrm{gm} \mathrm{PQRS}) \ldots \ldots(3) $$
Now $$ [\text { ar }(\| g m A B C D) / \text { ar }(\| g m P Q R S)]=[(B C \times h e i g h t) /(P Q \times h e i g h t)]= $$
$$ [(3 P Q \times h e i g h t) /(1 P Q \times h i g h t)] $$
$$ \operatorname{ar}(\| g m P Q R S)=1 / 3 \operatorname{ar}(\| g m A B C D) \ldots .(4) $$
by using ( 2 ), ( 3 ), (4), we get $$ \operatorname{ar}(\Delta \mathrm{APQ})=\operatorname{ar}(\Delta \mathrm{DPQ}) $$
$$\begin{array}{l}=1 / 2 \text { ar }(\| g m P Q R S) \\=1 / 2 \times 1 / 3 \text { ar }(\| g m A B C D) \\=1 / 6 \text { ar }(\| g m A B C D)\end{array}$$
Hence proved.
If E,F,G and H are the midpoint of the point of the sides AB,BC,CD and DA respectively of a parallelogram ABCD, prove that area of || gm ABCD.
Given:In parallelogram ABCD, E, F, G, H are the mid-points of its sides AB, BC, CD and DA. Join EF, FG, GH and HE.
To prove:
area of quad. EFGH $$ =1 / 2 $$ area of $$ \| $$ gm $$ \mathrm{ABCD} $$
Proof:
Let us join EG. We know that,
$$ \mathrm{E} $$ and $$ \mathrm{G} $$ are mid-points of $$ \mathrm{AB} $$ and $$ \mathrm{CD} $$.
$$ \mathrm{EG}\|\mathrm{AD}\| \mathrm{BC} $$
AEGD and EBCG are parallelogram Now,
$$ \| \mathrm{gm} \mathrm{AEGD} $$ and $$ \Delta \mathrm{EHG} $$ are on the same base and between the parallel lines.
ar $$ \triangle \mathrm{EHG}=1 / 2 \text { ar } \| \mathrm{gm} \text { AEGD } \ldots \text { ( } 1) $$
Similarly, ar $$ \Delta \mathrm{EFG}=1 / 2 \text { ar } \| \mathrm{gm} \text { EBCG } \ldots \text { ( } 2) $$
Now by adding (1) and (2) ar $$ \Delta \mathrm{EHG}+ $$ ar $$ \Delta \mathrm{EFG}=1 / 2 $$ ar $$ \| \mathrm{gm} \mathrm{AEGD}+1 / 2 $$ ar $$ \| \mathrm{gm} $$ EBCG
area quad. EFGH $$ =1 / 2 $$ ar $$ \| $$ gm $$ \mathrm{ABCD} $$
Hence proved.
To prove:
area of quad. EFGH $$ =1 / 2 $$ area of $$ \| $$ gm $$ \mathrm{ABCD} $$
Proof:
Let us join EG. We know that,
$$ \mathrm{E} $$ and $$ \mathrm{G} $$ are mid-points of $$ \mathrm{AB} $$ and $$ \mathrm{CD} $$.
$$ \mathrm{EG}\|\mathrm{AD}\| \mathrm{BC} $$
AEGD and EBCG are parallelogram Now,
$$ \| \mathrm{gm} \mathrm{AEGD} $$ and $$ \Delta \mathrm{EHG} $$ are on the same base and between the parallel lines.
ar $$ \triangle \mathrm{EHG}=1 / 2 \text { ar } \| \mathrm{gm} \text { AEGD } \ldots \text { ( } 1) $$
Similarly, ar $$ \Delta \mathrm{EFG}=1 / 2 \text { ar } \| \mathrm{gm} \text { EBCG } \ldots \text { ( } 2) $$
Now by adding (1) and (2) ar $$ \Delta \mathrm{EHG}+ $$ ar $$ \Delta \mathrm{EFG}=1 / 2 $$ ar $$ \| \mathrm{gm} \mathrm{AEGD}+1 / 2 $$ ar $$ \| \mathrm{gm} $$ EBCG
area quad. EFGH $$ =1 / 2 $$ ar $$ \| $$ gm $$ \mathrm{ABCD} $$
Hence proved.
In the adjoining figure, the lines $$l,\ m,$$ and n are parallel to each other, and G is mid-point of CD. Calculate: $$ED$$ if $$FD=4.4cm$$
Angles of a quadrilateral are $$(4x)^o, 5(x+2)^o, (7x-20)^o$$ and $$6(x+3)^o$$. Find
Each angle of the quadrilateral.
In quadrilateral $$ABCD$$, side $$AB$$ is parallel to side $$DC$$. If $$\angle A: \angle D =1:2$$ and $$\angle C: \angle B =4:5$$
Calculate each angle of the quadrilateral.
Given
$$\angle A: \angle D=1:2$$
Let us consider $$\angle A=x$$ and $$\angle D=2x$$
$$\angle C:\angle B=4:5$$
Let us consider $$\angle C=4y$$ and $$\angle B=5y$$
Also, given
$$AB || DC$$ and the sum of opposite angles of quadrilateral is $$180^o$$
So.
$$\angle A+\angle D=180^o$$
$$x+2x=180^o$$
$$3x=180^o$$
We get,
$$x=60^o$$
Therefore, $$\angle A=60^o$$
$$\angle D=2x$$
$$2\times 60^o$$
$$=120^o$$
Therefore, $$\angle D=120^o$$
Now,
$$\angle B+\angle C=180^o$$
$$5y+4y=180^o$$
$$9y=180^o$$
We get,
$$y=20^o$$
$$\angle B=5y=5\times 20^o$$
$$=100^o$$
$$\angle C=4y=4\times 20^o$$
$$=80^o$$
Therefore, $$\angle A=60^o, \angle B=100^o, \angle C=80^o$$ and $$\angle D=120^o$$
Use the information given in the following figure to find:
$$\angle B$$ and $$\angle C$$
Use the information given in the following figure to find:
$$x$$
From the following figure find:
$$x$$
Two angles of a quadrilateral are $$89^o$$ and $$113^o$$. If the other angles are equal;
find equal angles.
One angle of a quadrilateral is $$90^o$$ and all other angles are equal; find each equal angle
If angles of quadrilateral are in the ratio $$4:5:3:6$$; find each angle of the quadrilateral.
Angles of a quadrilateral are $$(4x)^o, 5(x+2)^o, (7x-20)^o$$ and $$6(x+3)^o$$. Find
The value of $$x$$.
Two angles of a quadrilateral are $$68^o$$ and $$76^o$$. If other two angles are in the ratio $$5:7$$, find the measure of each of them.
In the given figure
$$\angle b=2a+15$$
And $$\angle c=3a+5$$; find the value of $$b$$ and $$c$$
Three angles of a quadrilateral are equal. If the fourth angle is $$69^o$$; find the measure of equal angles.
Given: In quadrilateral $$ABCD; \angle C=40^o, \angle D= \angle C$$$$ -8^o , \angle A=5(a+2)^o$$ and $$\angle B=2(2a+7)^o$$.
Each angle of a quadrilateral is $$x+5^o$$. Find
The value of $$x$$
In the quadrilateral $$PQRS, \angle P: \angle Q: \angle R: \angle S=3:4:6:7$$,
Calculate to each other. Is $$PS$$ also parallel to $$QR$$?
From the following figure find:
$$\angle ABC$$
From the following figure find:
$$\angle ACD$$
The following figure shows a quadrilateral in which sides $$AB$$ and $$DC$$ are parallel. If $$\angle A:\angle D=4:5, \angle B=(3x-15)^o$$ and $$\angle C=(4x+20)^o$$, find each angle of the quadrilateral $$ABCD$$.
Two diagonals of an isosceles trapezium are $$x\ cm$$ and $$(3x-8) cm$$. Find the value of $$x$$.
Use the information given in the following figure to find the value of $$x$$.
$${\textbf{Step-1: Draw a given diagram find their missing angles.}}$$
$${\textbf{}}$$
$${\text{Take A, B, C, D as the vertices of quadrilateral and BA is produced to E(say).}}$$
$${\text{Observe given figure.}}$$
$$\Rightarrow \angle EAD = 70^0$$
$$\Rightarrow \angle EAD + \angle DAB = 180^0$$
$$\Rightarrow \angle DAB = 180^0 - 70^0 =110^0$$ $$\boldsymbol{[\because{\text{EAB is a straight line and AD stands on it.]}}}$$
$${\textbf{Step-2: Apply the sum of interior angles of a quadrilateral are}}$$ $${\mathbf{360^0.}}$$
$$\Rightarrow \angle DAB + \angle ADC + \angle DCB + \angle CBA = 360^0$$
$$\Rightarrow 110^0 + 80^0 + 56^0 +3x-6^0 = 360^0$$
$$\Rightarrow 3x-6^0 = 360^0 -110^0 - 80^0 - 56^0$$
$$\Rightarrow 3x- 6^0 = 114$$
$$\Rightarrow 3x = 120^0$$
$$\Rightarrow x = 40^0$$
$${\textbf{Thus, the value of }}$$ $${\mathbf{x = 40^0.}}$$
Each angle of a quadrilateral is $$x+5^o$$. Find
Give the special name of the quadrilateral taken.
In parallelogram $$ABCD,E$$ is a mid-point of side $$AB$$ and $$CE$$ bisect angle $$BCD.$$ Prove that
Angle $$DEC$$ is a right angle.
Given : ||gm $$ABCD$$ in which $$E$$ is mid-point of $$AB$$ and $$CE$$ bisects $$\angle BCD.$$
To prove: (i) $$\angle DEC=90^{o}$$
Const. join $$DE$$
Proof: (i) $$AB\parallel CD$$ (Given)
And $$CE$$ bisect it.
$$\angle 1=\angle 3$$ (Alternate $$\angle S$$).....(i)
But $$\angle 1=\angle 2$$ (Given)......(ii)
From (i)&(ii)
$$\angle 2=\angle 3$$
$$BC=BE$$ (Sides opp. to equal angles)
But $$BC=AD$$ (opp. sides of ||gm)
And $$BE=AE$$ (Given)
$$AD=AE$$
$$\angle 4=\angle 5$$ ($$\angle S$$ opp. to equal sides)
But $$\angle 5=\angle 6$$ (Alternate $$\angle S$$)
$$\Rightarrow \angle 4=\angle 6$$
$$DE$$ bisects $$\angle ADC.$$
Now $$AD//BC$$
$$\Rightarrow \angle D+\angle C=180^{o}$$
$$2\angle 6+2\angle 1=180^{o}$$
$$DE$$ and $$CE$$ are bisectors.
$$\angle 6+\angle 1=\dfrac{180^{0}}{2}$$
$$\angle 6+\angle 1=90^{0}$$
But $$\angle DEC+\angle 6+\angle 1=180^{o}$$
$$\angle DEC+90^{o}=180^{o}$$
$$\angle DEC=180^{o}-90^{o}$$
$$\angle DEC =90^{o}$$
Hence the result.
To prove: (i) $$\angle DEC=90^{o}$$
Const. join $$DE$$
Proof: (i) $$AB\parallel CD$$ (Given)
And $$CE$$ bisect it.
$$\angle 1=\angle 3$$ (Alternate $$\angle S$$).....(i)
But $$\angle 1=\angle 2$$ (Given)......(ii)
From (i)&(ii)
$$\angle 2=\angle 3$$
$$BC=BE$$ (Sides opp. to equal angles)
But $$BC=AD$$ (opp. sides of ||gm)
And $$BE=AE$$ (Given)
$$AD=AE$$
$$\angle 4=\angle 5$$ ($$\angle S$$ opp. to equal sides)
But $$\angle 5=\angle 6$$ (Alternate $$\angle S$$)
$$\Rightarrow \angle 4=\angle 6$$
$$DE$$ bisects $$\angle ADC.$$
Now $$AD//BC$$
$$\Rightarrow \angle D+\angle C=180^{o}$$
$$2\angle 6+2\angle 1=180^{o}$$
$$DE$$ and $$CE$$ are bisectors.
$$\angle 6+\angle 1=\dfrac{180^{0}}{2}$$
$$\angle 6+\angle 1=90^{0}$$
But $$\angle DEC+\angle 6+\angle 1=180^{o}$$
$$\angle DEC+90^{o}=180^{o}$$
$$\angle DEC=180^{o}-90^{o}$$
$$\angle DEC =90^{o}$$
Hence the result.
Each angle of a quadrilateral is $$x+5^o$$. Find
Each angle of the quadrilateral
In parallelogram $$ABCD,E$$ is a mid-point of side $$AB$$ and $$CE$$ bisect angle $$BCD.$$ Prove that
$$DE$$ bisects and $$\angle ADC$$ and
Given : ||gm $$ABCD$$ in which $$E$$ is mid-point of $$AB$$ and $$CE$$ bisects $$\angle BCD.$$
To prove: (i) $$DE$$ bisects $$\angle ADC$$
Const. join $$DE$$
Proof: (i) $$AB\parallel CD$$ (Given)
And $$CE$$ bisect it.
$$\angle 1=\angle 3$$ (Alternate $$\angle S$$).....(i)
But $$\angle 1=\angle 2$$ (Given)......(ii)
From (i)&(ii)
$$\angle 2=\angle 3$$
$$BC=BE$$ (Sides opp. to equal angles)
But $$BC=AD$$ (opp. sides of ||gm)
And $$BE=AE$$ (Given)
$$AD=AE$$
$$\angle 4=\angle 5$$ ($$\angle S$$ opp. to equal sides)
But $$\angle 5=\angle 6$$ (Alternate $$\angle S$$)
$$\Rightarrow \angle 4=\angle 6$$
$$DE$$ bisects $$\angle ADC.$$ proved.
To prove: (i) $$DE$$ bisects $$\angle ADC$$
Const. join $$DE$$
Proof: (i) $$AB\parallel CD$$ (Given)
And $$CE$$ bisect it.
$$\angle 1=\angle 3$$ (Alternate $$\angle S$$).....(i)
But $$\angle 1=\angle 2$$ (Given)......(ii)
From (i)&(ii)
$$\angle 2=\angle 3$$
$$BC=BE$$ (Sides opp. to equal angles)
But $$BC=AD$$ (opp. sides of ||gm)
And $$BE=AE$$ (Given)
$$AD=AE$$
$$\angle 4=\angle 5$$ ($$\angle S$$ opp. to equal sides)
But $$\angle 5=\angle 6$$ (Alternate $$\angle S$$)
$$\Rightarrow \angle 4=\angle 6$$
$$DE$$ bisects $$\angle ADC.$$ proved.
In the given figure, $$M$$ is mid-point of $$AB$$ and $$DE$$, whereas $$N$$ is mid-point of $$BC$$ and $$DF$$. Show that: $$EF=AC$$.
$$D, E$$ and $$F$$ are the mid-points of the sides $$AB, BC$$ and $$CA$$ of an isosceles triangle $$DEF$$ is also isoceles.
The figure is shown below
Given that $$ABC$$ is an isosceles triangle where $$AB=AC$$.
Since $$D, E, F$$ are midpoints of $$AB, BC, CA$$ therefore
$$DE=\dfrac{AC}{2}$$ and $$EF=\dfrac{AB}{2}$$ (as per mid-point theorem)
Given that $$ABC$$ is an isosceles triangle where $$AB=AC$$.
Since $$D, E, F$$ are midpoints of $$AB, BC, CA$$ therefore
$$DE=\dfrac{AC}{2}$$ and $$EF=\dfrac{AB}{2}$$ (as per mid-point theorem)
$$\implies 2DE=AC$$ and $$2EF=AB$$
$$\implies 2DE=2DF$$ $$(\because AB=AC)$$
$$\implies DE=EF$$
Therefore $$DEF$$ is an isosceles triangle
Hence proved
Therefore $$DEF$$ is an isosceles triangle
Hence proved
In parallelogram $$ABCD,E$$ is a mid-point of side $$AB$$ and $$CE$$ bisect angle $$BCD.$$ Prove that
$$AE=AD.$$
Given : ||gm $$ABCD$$ in which $$E$$ is mid-point of $$AB$$ and $$CE$$ bisects $$\angle BCD.$$
To prove: (i) $$AE=AD$$
Const. join $$DE$$
Proof: (i) $$AB\parallel CD$$ (Given)
And $$CE$$ bisect it.
$$\angle 1=\angle 3$$ (Alternate $$\angle S$$).....(i)
But $$\angle 1=\angle 2$$ (Given)......(ii)
From (i)&(ii)
$$\angle 2=\angle 3$$
$$BC=BE$$ (Sides opp. to equal angles)
But $$BC=AD$$ (opp. sides of ||gm)
And $$BE=AE$$ (Given)
$$AE=AD$$ proved.
To prove: (i) $$AE=AD$$
Const. join $$DE$$
Proof: (i) $$AB\parallel CD$$ (Given)
And $$CE$$ bisect it.
$$\angle 1=\angle 3$$ (Alternate $$\angle S$$).....(i)
But $$\angle 1=\angle 2$$ (Given)......(ii)
From (i)&(ii)
$$\angle 2=\angle 3$$
$$BC=BE$$ (Sides opp. to equal angles)
But $$BC=AD$$ (opp. sides of ||gm)
And $$BE=AE$$ (Given)
$$AE=AD$$ proved.
In rhombus $$ABCD:$$
If $$\angle A=74^{\circ};$$ find $$\angle B$$ and $$\angle C.$$
$$AD\parallel BC$$
$$\angle A+\angle B=180^{\circ}$$
$$74^{\circ}+\angle B=180^{\circ}$$
$$\angle B=180^{\circ}-74^{\circ}=106^{\circ}$$
Opposite angles of Rhombus are equal.
$$\angle A=\angle C=74^{\circ}$$
Sides of Rhombus are equal.
$$BC=CD=AD=7.5\ cm$$
$$\angle B=106^{\circ};\angle C=74^{\circ}$$
In parallelogram $$ABCD,\angle A=3$$ times $$\angle B.$$ Find all the angles of the parallelogram. In the same parallelogram if $$AB=5x-7$$ and $$CD=3x+1;$$ find the length of $$CD.$$
Let $$\angle B =x$$
$$\angle A=3 \angle B=3x$$
$$AD || BC$$
$$\angle A+ \angle B=180^o$$
$$3x+x=180^o$$
$$\Rightarrow 4x=180^o$$
$$\Rightarrow x=45^o$$
$$\angle B=45^o$$
$$\angle A=3x=3\times 45=135^o$$
And $$\angle B=\angle D =45^o$$
Opposite angles of parallelogram are equal
$$\angle A= \angle C=135^o$$
Opposite sides of parallelogram are equal.
$$AB=CD$$
$$5x-7=3x+1$$
$$\Rightarrow 5x-3x=1+7$$
$$\Rightarrow 2x=8$$
$$\Rightarrow x=4$$
$$CD=3\times 4+1=13$$
Hence $$135^o, 45^o, 135^o$$ and $$45^o.$$
Given: Parallelogram $$ABCD$$ in which diagonals $$AC$$ and $$BD$$ intersect at $$M.$$ Prove: $$M$$ is mid-point of $$LN.$$
Proof:Diagonals of parallelogram bisect each other
$$MD=MB$$
Also $$\angle ADB=\angle DBN$$ (Alternate $$\angle S$$)
$$\angle DML=\angle BMN$$ (ert.opp.$$\angle S$$)
$$\triangle DML=\triangle BMN$$
$$LM=MN$$
$$M$$ is mid-point of $$LN.$$
Hence proved.
$$MD=MB$$
Also $$\angle ADB=\angle DBN$$ (Alternate $$\angle S$$)
$$\angle DML=\angle BMN$$ (ert.opp.$$\angle S$$)
$$\triangle DML=\triangle BMN$$
$$LM=MN$$
$$M$$ is mid-point of $$LN.$$
Hence proved.
In trapezium $$ABCD$$, sides $$AB$$ and $$DC$$ are parallel to each other. $$E$$ is mid-point of $$AD$$ and $$F$$ is the mid-point of $$BC$$.
Prove that: $$AB+DC=2EF$$
Consider trapezium $$ABCD$$.
Given $$E$$ and $$F$$ are midpoint on sides $$AD$$ and $$BC$$, respectively.
We know that $$AB=GH=IJ$$
From midpoint theorem,
$$EG=\dfrac{1}{2}DI, HF=\dfrac{1}{2}JC$$
Consider $$LHS$$,
$$AB+CD=AB+CJ+JI+ID=AB+2HF+AB+2EG$$
So $$AB+CD=2(AB+HF+EG)=2(EG+GH+HF)=2EF$$
$$AB+CD=2EF$$
Hence Proved.
Given $$E$$ and $$F$$ are midpoint on sides $$AD$$ and $$BC$$, respectively.
We know that $$AB=GH=IJ$$
From midpoint theorem,
$$EG=\dfrac{1}{2}DI, HF=\dfrac{1}{2}JC$$
Consider $$LHS$$,
$$AB+CD=AB+CJ+JI+ID=AB+2HF+AB+2EG$$
So $$AB+CD=2(AB+HF+EG)=2(EG+GH+HF)=2EF$$
$$AB+CD=2EF$$
Hence Proved.
In parallelogram $$ABCD, E$$ is the mid-point of $$AB$$ and $$AP$$ is parallel to $$EC$$ which meets $$DC$$ at point $$O$$ and $$BC$$ produced at $$P$$. Prove that: $$BP=2AD$$
In the following figure, $$AC || PS || QR || SR$$.
Prove that:
Area of quadrilateral $$PQRS=2\times $$ Area of quad $$ABCD$$.
In each of the following figures ,ABCD is a parallelogram in each case find the value of x and y.
In the following figure, $$BD$$ is parallel to $$CA,E$$ is midpoint of $$CA$$ and $$BD=\dfrac{1}{2}CA.$$ Prove that:
$$Ar.(\triangle ABC)=2\times Ar.(\triangle DBC)$$
$$E,F,G$$ and $$H$$ are the midpoints of the sides of a parallelogram $$ABCD.$$ SHow that area of quadrilateral $$EFGH$$ is half of the area of parallelogram $$ABCD.$$
Also given that ,
$$E,F, G$$ and $$H$$ are respectively the midpoints of the sides $$AB,BC,CD$$ and $$AD$$ of a ||gm $$ABCD.$$
To prove: $$ar(EFGH)=\dfrac{1}{2} ar(ABCD).$$
Proof:
Join $$FH,$$ such that $$FH\parallel AB\parallel CD.$$
Since, $$FH\parallel AB$$ and $$AH\parallel BF.$$
So, $$ABFH$$ is a parallelogram.
$$\triangle EFH$$ and parallelogram $$ABFH$$ are on same base $$FH$$ and between the same parallel lines.
Hence, $$ar(\triangle EFH)=\dfrac{1}{2}ar(\ parallelogram ABFH).....(1).$$
Since, $$FH\parallel CD$$ and $$DH\parallel FC.$$
$$DCFH$$ is a parallelogram.
we Know that,
$$\triangle FGH$$ and parallelogram $$DCFH$$ are on same base $$FH$$ and between the same parallel lines.
$$ar(\triangle FGH)=\dfrac{1}{2}ar(\ parallelogram\ DCFH).....(2).$$
Adding equation (1) and (2), we get
$$ar(\triangle EFH)+ar(\triangle FGH)=\dfrac{1}{2}ar(\ parallelogram\ ABFH)+\dfrac{1}{2}ar(\ parallelogram\ DCFH).$$
$$ar(\triangle EFH)+ar(\triangle FGH)=\dfrac{1}{2}[ar(\ parallelogram\ ABFH)+\dfrac{1}{2}ar(\ parallelogram\ DCFH)].$$
$$ar(EFGH)=\dfrac{1}{2}ar(ABCD).$$
Hence proved.
$$E,F, G$$ and $$H$$ are respectively the midpoints of the sides $$AB,BC,CD$$ and $$AD$$ of a ||gm $$ABCD.$$
To prove: $$ar(EFGH)=\dfrac{1}{2} ar(ABCD).$$
Proof:
Join $$FH,$$ such that $$FH\parallel AB\parallel CD.$$
Since, $$FH\parallel AB$$ and $$AH\parallel BF.$$
So, $$ABFH$$ is a parallelogram.
$$\triangle EFH$$ and parallelogram $$ABFH$$ are on same base $$FH$$ and between the same parallel lines.
Hence, $$ar(\triangle EFH)=\dfrac{1}{2}ar(\ parallelogram ABFH).....(1).$$
Since, $$FH\parallel CD$$ and $$DH\parallel FC.$$
$$DCFH$$ is a parallelogram.
we Know that,
$$\triangle FGH$$ and parallelogram $$DCFH$$ are on same base $$FH$$ and between the same parallel lines.
$$ar(\triangle FGH)=\dfrac{1}{2}ar(\ parallelogram\ DCFH).....(2).$$
Adding equation (1) and (2), we get
$$ar(\triangle EFH)+ar(\triangle FGH)=\dfrac{1}{2}ar(\ parallelogram\ ABFH)+\dfrac{1}{2}ar(\ parallelogram\ DCFH).$$
$$ar(\triangle EFH)+ar(\triangle FGH)=\dfrac{1}{2}[ar(\ parallelogram\ ABFH)+\dfrac{1}{2}ar(\ parallelogram\ DCFH)].$$
$$ar(EFGH)=\dfrac{1}{2}ar(ABCD).$$
Hence proved.
State 'true' or 'false'
The diagonals of a quadrilateral bisect each other.
False
This is not true for any random quadrilateral .Observe the quadrilateral shown below.
Clearly the diagonals of the given quadrilateral do not bisect each other . However , if the quadrilateral was a special quadrilateral like a parallelogram , this would hold true.
This is not true for any random quadrilateral .Observe the quadrilateral shown below.
Clearly the diagonals of the given quadrilateral do not bisect each other . However , if the quadrilateral was a special quadrilateral like a parallelogram , this would hold true.
In parallelogram ABCD , AP and AQ are perpendicular from vertex of obtuse angle A as shown .If angles x : y = 2 : 1 ; find the angle of the parallelogram.
In triangle $$ABC,E$$ and $$F$$ are the mid-points of sides $$AB$$ and $$AC$$ respectively. If $$BF$$ and $$CE$$ intersect each other at point $$O,$$ prove that the triangle $$OBC$$ and quadrilateral $$AEOF$$ are equal in area.
$$E$$ and $$F$$ are the midpoints of the sides $$AB$$ and $$AC.$$
Consider the following figure.
Therefore, by midpoint theorem, we have $$EF\parallel BC$$
Triangles $$BEF$$ and $$CEF$$ lie on the common base $$EF$$ and between the parallels, $$EF$$ and $$BC$$
Therefore, $$Ar.(\triangle BEF)=Ar.(\triangle CEF)$$
$$\Rightarrow Ar.(\triangle BOE)+Ar.(\triangle EOF)=Ar.(\triangle EOF)+Ar.(\triangle COF)$$
$$\Rightarrow Ar.(\triangle BOE)=Ar.(\triangle COF)$$
Now $$BF$$ and $$CE$$ are the medians of the triangle $$ABC$$ {mid points}
Medians of the triangle divides it into two equal areas of triangles.
Thus, we have $$Ar.\triangle ABF=Ar.\triangle CBF$$
Subtracting $$Ar.\triangle BOE$$ on the both the sides, we have
$$Ar.\triangle ABF-Ar.\triangle BOE=Ar.\triangle CBF-Ar.\triangle BOE$$
Since, $$Ar.(\triangle BOE)=Ar.(\triangle COF)$$
$$Ar.\triangle ABF-Ar.\triangle BOE=Ar.\triangle CBF-Ar.\triangle COF$$
$$Ar.(quad. AEOF)=Ar.(\triangle OBC),$$ hence proved
E is midpoint of sides AB and F is the midpoint of side DC pf parallelogram ABCD . Prove that AEFD is a parallelogram,
Let us draw a parallelogram ABCD, where F is the midpoint of side DC and E is the midpoint of the side AB
To prove : AEFD is a parallelogram
Proof :
AB $$||$$ DC
To prove : AEFD is a parallelogram
Proof :
AB $$||$$ DC
AB $$=$$ CD
$$\implies \dfrac{1}{2} \text{AB} = \dfrac{1}{2} \text{CD}$$
$$\implies$$ AE $$=$$ DF -----(1)
since, AB $$||$$ DC
$$\implies \dfrac{1}{2} \text{AB} = \dfrac{1}{2} \text{CD}$$
$$\implies$$ AE $$=$$ DF -----(1)
since, AB $$||$$ DC
AE $$||$$ DF ( AE and DF are the parts of sides AB and CD ) ----(2)
From (1) and (2), a pair of sides of the quadrilateral AEFD is equal and parallel
Therefore, AEFD is a parallelogram.
$$ABCD$$ is a parallegram. $$E,\ F$$ are the midpoints of $$BC$$
and $$CD$$ respectively. $$AE,\ AF$$ meet the diagonal $$BD$$ at
$$Q$$ and $$P$$ respectively. Show that $$P$$ and $$Q$$ trisect
$$DB.$$
Prove that the line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.
Let $$ABCD$$ be a quadrilateral and $$P, Q, R, S$$ be the midpoints of the sides $$AB, BC, CD$$, and $$DA$$ respectively.
Let $$\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{p}, \bar{q}, \bar{r}$$ and $$\bar{s}$$ be the position vectors of the points $$A, B, C, D, P, Q, R$$ and $$S$$ respectively.
Since $$P, Q, R$$ and $$S$$ are the midpoints of the sides $$AB, BC, CD$$ and $$DA$$ respectively,
$$\bar{p}=\dfrac{\bar{a}+\bar{b}}{2}, \bar{q}=\dfrac{\bar{b}+\bar{c}}{2}, \bar{r}=\dfrac{\bar{c}+\bar{d}}{2}$$ and $$\bar{s}=\dfrac{\bar{d}+\bar{a}}{2}$$
$$\therefore \bar{PQ}=\bar{q}-\bar{p}$$
$$=\left(\dfrac{\bar{c}+\bar{d}}{2}\right)-\left(\dfrac{\bar{d}+\bar{a}}{2}\right)$$
$$=\dfrac{1}{2}(\bar{c}+\bar{d}-\bar{d}-\bar{a})=\dfrac{1}{2}(\bar{c}-\bar{a})$$
$$\therefore \bar{PQ}=\bar{SR}$$
$$\therefore \bar{PQ}\parallel \bar{SR}$$
Similarly $$\bar{QR}\parallel \bar{PS}$$
$$\therefore PQRS$$ is a parallelogram.
Let $$\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{p}, \bar{q}, \bar{r}$$ and $$\bar{s}$$ be the position vectors of the points $$A, B, C, D, P, Q, R$$ and $$S$$ respectively.
Since $$P, Q, R$$ and $$S$$ are the midpoints of the sides $$AB, BC, CD$$ and $$DA$$ respectively,
$$\bar{p}=\dfrac{\bar{a}+\bar{b}}{2}, \bar{q}=\dfrac{\bar{b}+\bar{c}}{2}, \bar{r}=\dfrac{\bar{c}+\bar{d}}{2}$$ and $$\bar{s}=\dfrac{\bar{d}+\bar{a}}{2}$$
$$\therefore \bar{PQ}=\bar{q}-\bar{p}$$
$$=\left(\dfrac{\bar{b}+\bar{c}}{2}\right)-=\left(\dfrac{\bar{a}+\bar{b}}{2}\right)$$
$$=-\dfrac{1}{2}(\bar{b}+\bar{c}-\bar{a}-\bar{b})=\dfrac{1}{2}(\bar{c}-\bar{a})$$
$$\bar{SR}=\bar{r}-\bar{s}$$$$=\left(\dfrac{\bar{c}+\bar{d}}{2}\right)-\left(\dfrac{\bar{d}+\bar{a}}{2}\right)$$
$$=\dfrac{1}{2}(\bar{c}+\bar{d}-\bar{d}-\bar{a})=\dfrac{1}{2}(\bar{c}-\bar{a})$$
$$\therefore \bar{PQ}=\bar{SR}$$
$$\therefore \bar{PQ}\parallel \bar{SR}$$
Similarly $$\bar{QR}\parallel \bar{PS}$$
$$\therefore PQRS$$ is a parallelogram.
$$ \square \mathrm{ABCD} $$ is a parallelogram, $$ \mathrm{P} $$ and $$ \mathrm{Q} $$ are midpoints of side $$ \mathrm{AB} $$ and $$ \mathrm{DC} $$ respectively, then prove $$ \square \mathrm{APCQ} $$ is a parallelogram.
SL = LR
Given that $$M$$ is the midpoint of $$PR$$.......(1)
$$\angle PSR = \angle MLR = 90^o$$ ....(Since $$\square PQRS$$ and $$\square MNRL$$ are rectangles)
Thus, $$LM || SP$$ (corresponding angles are equal) ......(2)
From (1) and (2) we have
$$L$$ is the midpoint of $$SR$$ .......(Converse of midpoint theorem)
Thus, $$SL = LR$$
$$\angle PSR = \angle MLR = 90^o$$ ....(Since $$\square PQRS$$ and $$\square MNRL$$ are rectangles)
Thus, $$LM || SP$$ (corresponding angles are equal) ......(2)
From (1) and (2) we have
$$L$$ is the midpoint of $$SR$$ .......(Converse of midpoint theorem)
Thus, $$SL = LR$$
In a parallelogram $$ABCD$$, If $$\angle A = (3x + 2)^o, \angle B = (2x - 32)^o$$ then find the value of $$x$$ and then find the measures of $$\angle C$$ and $$\angle D$$.
Interior angles on the same side of the parallelogram are supplementary.
$$\angle A + \angle B = 180^o$$
$$\Rightarrow (3x + 2)^o + (2x - 32)^o = 180^o$$
$$\Rightarrow 3x + 2 + 2x - 32 = 180^o$$
$$\Rightarrow 5x - 30 = 180^o$$
$$\Rightarrow 5x = 210$$
$$\Rightarrow x = 42^o$$
Thus,
$$\angle A + \angle B = 180^o$$
$$\Rightarrow (3x + 2)^o + (2x - 32)^o = 180^o$$
$$\Rightarrow 3x + 2 + 2x - 32 = 180^o$$
$$\Rightarrow 5x - 30 = 180^o$$
$$\Rightarrow 5x = 210$$
$$\Rightarrow x = 42^o$$
Thus,
$$\angle A = 3x + 2 \\= 3 \times 42^o + 2\\ = 128^o$$
Also,
$$\angle B = 2x - 32 \\= 2 \times 42^o - 32 \\= 52^o$$
Also, opposite angles of a parallelogram are congruent.
$$\angle A = \angle C = 128^o$$
$$\angle B = \angle D = 52^o$$
$$\angle B = 2x - 32 \\= 2 \times 42^o - 32 \\= 52^o$$
Also, opposite angles of a parallelogram are congruent.
$$\angle A = \angle C = 128^o$$
$$\angle B = \angle D = 52^o$$
In the given figure, $$\Delta$$ABC is an equilateral triangle. Points F, D and E are the midpoints of sides AB, BC and AC respectively. Show that $$\Delta$$ FED is an equilateral triangle.
In the given figure, points X, Y, Z are the midpoints of side AB, side BC and side AC of $$\Delta$$ABC respectively. AB = 5 cm, AC = 9 cm and BC = 11 cm. Find the length of XY, YZ, XZ.
$$\square ABCD$$ is a parallelogram, ratio of $$\angle A$$ to $$\angle B$$ of this parallelogram is $$5 : 4$$. Find the measure of $$\angle B$$.
LN = $$\frac{1}{2} $$ SQ
Given that M is the midpoint of PR.......(1)
$$\angle PSR = \angle MLR = 90^o$$ ....(Since $$\square PQRS$$ and $$\square MNRL$$ are rectangles)
Thus, $$LM || SP$$ (corresponding angles are equal) ......(2)
From $$(1)$$ and $$(2)$$ we have
$$L$$ is the midpoint of $$SR$$ .......(Converse of midpoint theorem)
$$\angle RNM = \angle RQP = 90^o$$ ....(Since $$\square PQRS$$ and $$\square MNRL$$ are rectangles)
Thus, $$MN || PQ$$ (corresponding angles are equal) ......(4)
From $$(1)$$ and $$(4)$$ we have
$$N$$ is the midpoint of $$RQ$$ .......(Converse of midpoint theorem)
Join $$LN$$ and $$SQ$$
By midpoint theorem,
$$LN || SQ$$ and $$LN$$ $$= \dfrac{1}{2} SQ$$
$$\angle PSR = \angle MLR = 90^o$$ ....(Since $$\square PQRS$$ and $$\square MNRL$$ are rectangles)
Thus, $$LM || SP$$ (corresponding angles are equal) ......(2)
From $$(1)$$ and $$(2)$$ we have
$$L$$ is the midpoint of $$SR$$ .......(Converse of midpoint theorem)
$$\angle RNM = \angle RQP = 90^o$$ ....(Since $$\square PQRS$$ and $$\square MNRL$$ are rectangles)
Thus, $$MN || PQ$$ (corresponding angles are equal) ......(4)
From $$(1)$$ and $$(4)$$ we have
$$N$$ is the midpoint of $$RQ$$ .......(Converse of midpoint theorem)
Join $$LN$$ and $$SQ$$
By midpoint theorem,
$$LN || SQ$$ and $$LN$$ $$= \dfrac{1}{2} SQ$$
If the ratio of measures of two adjacent angles of a parallelogram is $$1:2$$, find the measures of all angles of the parallelogram.
In a quadrilateral, $$\mathrm{IJKL},$$ $$IJ\parallel KL$$, $$ \angle I=108^{\circ}, \angle \mathrm{K}=53^{\circ} \text { then find the measure of } \angle \mathrm{J} \text { and } \angle \mathrm{L}$$.
Given: $$IJ \|{KL} $$
So, the interior angles on the same side of the transversal will be supplementary.
$$ \angle I+\angle L=180^{\circ} $$
$$ 108^{\circ}+\angle \mathrm{L}=180^{\circ} $$
$$ \Rightarrow \angle L=72^{\circ} $$
$$ \angle I+\angle L=180^{\circ} $$
$$ 108^{\circ}+\angle \mathrm{L}=180^{\circ} $$
$$ \Rightarrow \angle L=72^{\circ} $$
Similarly, $$ \angle \mathrm{K}+\angle \mathrm{J}=180^{\circ} $$
$$ 53^{\circ}+\angle \mathrm{J}=180^{\circ} $$
$$ \Rightarrow \angle J=127^{\circ} $$
$$ 53^{\circ}+\angle \mathrm{J}=180^{\circ} $$
$$ \Rightarrow \angle J=127^{\circ} $$
ABCD is a quadrilateral in which AD = BC and $$\angle DAB = \angle CBA $$. Prove that
(1) $$\triangle ABD \cong \triangle BAC$$
(2) BD = AC
(3) $$\angle ABD = \angle BAC $$
Data: $$ABCD$$ is a quadrilateral.
$$AD = BC$$ and $$\angle DAB = \angle CBA$$.
To Prove:
(1) $$\Delta ABD \cong \Delta BAC$$
(2) $$BD = AC$$
(3) $$\angle ABD = \angle BAC$$
Proof: (i) In $$\Delta ABD$$ and $$\Delta BAC$$
$$AD = BC$$ (Data)
$$\angle DAB = \angle CBA$$ (Data)
$$AB$$ is common $$SAS$$ postulate.
$$\therefore \ \ \Delta ABD \cong \Delta BAC$$.
(ii) $$\Delta ABD \cong \Delta BAC$$
$$\because$$ Corresponding sides are equal.
$$\therefore \ \ BD = AC$$.
(iii) $$\Delta ABD \cong \Delta BAC$$. (Proved)
Equal sides of Adjacent angles are equal.
As we have $$AD = BC$$,
Adjacent angle for $$AD$$ is $$ABD$$
Adjacent angle for $$BC$$ is $$BAC$$
$$\therefore \ \ \angle ABD = \angle BAC$$.
$$AD = BC$$ and $$\angle DAB = \angle CBA$$.
To Prove:
(1) $$\Delta ABD \cong \Delta BAC$$
(2) $$BD = AC$$
(3) $$\angle ABD = \angle BAC$$
Proof: (i) In $$\Delta ABD$$ and $$\Delta BAC$$
$$AD = BC$$ (Data)
$$\angle DAB = \angle CBA$$ (Data)
$$AB$$ is common $$SAS$$ postulate.
$$\therefore \ \ \Delta ABD \cong \Delta BAC$$.
(ii) $$\Delta ABD \cong \Delta BAC$$
$$\because$$ Corresponding sides are equal.
$$\therefore \ \ BD = AC$$.
(iii) $$\Delta ABD \cong \Delta BAC$$. (Proved)
Equal sides of Adjacent angles are equal.
As we have $$AD = BC$$,
Adjacent angle for $$AD$$ is $$ABD$$
Adjacent angle for $$BC$$ is $$BAC$$
$$\therefore \ \ \angle ABD = \angle BAC$$.
In quadrilateral ACBD, AC = AD and AB bisects $$\angle A$$. Show that $$\triangle ABC = \triangle ABD$$. What can you say about BC and BD?
Data : In a quadrilateral $$ABCD, $$
$$AC = AD$$ and $$AB$$ bisects $$\angle A$$.
To Prove: $$\triangle ABC = \triangle ABD$$
Proof : In $$\triangle ABC$$ and $$\triangle ABD $$
$$AC = AD (Data) $$
$$\angle CAB = \angle DAB$$ $$\therefore$$ $$\angle A$$ bisected
$$ AB$$ is common.
Here Side, angle, side rule is there.
$$\therefore \triangle ABC = \triangle ABD$$.
$$AC = AD$$ and $$AB$$ bisects $$\angle A$$.
To Prove: $$\triangle ABC = \triangle ABD$$
Proof : In $$\triangle ABC$$ and $$\triangle ABD $$
$$AC = AD (Data) $$
$$\angle CAB = \angle DAB$$ $$\therefore$$ $$\angle A$$ bisected
$$ AB$$ is common.
Here Side, angle, side rule is there.
$$\therefore \triangle ABC = \triangle ABD$$.
$$c.p.c.t...BC=BD$$
In $$ \square \mathrm{ABCD}, $$ side $$ \mathrm{BC}< $$ side $$ \mathrm{AD} $$ in following figure.side $$ \mathrm{BC} \| $$ side $$ \mathrm{AD} $$ and if side $$ \mathrm{BA} \cong \operatorname{side} \mathrm{CD} $$ then prove that $$ \angle \mathrm{ABC} \cong \angle \mathrm{DCB} $$
Construction: Draw a line $$ \mathrm{CE} \| \mathrm{AB} $$
CE $$ \| \mathrm{AB} \quad $$ (By construction)
$$ A E \| B C $$(Given)
So, ABCE is a parallelogram.
Since, AB $$ \| $$ CE
$$ \Rightarrow \angle B A E=\angle C E D=\angle X $$(Corresponding angles)....(1)
Now, $$ \mathrm{BA} \cong \mathrm{CE} \quad $$ (ABCE is a parallelogram)
and $$ \mathrm{BA} \cong \mathrm{CD} \quad$$(Given)
$$ \mathrm{So}, \mathrm{CE} \cong \mathrm{CD} $$
$$ \Rightarrow \angle C E D=\angle C D E=\angle X $$(Isosceles triangle)....(2)
From (1) and (2) we have
$$ \angle \mathrm{BAE}=\angle \mathrm{CDE}=\angle \mathrm{x} \quad$$...(3)
$$ \angle C E A=180^{\circ}-\angle C E D=180^{\circ}-\angle X $$
In a parallelogram, the opposite angles are equal.
$$ \mathrm{So}, \angle \mathrm{ABC}=\angle \mathrm{CEA}=180^{\circ}-\angle \mathrm{x} \quad$$...(4)
Similarly, $$ \angle \mathrm{BAE}=\angle \mathrm{BCE}=\angle \mathrm{x} $$
In $$ \Delta \mathrm{CED} $$
$$ \angle \mathrm{ECD}+\angle \mathrm{CED}+\angle \mathrm{EDC}=180^{\circ} $$
$$ \Rightarrow \angle \mathrm{ECD}+\mathrm{x}+\mathrm{x}=180^{\circ} $$
$$ \Rightarrow \angle \mathrm{ECD}=180^{\circ}-2 \mathrm{x} $$
Now $$ \angle \mathrm{DCB}=\angle \mathrm{BCE}+\angle \mathrm{ECD} $$
$$ =\mathrm{x}+180^{\circ}-2 \mathrm{x} $$
$$ =180^{\circ}-\mathrm{x} $$
Thus, $$ \angle \mathrm{ABC} \cong \angle \mathrm{DCB} $$
CE $$ \| \mathrm{AB} \quad $$ (By construction)
$$ A E \| B C $$(Given)
So, ABCE is a parallelogram.
Since, AB $$ \| $$ CE
$$ \Rightarrow \angle B A E=\angle C E D=\angle X $$(Corresponding angles)....(1)
Now, $$ \mathrm{BA} \cong \mathrm{CE} \quad $$ (ABCE is a parallelogram)
and $$ \mathrm{BA} \cong \mathrm{CD} \quad$$(Given)
$$ \mathrm{So}, \mathrm{CE} \cong \mathrm{CD} $$
$$ \Rightarrow \angle C E D=\angle C D E=\angle X $$(Isosceles triangle)....(2)
From (1) and (2) we have
$$ \angle \mathrm{BAE}=\angle \mathrm{CDE}=\angle \mathrm{x} \quad$$...(3)
$$ \angle C E A=180^{\circ}-\angle C E D=180^{\circ}-\angle X $$
In a parallelogram, the opposite angles are equal.
$$ \mathrm{So}, \angle \mathrm{ABC}=\angle \mathrm{CEA}=180^{\circ}-\angle \mathrm{x} \quad$$...(4)
Similarly, $$ \angle \mathrm{BAE}=\angle \mathrm{BCE}=\angle \mathrm{x} $$
In $$ \Delta \mathrm{CED} $$
$$ \angle \mathrm{ECD}+\angle \mathrm{CED}+\angle \mathrm{EDC}=180^{\circ} $$
$$ \Rightarrow \angle \mathrm{ECD}+\mathrm{x}+\mathrm{x}=180^{\circ} $$
$$ \Rightarrow \angle \mathrm{ECD}=180^{\circ}-2 \mathrm{x} $$
Now $$ \angle \mathrm{DCB}=\angle \mathrm{BCE}+\angle \mathrm{ECD} $$
$$ =\mathrm{x}+180^{\circ}-2 \mathrm{x} $$
$$ =180^{\circ}-\mathrm{x} $$
Thus, $$ \angle \mathrm{ABC} \cong \angle \mathrm{DCB} $$
In the adjacent figure, $$ \square \mathrm{ABCD} $$ is a trapezium $$ \mathrm{AB} \| \mathrm{DC} $$. Points $$ \mathrm{M} $$ and $$ \mathrm{N} $$ are midpoints of diagonal $$ \mathrm{AC} $$ and $$ \mathrm{DB} $$ respectively then prove that $$ \mathrm{MN} \| \mathrm{AB} $$.
Construction: Join $$MN$$. Also, join $$DM$$ and produce it to $$AB$$ in $$P.$$
To prove: $$ \mathrm{MN} \| \mathrm{AB} $$
Proof: $$AB$$ $$ \| D C $$ and $$ A C $$ is the transversal.
$$ \mathrm{So}, \angle 1=\angle 2 \quad $$ (Alternate angles are equal)
$$ \ln \Delta \mathrm{AMP} $$ and $$ \Delta \mathrm{CMD} $$
$$ \angle 1=\angle 2 \quad $$ (Alternate angles are equal)
$$ \angle 3=\angle 4 \quad $$ (Vertically opposite angles)
$$ \mathrm{AM}=\mathrm{MC} \quad(\mathrm{M} $$ is midpoint of $$ \mathrm{AC}) $$
$$ \mathrm{So}, \Delta \mathrm{AMP} \cong \Delta \mathrm{CMD} \quad $$ (ASA congruency criteria)
By $$ \mathrm{C.P.C.T},$$
$$ \mathrm{DM}=\mathrm{MP} $$
So, $$ M $$ is the midpoint of $$DP.$$
To prove: $$ \mathrm{MN} \| \mathrm{AB} $$
Proof: $$AB$$ $$ \| D C $$ and $$ A C $$ is the transversal.
$$ \mathrm{So}, \angle 1=\angle 2 \quad $$ (Alternate angles are equal)
$$ \ln \Delta \mathrm{AMP} $$ and $$ \Delta \mathrm{CMD} $$
$$ \angle 1=\angle 2 \quad $$ (Alternate angles are equal)
$$ \angle 3=\angle 4 \quad $$ (Vertically opposite angles)
$$ \mathrm{AM}=\mathrm{MC} \quad(\mathrm{M} $$ is midpoint of $$ \mathrm{AC}) $$
$$ \mathrm{So}, \Delta \mathrm{AMP} \cong \Delta \mathrm{CMD} \quad $$ (ASA congruency criteria)
By $$ \mathrm{C.P.C.T},$$
$$ \mathrm{DM}=\mathrm{MP} $$
So, $$ M $$ is the midpoint of $$DP.$$
$$ \ln \Delta \mathrm{DPB} $$
$$ \mathrm{M} $$ is the midpoint of $$DP$$ and $$ \mathrm{N} $$ is the midpoint of $$DB$$.
$$ \mathrm{M} $$ is the midpoint of $$DP$$ and $$ \mathrm{N} $$ is the midpoint of $$DB$$.
By, midpoint theorem $$ \mathrm{MN} \| \mathrm{AB} $$.
In the adjacent figure, seg $$ \mathrm{AB} \| $$ seg $$ \mathrm{PQ}, $$ seg $$ \mathrm{AB} \cong \operatorname{seg} \mathrm{PQ}, $$ seg $$ \mathrm{AC} \| $$ seg $$ \mathrm{PR}, $$ seg $$ \mathrm{AC} \cong \operatorname{seg} \mathrm{PR} ,$$ then prove that, seg $$ \mathrm{BC} \| $$ seg $$ \mathrm{QR} $$ and $$ \operatorname{seg} B C \cong \operatorname{seg} $$ $$QR$$.
Given: seg $$ \mathrm{AB} \| $$ seg $$ \mathrm{PQ}, $$ seg $$ \mathrm{AB} \cong \operatorname{seg} \mathrm{PQ} $$
So, $$ABQP$$ is a parallelogram as a pair of opposite sides are congruent and parallel.
Thus, seg $$AP$$ $$ \| $$ seg $$ \mathrm{BQ}, $$ seg $$ \mathrm{AP} \cong \operatorname{seg} \mathrm{BQ} $$ $$\ldots . .(1)$$
So, $$ABQP$$ is a parallelogram as a pair of opposite sides are congruent and parallel.
Thus, seg $$AP$$ $$ \| $$ seg $$ \mathrm{BQ}, $$ seg $$ \mathrm{AP} \cong \operatorname{seg} \mathrm{BQ} $$ $$\ldots . .(1)$$
Similarly, seg $$ \mathrm{AC} \| $$ seg $$ \mathrm{PR}, \operatorname{seg} \mathrm{AC} \cong \operatorname{seg} \mathrm{PR} $$
So, $$APRC$$ is a parallelogram.
Thus, seg $$AP \| \operatorname{seg} C R, \operatorname{seg} A P \cong \operatorname{seg} C R \ldots . .(2)$$
So, $$APRC$$ is a parallelogram.
Thus, seg $$AP \| \operatorname{seg} C R, \operatorname{seg} A P \cong \operatorname{seg} C R \ldots . .(2)$$
From $$( 1 )$$ and $$(2)$$ we have
seg $$BQ \| $$ seg $$ C R $$
Also, seg $$ \mathrm{BQ} \cong \operatorname{seg} \mathrm{CR} $$
Thus, $$BQRC$$ is a parallelogram as a pair of opposite sides are congruent and parallel.
seg $$BQ \| $$ seg $$ C R $$
Also, seg $$ \mathrm{BQ} \cong \operatorname{seg} \mathrm{CR} $$
Thus, $$BQRC$$ is a parallelogram as a pair of opposite sides are congruent and parallel.
Therefore, seg $$BC \| $$ seg $$ \mathrm{QR} $$ and $$ \operatorname{seg} \mathrm{BC} \cong \operatorname{seg} \mathrm{QR} $$ as $$ \mathrm{BQRC} $$ is a parallelogram.
Give a definition for each of the following terms. Are there other terms that need to be defined first ? What are they, and how might you define them ?
Square:
Square is a figure in which all sides and angles are equal.
$$AB = BC = CD = DA = 4 \ cms$$.
$$\angle A = \angle B = \angle C = \angle D = 90^0$$.
Necessary terms - Quadrilateral, angle.
Quadrilateral In a plane, if 3 points in 4 points are non-collinear and meet in different points, such figure is called quadrilateral.
Angle: $$'O'$$ is common point between $$\overrightarrow{OA}$$ and $$\overrightarrow{OB}$$ Such figure is called an angle.
$$AB = BC = CD = DA = 4 \ cms$$.
$$\angle A = \angle B = \angle C = \angle D = 90^0$$.
Necessary terms - Quadrilateral, angle.
Quadrilateral In a plane, if 3 points in 4 points are non-collinear and meet in different points, such figure is called quadrilateral.
Angle: $$'O'$$ is common point between $$\overrightarrow{OA}$$ and $$\overrightarrow{OB}$$ Such figure is called an angle.
In $$ \square \mathrm{ABCD}, $$ side $$ \mathrm{BC} \| $$ side $$ \mathrm{AD}, $$ side $$ \mathrm{AB} \cong $$ side $$ \mathrm{DC} $$ If $$ \angle \mathrm{A}=72^{\circ} $$ then find the measure of $$ \angle \mathrm{B}, $$ and $$ \angle \mathrm{D} $$
side $$ \mathrm{BC} \| $$ side $$ \mathrm{AD} $$
Interior angles on the same side of the transversal are supplementary.
$$ \angle A+\angle B=180^{\circ} $$
$$ \Rightarrow 72^{\circ}+\angle \mathrm{B}=180^{\circ} $$
$$ \Rightarrow \angle B=180^{\circ}-72^{\circ} $$
$$ \Rightarrow \angle B=108^{\circ} $$
Construction:
Draw BP $$ \| \subset D $$
$$ \mathrm{So}, \mathrm{BC} \| \mathrm{AD} $$ and $$ \mathrm{BP} \| \mathrm{CD} $$
$$ \Rightarrow \mathrm{PBCD} $$ is a parallelogram
$$ \Rightarrow C D \cong B P $$
Now $$ C D=B P $$ and $$ C D=A B $$
$$ \Rightarrow \mathrm{BP}=\mathrm{AB} $$
$$ \Rightarrow \angle B A P=\angle B P A=72^{\circ} $$
$$ \mathrm{BP} \| \mathrm{CD \ } \mathrm{so}, $$
$$ \angle C D P=\angle B P A=72^{\circ} $$(corresponding angles are equal)
$$ \angle B=108^{\circ} $$
$$ \angle D=72^{\circ} $$
Interior angles on the same side of the transversal are supplementary.
$$ \angle A+\angle B=180^{\circ} $$
$$ \Rightarrow 72^{\circ}+\angle \mathrm{B}=180^{\circ} $$
$$ \Rightarrow \angle B=180^{\circ}-72^{\circ} $$
$$ \Rightarrow \angle B=108^{\circ} $$
Construction:
Draw BP $$ \| \subset D $$
$$ \mathrm{So}, \mathrm{BC} \| \mathrm{AD} $$ and $$ \mathrm{BP} \| \mathrm{CD} $$
$$ \Rightarrow \mathrm{PBCD} $$ is a parallelogram
$$ \Rightarrow C D \cong B P $$
Now $$ C D=B P $$ and $$ C D=A B $$
$$ \Rightarrow \mathrm{BP}=\mathrm{AB} $$
$$ \Rightarrow \angle B A P=\angle B P A=72^{\circ} $$
$$ \mathrm{BP} \| \mathrm{CD \ } \mathrm{so}, $$
$$ \angle C D P=\angle B P A=72^{\circ} $$(corresponding angles are equal)
$$ \angle B=108^{\circ} $$
$$ \angle D=72^{\circ} $$
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that $$\angle A> \angle C$$ and $$ \angle B > \angle D$$.
Data : $$AB$$ and $$CD$$ are respectively the smallest and longest sides of a quadrilateral $$ABCD$$.
To Prove :
(i) $$\angle A > \angle C$$
(ii) $$\angle B > \angle D$$
Construction: Join $$AC$$ and $$BD$$
Proof: In $$\Delta ABC, AB < BC$$ (data)
$$\therefore \ \ \angle 3 < \angle 4......(i)$$
In $$\Delta ADC, AD < CD$$ (data)
$$\angle 2 < \angle 1.....(ii)$$
Adding (i) and (ii),
$$\angle 3 + \angle 2 < \angle 4 + \angle 1$$
$$\angle C < \angle A$$
$$\angle A > \angle C$$
By adding $$B$$,
In $$\Delta ABD, AB < AD$$
$$\angle 5 < \angle 8$$
In $$\Delta BCD, BC < CD$$
$$\angle 6 < \angle 7$$
$$\therefore \ \ \angle 5 + \angle 6 < \angle 8 + \angle 7$$
$$\angle D < \angle B$$
OR $$\angle B > \angle D$$.
To Prove :
(i) $$\angle A > \angle C$$
(ii) $$\angle B > \angle D$$
Construction: Join $$AC$$ and $$BD$$
Proof: In $$\Delta ABC, AB < BC$$ (data)
$$\therefore \ \ \angle 3 < \angle 4......(i)$$
In $$\Delta ADC, AD < CD$$ (data)
$$\angle 2 < \angle 1.....(ii)$$
Adding (i) and (ii),
$$\angle 3 + \angle 2 < \angle 4 + \angle 1$$
$$\angle C < \angle A$$
$$\angle A > \angle C$$
By adding $$B$$,
In $$\Delta ABD, AB < AD$$
$$\angle 5 < \angle 8$$
In $$\Delta BCD, BC < CD$$
$$\angle 6 < \angle 7$$
$$\therefore \ \ \angle 5 + \angle 6 < \angle 8 + \angle 7$$
$$\angle D < \angle B$$
OR $$\angle B > \angle D$$.
In the given Figure, $$ \mathrm{ABCD} $$ is a trapezium. $$AB\parallel \mathrm{DC} $$. Points $$ \mathrm{P} $$ and $$ \mathrm{Q} $$ are midpoints of seg $$ \mathrm{AD} $$ and seg $$ \mathrm{BC} $$ respectively. Then prove that, $$ \mathrm{PQ} \| \mathrm{AB} $$ and $$ \mathrm{PQ}=\dfrac{1}{2}(A B+D C) $$.
Construction: Join $$PB$$ and extend it to meet $$CD$$ produced at $$R.$$
To prove: $$ \mathrm{PQ} \| \mathrm{AB} $$ and $$ \mathrm{PQ}=\dfrac{1}{2}(\mathrm{AB}+\mathrm{DC}) $$
Proof : In $$ \triangle \mathrm{ABP} $$ and $$ \Delta \mathrm{DRP}, $$
$$ \angle \mathrm{APB}=\angle \mathrm{DPR} \quad $$ (Vertically opposite angles)
$$ \angle \mathrm{PDR}=\angle \mathrm{PAB} \quad $$ (Alternate interior angles are equal)
$$ \mathrm{AP}=\mathrm{PD} \quad(\mathrm{P} $$ is the mid point of $$ \mathrm{AD}) $$
Thus, by $$ASA$$ congruency,
To prove: $$ \mathrm{PQ} \| \mathrm{AB} $$ and $$ \mathrm{PQ}=\dfrac{1}{2}(\mathrm{AB}+\mathrm{DC}) $$
Proof : In $$ \triangle \mathrm{ABP} $$ and $$ \Delta \mathrm{DRP}, $$
$$ \angle \mathrm{APB}=\angle \mathrm{DPR} \quad $$ (Vertically opposite angles)
$$ \angle \mathrm{PDR}=\angle \mathrm{PAB} \quad $$ (Alternate interior angles are equal)
$$ \mathrm{AP}=\mathrm{PD} \quad(\mathrm{P} $$ is the mid point of $$ \mathrm{AD}) $$
Thus, by $$ASA$$ congruency,
$$ \Delta \mathrm{ABP} \cong \Delta DRP.$$
By $$ C P C T, P B=P R $$ and $$ A B=R D $$
$$ \ln \Delta \mathrm{BRC} $$
$$ \mathrm{Q} $$ is the mid point of $$ \mathrm{BC} $$ (Given)
$$ \mathrm{P} $$ is the mid point of $$ \mathrm{BR} $$
$$ (\mathrm{As}\ \mathrm{PB}=\mathrm{PR}) $$
So, by midpoint theorem, PQ $$ \| \mathrm{RC} $$
$$ \Rightarrow \mathrm{PQ} \| \mathrm{DC} $$
But $$ \mathrm{AB} \| \mathrm{DC} $$(Given)
$$ \mathrm{So}, \mathrm{PQ} \| \mathrm{AB} $$
$$ \mathrm{Also}, \mathrm{PQ}=\dfrac{1}{2}(\mathrm{RC}) \ldots . $$ (using midpoint theorem)
$$ \mathrm{PQ}=\dfrac{1}{2}(\mathrm{RD}+\mathrm{DC}) $$
$$ \mathrm{PQ}=\dfrac{1}{2}(\mathrm{AB}+\mathrm{DC}) \quad(\because \mathrm{AB}=\mathrm{RD}) $$
By $$ C P C T, P B=P R $$ and $$ A B=R D $$
$$ \ln \Delta \mathrm{BRC} $$
$$ \mathrm{Q} $$ is the mid point of $$ \mathrm{BC} $$ (Given)
$$ \mathrm{P} $$ is the mid point of $$ \mathrm{BR} $$
$$ (\mathrm{As}\ \mathrm{PB}=\mathrm{PR}) $$
So, by midpoint theorem, PQ $$ \| \mathrm{RC} $$
$$ \Rightarrow \mathrm{PQ} \| \mathrm{DC} $$
But $$ \mathrm{AB} \| \mathrm{DC} $$(Given)
$$ \mathrm{So}, \mathrm{PQ} \| \mathrm{AB} $$
$$ \mathrm{Also}, \mathrm{PQ}=\dfrac{1}{2}(\mathrm{RC}) \ldots . $$ (using midpoint theorem)
$$ \mathrm{PQ}=\dfrac{1}{2}(\mathrm{RD}+\mathrm{DC}) $$
$$ \mathrm{PQ}=\dfrac{1}{2}(\mathrm{AB}+\mathrm{DC}) \quad(\because \mathrm{AB}=\mathrm{RD}) $$
Hence, Proved.
$$D, E$$ and $$F$$ are respectively the midpoints of the sides $$BC, CA$$ and $$AB$$ of a $$\triangle ABC$$. Show that
$$BDEF$$ is a parallelogram.
In figure $$ABC$$ and $$BDE$$ are two equilateral triangles such thta $$D$$ is the mid-point of $$BC.AE$$ intersects $$BC$$ in $$F$$. Prove that
$$ar(\triangle BDE)=\dfrac {1}{4}ar(\triangle ABC)$$
$$ar. (\triangle ABC)=2 ar.(\triangle BEC)$$
Complete the hexagonal and star shaped Rangolies by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles ?
Fig. (i): Regular Hexagonal with side 5 cm. is constructed. It has 6 equal sides. Measure of each side is 5 cm.
Number of equilateral $$\Delta $$ with 1 cm side
= $$\frac{\ Area \ of \ whole \ Hexagonal}{\ Area \ of \ equilateral \Delta with side 1 \ cm}$$.
= $$\frac{6 \times \ equilateral \ triangles}{\frac{\sqrt3}{4}a^2}$$
= $$\frac{6 \times \frac{\sqrt3}{4}a^2}{\frac{sqrt3}{4}a^2}$$
= $$\frac{6(5)^2}{(1)^2}$$
= $$6 \times 25$$
= $$150 \ triangles$$.
In Fig. (ii) : there are 12 equilateral triangles with side 5 cm
$$\therefore$$ Number of equilateral $$A$$ with 1 cm side
= $$\frac{\ Area \ of \ 12 \ equilateral \ triangles \ with \ side \ 5 \ cm.}{\ Area \ of \ equilateral \ triangle \ with \ side 1 \ cm}$$.
= $$\frac{12 \frac{\sqrt3}{4}a^2}{1 \times \frac{\sqrt3}{4}a^2}$$
= $$\frac{12(5)^2}{(1)^2}$$
$$= 12 \times 25$$
= $$300 \ triangles$$.
Number of equilateral $$\Delta $$ with 1 cm side
= $$\frac{\ Area \ of \ whole \ Hexagonal}{\ Area \ of \ equilateral \Delta with side 1 \ cm}$$.
= $$\frac{6 \times \ equilateral \ triangles}{\frac{\sqrt3}{4}a^2}$$
= $$\frac{6 \times \frac{\sqrt3}{4}a^2}{\frac{sqrt3}{4}a^2}$$
= $$\frac{6(5)^2}{(1)^2}$$
= $$6 \times 25$$
= $$150 \ triangles$$.
In Fig. (ii) : there are 12 equilateral triangles with side 5 cm
$$\therefore$$ Number of equilateral $$A$$ with 1 cm side
= $$\frac{\ Area \ of \ 12 \ equilateral \ triangles \ with \ side \ 5 \ cm.}{\ Area \ of \ equilateral \ triangle \ with \ side 1 \ cm}$$.
= $$\frac{12 \frac{\sqrt3}{4}a^2}{1 \times \frac{\sqrt3}{4}a^2}$$
= $$\frac{12(5)^2}{(1)^2}$$
$$= 12 \times 25$$
= $$300 \ triangles$$.
In a triangle $$ABC, E$$ is the mid-point of median $$AD$$. Show that $$ar.(\triangle BED)=\dfrac{1}{4} ar.(\triangle ABC)$$.
If $$E, F, G$$ and $$H$$ are respectively the mid-points of the sides of a parallelogram $$ABCD$$, show that
$$D, E$$ and $$F$$ are respectively the midpoints of the sides $$BC, CA$$ and $$AB$$ of a $$\triangle ABC$$. Show that
$$ar.(\triangle DEF)=\dfrac{1}{4}\ ar.(\triangle ABC)$$
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.
$$\triangle PDC$$ and quadrilateral $$ABCD$$ are on same base $$DC$$. They are in between $$AD \parallel DC$$
In figure $$ABC$$ and $$BDE$$ are two equilateral triangles such thta $$D$$ is the mid-point of $$BC.AE$$ intersects $$BC$$ in $$F$$. Prove that
$$ar(\triangle BDE)=\dfrac {1}{4}ar(\triangle ABC)$$
$$ar. (\triangle FED)=\dfrac 18 ar.(\triangle AFC)$$
In figure $$ABC$$ and $$BDE$$ are two equilateral triangles such thta $$D$$ is the mid-point of $$BC.AE$$ intersects $$BC$$ in $$F$$. Prove that
$$ar(\triangle BDE)=\dfrac {1}{4}ar(\triangle ABC)$$
$$ar. (\triangle BDE)=\dfrac 12 ar.(\triangle BAE)$$
In figure $$ABC$$ and $$BDE$$ are two equilateral triangles such thta $$D$$ is the mid-point of $$BC.AE$$ intersects $$BC$$ in $$F$$. Prove that
$$ar(\triangle BDE)=\dfrac {1}{4}ar(\triangle ABC)$$
$$ar. (\triangle BEF)=ar.(\triangle AFD)$$
Complete the following:
A quadrilateral has ___ diagonals
$$P$$ and $$Q$$ are respectively the mid points of sides $$AB$$ and $$BC$$ of a triangle $$AABC$$ and $$R$$ is the mid point of $$AP$$, show that
$$ar. (\triangle RQC)=\dfrac 38 ar. (\triangle ABC)$$
In fig. $$ABCDE$$ is a pentagon. A line through $$B$$ parallel to $$AC$$ meets $$DC$$ produced at $$F$$. Show that
$$ar.(\triangle ACB)=ar.(\triangle ACF)$$
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decide to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot.
Explain how this proposal with be implemented.
$$BACD$$ is a trapezium with $$AB\parallel DC$$. A line parallel to $$AC$$ intersects $$AB$$ at $$X$$ and $$BC$$ at $$Y$$. Prove that area $$(ADx)=$$ area $$(ACy)$$. [Hint: Join $$Cx$$.)
In the figure ABCD is a square . M , N , O & P are the midpoint of sides AB , BC CD and DA respectively . Identify the congruent triangles.
In fig. $$ABCDE$$ is a pentagon. A line through $$B$$ parallel to $$AC$$ meets $$DC$$ produced at $$F$$. Show that
$$ar.(\triangle EDF)=ar.(ABCDE)$$.
Complete the following:
A quadrilateral has _____ sides.
$$D, E$$ and $$F$$ are respectively the midpoints of the sides $$BC, CA$$ and $$AB$$ of a $$\triangle ABC$$. Show that
$$ar.(BDEF)=\dfrac{1}{2}\ ar.(\triangle ABC)$$.
$$P$$ and $$Q$$ are respectively the mid points of sides $$AB$$ and $$BC$$ of a triangle $$AABC$$ and $$R$$ is the mid point of $$AP$$, show that
$$ar. (\triangle PRQ)=\dfrac 12 ar. (\triangle ARC)$$
In any quadrilateral $$ABCD$$, $$\angle A+\angle D={90}^{o}$$. Prove that $${AC}^{2}+{BD}^{2}={AD}^{2}+{BC}^{2}$$
Proof: $$\angle A+\angle D={90}^{o}$$
In triangle $$ADE$$
$$\angle A+\angle D+\angle E={180}^{o}$$
$$\Rightarrow$$ $${90}^{o}+\angle E={180}^{o}$$
$$\Rightarrow$$ $$\angle E={180}^{o}-{90}^{o}={90}^{o}$$
In right triangle BEC
$${BC}^{2}={BE}^{2}+{CE}^{2}$$
(By Pythagoras theorem)
In triangle $$ADE$$
$$\angle A+\angle D+\angle E={180}^{o}$$
$$\Rightarrow$$ $${90}^{o}+\angle E={180}^{o}$$
$$\Rightarrow$$ $$\angle E={180}^{o}-{90}^{o}={90}^{o}$$
In right triangle BEC
$${BC}^{2}={BE}^{2}+{CE}^{2}$$
(By Pythagoras theorem)
In right triangle AED
$${AD}^{2}={AE}^{2}+{DE}^{2}$$...(ii)
Adding (i) and (ii)
$${BC}^{2}+{AD}^{2}={BE}^{2}+{AE}^{2}+{CE}^{2}+{DE}^{2}$$....(iii)
In right $$\triangle AEC$$
$${AC}^{3}={AE}^{2}+{CE}^{2}$$...(iv)
In right angled triangle $$BED$$
$${BD}^{2}={BE}^{2}+{DE}^{2}$$...(v)
From equation (iv) and (v)
$${AC}^{2}+{BD}^{2}={BE}^{2}+{AE}^{2}+{CE}^{2}+{DE}^{2}$$....(vi)
From equations (iii) and (iv)
$${BC}^{2}+{AD}^{2}={AC}^{2}+{BD}^{2}$$
$$\Rightarrow {AC}^{2}+{BD}^{2}={AD}^{2}+{BC}^{2}$$
$${AD}^{2}={AE}^{2}+{DE}^{2}$$...(ii)
Adding (i) and (ii)
$${BC}^{2}+{AD}^{2}={BE}^{2}+{AE}^{2}+{CE}^{2}+{DE}^{2}$$....(iii)
In right $$\triangle AEC$$
$${AC}^{3}={AE}^{2}+{CE}^{2}$$...(iv)
In right angled triangle $$BED$$
$${BD}^{2}={BE}^{2}+{DE}^{2}$$...(v)
From equation (iv) and (v)
$${AC}^{2}+{BD}^{2}={BE}^{2}+{AE}^{2}+{CE}^{2}+{DE}^{2}$$....(vi)
From equations (iii) and (iv)
$${BC}^{2}+{AD}^{2}={AC}^{2}+{BD}^{2}$$
$$\Rightarrow {AC}^{2}+{BD}^{2}={AD}^{2}+{BC}^{2}$$
Fill in the blanks:
All squares are _____.
All squares are similar.
Let $$ABCD$$ be a quadrilateral in which $$AB$$ is the smallest side and $$CD$$ is a largest side $$\angle A > \angle C$$ and $$\angle B > \angle D$$ ( Hint: join $$AC$$ and $$BD$$)
Join $$AC$$ and $$BD$$ In $$\triangle ABD, AB$$ is the smallest side
$$AD > AB$$
$$\angle ABD > \angle APB$$
[ Angle opposite to larger side ] ....(i)
In $$\Delta CBD, CD$$ is the largest side
$$\therefore CD > CB$$
$$\therefore [ CBD > \angle CDB$$ [ Angle opposite to larger side ] .... (ii)
By adding (i) and (ii)
$$\angle ABD+\angle CBD > \angle ADB +\angle CDB$$
$$\angle ABC > \angle ADC$$
$$\therefore \angle B > \angle D$$
In $$\triangle ABC, AB$$ is the smallest side ] ......(iv)
$$\therefore BC > AB \angle BAC > \angle ACB$$ [ Angle opposite to larger side ] .....(iii)
In $$\Delta ADC, CD$$ is the largest side
$$\therefore CD > AD$$
$$\angle CAD > \angle ACD$$ [ Angle opposite to larger side ] .....(iv)
By adding (iii) and (iv)
$$\angle BAC +\angle CAD > \angle ACB + \angle ACD$$
$$\angle BAD > \angle ACD$$
$$\angle A > \angle C$$
$$AD > AB$$
$$\angle ABD > \angle APB$$
[ Angle opposite to larger side ] ....(i)
In $$\Delta CBD, CD$$ is the largest side
$$\therefore CD > CB$$
$$\therefore [ CBD > \angle CDB$$ [ Angle opposite to larger side ] .... (ii)
By adding (i) and (ii)
$$\angle ABD+\angle CBD > \angle ADB +\angle CDB$$
$$\angle ABC > \angle ADC$$
$$\therefore \angle B > \angle D$$
In $$\triangle ABC, AB$$ is the smallest side ] ......(iv)
$$\therefore BC > AB \angle BAC > \angle ACB$$ [ Angle opposite to larger side ] .....(iii)
In $$\Delta ADC, CD$$ is the largest side
$$\therefore CD > AD$$
$$\angle CAD > \angle ACD$$ [ Angle opposite to larger side ] .....(iv)
By adding (iii) and (iv)
$$\angle BAC +\angle CAD > \angle ACB + \angle ACD$$
$$\angle BAD > \angle ACD$$
$$\angle A > \angle C$$
Let AB ,. CD be two segment such that $$AB | | CD $$ and $$AD | | BC $$ . Let E be the midpoint of BC and let DE extended meet AB in F. Prove that AB = BF
In $$\bigtriangleup BEF $$ and $$\bigtriangleup DEC$$
$$BE = EC$$ [data]
$$\angle BEF - \angle DEC$$ [vertically opposite angles ] \
$$\therefore \bigtriangleup BEF \cong \bigtriangleup DEC $$ [ASA postulate]
$$\therefore BF = DC $$ [corresponding sides]
But $$AB = DC $$ [data]
$$\therefore AB = BF $$
$$BE = EC$$ [data]
$$\angle BEF - \angle DEC$$ [vertically opposite angles ] \
$$\therefore \bigtriangleup BEF \cong \bigtriangleup DEC $$ [ASA postulate]
$$\therefore BF = DC $$ [corresponding sides]
But $$AB = DC $$ [data]
$$\therefore AB = BF $$
Suppose ABC is a triangle and D is the midpoint of BC . Assume that the perpendicular from D to AB and AC are of equal length . Prove that ABC is isosceles .
Let ABCD be a quadrilateral in which AD is the largest side and BC is the smallest side. Prove that $$ \angle A < \angle C$$ [Hint: join AC ]
In a triangle $$ABC$$, let $$D$$ be the midpoint of $$BC$$. Prove that $$AB+AC > 2AD$$ (what property of quadrilateral is needed here )
Let $$ABCD$$ be a quadrilateral and $$P, Q, R, S$$ be the midpoints of $$AB, BC, CD, DA$$ respectively prove that $$PQRS$$ is a parallelogram [ what extra result you need to prove this ] It can be proved if $$ABCD$$ is a rectangle
Prove that midpoint (C) of hypotaneous in a right-angled triangle AOB is situated at equal distance from vertices $$ 0, \mathrm{A} $$, and $$ \mathrm{B} $$ of triangle.
Let in right angled triangle $$ \mathrm{AOB} $$, the vertices $$ 0(0,0), \mathrm{A}(\mathrm{a}, 0) $$ and $$ \mathrm{B}(0, \mathrm{b}) $$ are shown in following.
Now co-ordinate of mid point C of hypotaneous
$$\begin{aligned}&=\left(\dfrac{0+a}{2}, \dfrac{0+b}{2}\right) \\&=\left(\dfrac{a}{2}, \dfrac{b}{2}\right) \\OC &=\sqrt{\left(\dfrac{a}{2}-0\right)^{2}+\left(\dfrac{b}{2}-0\right)^{2}} \\&=\sqrt{\dfrac{a^{2}}{4}+\dfrac{b^{2}}{4}} \\C A &=\sqrt{\left(a-\dfrac{a}{2}\right)^{2}+\left(-0-\dfrac{b}{2}\right)^{2}} \\&=\sqrt{\left(\dfrac{a}{2}\right)^{2}+\left(-\dfrac{b}{2}\right)^{2}} \\&=\sqrt{\dfrac{a^{2}}{4}+\dfrac{b^{2}}{4}}\end{aligned}$$
$$\begin{aligned}\mathrm{CB} &=\sqrt{\left(0-\dfrac{a}{2}\right)^{2}+\left(b-\dfrac{b}{2}\right)^{2}} \\&=\sqrt{\left(-\dfrac{a}{2}\right)^{2}+\left(\dfrac{b}{2}\right)^{2}} \\&=\sqrt{\dfrac{a^{2}}{4}+\dfrac{b^{2}}{4}}\end{aligned}$$
Clearly : $$ O C=C A=C B $$
Hence, the mid point $$ C $$ of hypotaneous in a right angled triangle $$ A O B $$ is situated at equal distance from vertices $$ 0, $$ A and $$ B $$ of the triangle.
Now co-ordinate of mid point C of hypotaneous
$$\begin{aligned}&=\left(\dfrac{0+a}{2}, \dfrac{0+b}{2}\right) \\&=\left(\dfrac{a}{2}, \dfrac{b}{2}\right) \\OC &=\sqrt{\left(\dfrac{a}{2}-0\right)^{2}+\left(\dfrac{b}{2}-0\right)^{2}} \\&=\sqrt{\dfrac{a^{2}}{4}+\dfrac{b^{2}}{4}} \\C A &=\sqrt{\left(a-\dfrac{a}{2}\right)^{2}+\left(-0-\dfrac{b}{2}\right)^{2}} \\&=\sqrt{\left(\dfrac{a}{2}\right)^{2}+\left(-\dfrac{b}{2}\right)^{2}} \\&=\sqrt{\dfrac{a^{2}}{4}+\dfrac{b^{2}}{4}}\end{aligned}$$
$$\begin{aligned}\mathrm{CB} &=\sqrt{\left(0-\dfrac{a}{2}\right)^{2}+\left(b-\dfrac{b}{2}\right)^{2}} \\&=\sqrt{\left(-\dfrac{a}{2}\right)^{2}+\left(\dfrac{b}{2}\right)^{2}} \\&=\sqrt{\dfrac{a^{2}}{4}+\dfrac{b^{2}}{4}}\end{aligned}$$
Clearly : $$ O C=C A=C B $$
Hence, the mid point $$ C $$ of hypotaneous in a right angled triangle $$ A O B $$ is situated at equal distance from vertices $$ 0, $$ A and $$ B $$ of the triangle.
Let $$ABCD$$ be a quadrilateral with diagonals $$AC$$ and $$BD$$. Prove the following statement
$$AB+BC+CD > AD$$
Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to the vertex of the right angle is equal to half the hypotenuse.
Let $$ABC$$ is a right triangle right angled at $$B$$.
Let $$P$$ be the mid-point of the hypotenuse $$AC$$.
Through $$P$$, draw $$PQ || BC$$
$$PQ || BC$$
$$\angle 1 = \angle B$$ (corresponding angle)
$$\angle 2 = \angle 1 = 90^{0}$$
In $$\Delta ABC, P$$ is the mid-point of $$AC$$ and $$PQ || BC$$
$$Q$$ must be mid-point of $$AB$$ (converse of mid-point theorem)
$$AQ = QB$$ …(i)
Now in $$\Delta AQP$$ and $$\Delta BPQ$$
$$AQ = QB$$
$$\angle 2 = \angle 1 = 90^{0}$$
$$PQ = PQ$$
$$\Delta AQP = \Delta BPQ$$ (by SAS congruency rule)
$$AP = PB$$ (by c.p.c.t)
Thus, $$AP = PB = PC$$
Hence in other words, we can say that the point $$P$$ is equi-distance from the vertices of a right angled triangle $$ABC$$.
Let $$P$$ be the mid-point of the hypotenuse $$AC$$.
Through $$P$$, draw $$PQ || BC$$
$$PQ || BC$$
$$\angle 1 = \angle B$$ (corresponding angle)
$$\angle 2 = \angle 1 = 90^{0}$$
In $$\Delta ABC, P$$ is the mid-point of $$AC$$ and $$PQ || BC$$
$$Q$$ must be mid-point of $$AB$$ (converse of mid-point theorem)
$$AQ = QB$$ …(i)
Now in $$\Delta AQP$$ and $$\Delta BPQ$$
$$AQ = QB$$
$$\angle 2 = \angle 1 = 90^{0}$$
$$PQ = PQ$$
$$\Delta AQP = \Delta BPQ$$ (by SAS congruency rule)
$$AP = PB$$ (by c.p.c.t)
Thus, $$AP = PB = PC$$
Hence in other words, we can say that the point $$P$$ is equi-distance from the vertices of a right angled triangle $$ABC$$.
Show that the line segment joining the mid-points of the consecutive sides of a square is also a square.
In figure, $$ABCD$$ is a parallelogram $$P$$ and $$Q$$ are the mid-points of opposite sides $$AB$$ and $$DC$$ of a parallelogram $$ABCD$$. Prove that $$PRQS$$ is a parallelogram.
In the given figure, $$ABCD$$ is a trapezium in which $$AB \parallel CD$$. $$M$$ and $$N$$ are mid-points of the diagonal $$AC$$ and $$BD$$ respectively. Prove that $$MN \parallel CD$$ and $$MN= \dfrac12 (CD- AB)$$.
One angle of a quadrilateral is $$108^\circ$$ and the other three angles are equal. Find the value of each of the equal angle.
In a quadrilateral $$ABCD, \angle A = 3x, \angle B= 5x, \angle C= 20x$$ and $$\angle D= 8 x$$. Find the value of $$x$$.
Can all the four angles a quadrilateral are obtuse? Give reason for your answer.
The three angles of a quadrilateral measure $$56^\circ, 100^\circ$$ and $$88^\circ$$. Find the measure of the fourth angle.
In figure, the diagonal $$AC$$ of a quadrilateral $$ABCD$$ bisects the angle $$A$$ and $$C$$. Prove that $$AB = AD$$ and $$CB = CD$$.
State whether the following statement is true or false, give a reason. The angles of a certain quadrilateral are $$ 50^{\circ}, 60^{\circ}, 112^{\circ}, 130^{\circ} $$
Let $$ABCD$$ be a unit square. Draw a quadrant of a circle with $$A$$ as the center and $$B,\ D$$ as endpoints of the arc. Similarly, draw a quadrant of a circle with $$B$$ as the center and $$A,\ C$$ as endpoints of the arc. Inscribe a circle $$\Gamma$$ touching the arc $$AC$$ externally, the arc $$BD$$ internally, and also touching the side $$AD.$$ Find the radius of the circle $$\Gamma$$.
Let $$O$$ be the center of $$\Gamma$$ and $$r$$ its radius.
Draw $$OP\, \perp\, AD$$ and $$OQ\, \perp\, AB$$.
Then,
$$OP =AQ= r$$
$$HA=1$$ [Arc $$BD$$ has been drawn with center $$A$$ and radius equal to $$1$$ unit]
$$\Rightarrow OA + OH = 1$$
$$\Rightarrow OA + r=1$$
$$\Rightarrow OA = 1 - r$$
Now apply Pythagoras theorem in $$\triangle OAQ,$$
$$AQ^2+OQ^2 = OA^2$$
$$\Rightarrow r^2 +OQ^2= (1 - r)^2$$
$$\Rightarrow OQ^2= (1 - r)^2-r^2$$
$$\Rightarrow {OQ^2} = \left( {1 - r - r} \right)\left( {1 - r + r} \right)$$
$$\Rightarrow {OQ^2} = \left( {1 - 2r} \right)$$
Now, in $$\triangle OBQ,$$
$$OB=OK+KO$$
$$\Rightarrow OB = 1 + r$$
$$BQ=AB-AQ$$
$$\Rightarrow BQ = 1 - r$$
By using Pythagoras theorem, we get
$$OB^2=BQ^2+OQ^2$$
$$\Rightarrow (1+ r)^2= (1 -r)^2 + 1 - 2r$$
$$\Rightarrow {\left( {1 + r} \right)^2} - {\left( {1 - r} \right)^2} = 1 - 2r$$
$$\Rightarrow \left( {1 + r - 1 + r} \right)\left( {1 + r + 1 - r} \right) = 1 - 2r$$
$$\Rightarrow 2r \times 2 = 1 - 2r$$
$$\Rightarrow 4r + 2r = 1$$
$$\Rightarrow 6r= 1$$
$$\Rightarrow r= \dfrac16$$ unit
Suppose $$EFGH$$ is a trapezium in which $$EF$$ is parallel to $$HG$$. Through $$X$$, the mid-point of $$EH$$, $$XY$$ is drawn parallel to $$EF$$ meeting $$FG$$ at $$Y$$. Prove that $$XY$$ bisects $$FG$$.
Given: $$EFGH$$ is a trapezium with $$EF | | HG$$. Through $$X$$ the mid-point of $$EH$$, is drawn parallel to $$EF$$.
To prove: $$XY$$ bisects $$FG$$.
Construction: Join $$HF$$ to intersect $$XY$$ at $$P$$
In $$Δ HEF, XP | | EF$$ (converse of mid-point theorem)
Therefore, $$XP$$ bisects $$HF, PY | | HG$$ ( as $$HY | | EF$$ and $$EF | | HG$$)
$$Δ HGF, PY$$ passes through mid-point of $$HF$$ and is parallel to $$HG$$ (converse of mid-point theorem)
Hence, $$XY$$ bisects $$FG$$.
Suppose $$X$$, $$Y$$ and $$Z$$ are the midpoints of the sides $$PQ$$, $$QR$$ and $$RP$$ respectively of a triangle $$PQR$$. Prove that $$XYRZ$$ is a parallelogram.
Given: $$x, y$$ and $$z$$ are the mid-point of the sides $$PQ, QR$$ and $$RP$$ of $$Δ QPR, xy$$ and $$xz$$ are joined.
To Prove: $$XYRZ$$ is a parallelogram.
Proof:
In $$\Delta PQR, x, y,$$ and $$z$$ are the mid-point of $$PQ,QR$$ and $$RP$$ respectively.
Also,
$$XY||PR$$ (mid-point theorem)
And,
$$ XY=\frac { 1 }{ 2 } PR$$ (mid-point theorem)
But we also know that,
$$ZR=\frac { 1 }{ 2 } PR$$ (Given)
Therefore, $$XYRZ$$ is a Parallelogram (Since, One pair of opposite sides are parallel and equal)
Hence proved.
The base of a pyramid is a rhombus whose diagonals are 6 m and 8 m long. The altitude of the pyramid passes through the point of intersection, of the diagonals of the rhombus and is 1 m long. Find the lateral area of the pyramid.
Prove that if the mid-points of the opposite sides of a quadrilateral are joined, they bisect each other.
Given: $$ABCD$$ is a quadrilateral $$P, Q, R$$ and $$S$$ are mid-points of $$AB, BC, CD$$ and $$DA$$ respectively.
To prove: $$PR$$ and $$SQ$$ bisect each other.
Construction: join $$AC$$.
In $$Δ ABC, PQ | | AC$$, and $$PQ=\frac { 1 }{ 2 } AC$$ (mid-point theorem) ……. (1)
In $$Δ ADC, SR | | AC$$, and $$SR=\frac { 1 }{ 2 } AC$$ (mid-point theorem) ……. (2)
Therefore, $$PQ | | SR$$ and $$PQ = SR$$ (from (1) and (2) opposite sides equal and parallel)
Thus, $$PQRS$$ is a parallelogram and $$PR$$ and $$SQ$$ are diagonals
Hence, $$PR$$ bisect $$SQ$$.
In a rectangle $$ABCD$$, $$P$$, $$Q$$, $$R$$ and $$S$$ are the midpoints of the sides $$AB$$, $$BC$$, $$CD$$ and $$DA$$ respectively. Find the area of $$PQRS$$ in terms of area of $$ABCD$$.
Given: $$ABCD$$ is a rectangle, $$P, Q, R$$ and $$S$$ are the midpoints of $$AB, BC, CD$$ and $$DA$$ respectively. $$PQ, QR, RS$$ and $$SP$$ are joined.
To find: the areas $$(PQRS)$$ in terms of area $$(ABCD)$$
$$S$$ and $$Q$$ are the mid-point of $$AD$$ and $$BC$$.
Therefore, $$SQ$$ divides rectangles $$ABCD$$ into two rectangles equal in area.
Area$$(ABQS) =$$ area$$(SQCD) =$$$$\frac { 1 }{ 2 }$$area$$(ABCD)$$
$$Δ SRQ$$ and parallelogram (rectangles) $$SQCD$$ stand on the same base $$SQ$$ and are between the same parallel.
Therefore, Area$$Δ SRQ =$$$$\frac { 1 }{ 2 }$$ area$$(SQCD)$$
Similarly Area$$Δ SPQ =$$$$\frac { 1 }{ 2 }$$ area$$(SABQ)$$
Adding, area $$Δ SRQ +$$ area $$Δ SPQ =$$$$\frac { 1 }{ 2 }$$ area $$(SQCD) +$$$$\frac { 1 }{ 2 }$$ area $$(SABQ)$$
i.e., area $$(PQRS) =$$$$\frac { 1 }{ 2 }$$ area $$(ABCD)$$
Hence, area $$(PQRS) =$$$$\frac { 1 }{ 2 }$$ area $$(ABCD)$$.
Prove the converse of the mid-point theorem following the guidelines given below: Consider a triangle $$ABC$$ with $$D$$ as the mid-point of $$AB$$. Draw $$DE \parallel BC$$ to intersect $$AC$$ in $$E$$. Let $${E}_{1}$$ be the mid-point of $$AC$$. Use mid-point theorem to get $$D{E}_{1} \parallel BC$$ and $$D{E}_{1} = {BC}/{2}$$. Conclude $$E = {E}_{1}$$ and hence $$E$$ is the mid-point of $$AC$$.
Given: In $$Δ ABC, D$$ is the mid-point of $$AB. DE | | BC$$
To prove: $$E$$ is the midpoint of $$AC$$
In $$Δ ABC, D$$ is the mid-point of $$AB. DE | | BC$$ (Given)
Let $$E$$, be the mid-point of $$AC$$.
Join $$DE, D$$ is the mid-point of $$AB$$ and $$E_1$$ is the mid-point of $$AC$$.
$$∴ DE_1 | | BC$$ (Mid-point theorem parallels)
But $$DE | | BC$$ (Given)
This is possible only if $$E$$ and $$E_1$$ consider (Through a given point, only one line can be drawn $$| |$$ to be a given line)
Therefore, $$E$$ and $$E_1$$ coincide.
i.e. $$DE_1$$ is the same as $$DE$$.
Thus a line drawn through the mid-point of a side of a triangle and parallel to another bisects the third side.
Hence, $$E=E_1$$ and $$E$$ is the mid-point of $$AC$$.
Prove that the sum of four angles of a quadrilateral ABCD is $$360^{0}$$.
Consider a quad quadrilateral ABCD.
Join QS.
To\quad prove:$$∠A+∠B+∠C+∠D=360º$$
Proof:Consider triangle ABD,
we have,$$∠A+∠ABD+∠ADB=180º\Longrightarrow (1)$$[Using Angle sum property of Triangle]
Similarly,triangle BCD,we have,$$∠DBC+∠C+∠BDC=180º\Longrightarrow (2)$$[Using Angle sum property of Triangle]
On adding (1) and (2),we get$$∠A+∠ABD+∠ADB+∠DBC+∠C+∠BDC=180º+180º$$
$$∠A+∠B+∠C+∠D=360º$$
Ina quadrilateral $$ABCD$$, $$AB=AD$$ and $$CB=CD$$. Prove that :(i) $$AC$$ bisects angle $$BAD$$.(ii) $$AC$$ is perpendicular bisector of $$BD$$.
In quadrilateral ABCD $$\angle$$B$$=90^o$$, $$\angle$$C $$-\angle$$D$$=60^o$$ and $$\angle$$A $$-\angle$$C$$-\angle$$D $$=10^o$$. Find $$\angle$$A, $$\angle$$C and $$\angle$$D.
In a parallelogram $$ABCD,$$ $$\angle A = x ^ { \circ } , \angle B = ( 3 x + 20 ) ^ { \circ }$$ . Find $$x$$ and $$\angle C$$ and $$\angle D$$
$$ABCD$$ is a parallelogram and $$\angle DAB=60^{o}$$. If the bisects $$AP$$ and $$BP$$ of angles $$A$$ and $$B$$ respectively, meet at $$P$$ on $$CD$$.Prove that $$P$$ is the midpoint of $$CD$$.
In the adjacent figure, $$ar(PEA)=ar(PAC)$$ and $$ar(RPC)=ar(KEA)$$ show that quadrilateral $$PARK$$ and $$PACE$$ are trapezium.
Given : $$ar (PEA ) = ar(PAC)$$ and $$ar(RPC) = ar (KEA)$$
we have to show $$PARK$$ & $$PACE$$ are trapezium
Given that: $$ar(PEA) = ar (PAC)$$
so $$ \triangle PEA$$ & $$ \triangle PAC $$ lie on same base $$PA$$ and are equal in area.
They lie between $$PA$$ & $$EC$$ Hence $$ PA \parallel EC$$
In $$PAEC$$ one pair of opposite sides of quadrilateral $$PAEC$$ are parallel.
$$ \therefore\boxed{ PACE \,is\, a\, trapezium}$$
Now take $$ar(PEA) = ar(PAC)$$ __ (2)
& $$ar(RPC) = ar (KEA)$$ __ (1)
subtract (1)-(2)
$$ar(RPC)-ar(PEA) = ar(KEA)-ar(PAC)$$
$$ar (RPA) = ar(KPA)$$
Now $$\triangle KPA$$ and $$ \triangle RPA $$ lie on same base $$PA$$ and equal in area. They lie on between $$KR$$ & $$PA$$
$$ \therefore KR \parallel PA$$
Since one pair of opposite sides are equal (parallel) in quadrilateral $$PARK.$$
Hence $$ \therefore \boxed {PARK \,is\, a\, trapezium}.$$
$$ABCD$$ is a rhombus and $$P, Q, R$$ and $$S$$ are the mid-points of the sides $$AB, BC, CD$$ and $$DA$$ respectively. Show that the quadrilateral $$PQRS$$ is a rectangle.
$${\textbf{Step1: Define and identify a rhombus.}}$$
$${\text{P is the midpoint of AB, Q is the midpoint of BC, R is Midpoint of CD,}}$$
$${\text{S is the midpoint of DA.}}$$
$$\Rightarrow$$ $${\text{join AC}}$$
$${\text{In}}$$ $$\Delta ABC$$ $$ \Rightarrow $$ $${\text{P is midpoint of AB, Q is midpoint of BC.}}$$
$$\therefore PQ||AC$$ $${\text{So}}$$ $$PQ = \dfrac{1}{2}AC……..(1)$$
$${\text{In}}$$ $$\Delta ADC$$ $$ \Rightarrow $$ $${\text{R is midpoint of CD, S is midpoint of AD.}}$$
$$\therefore RS||AC$$ $${\text{So}}$$ $$RS = \dfrac{1}{2}AC……..(2)$$
$${\text{From (1) and (2)}}$$
$$PQ||RS \Rightarrow AS$$ $${\text{opposite sides are equal.}}$$
$$\therefore $$ $${\text{PQRS is parallelogram.}}$$
$${\text{Now, we have to prove PQRS is a rectangle since ABCD is rhombus AB=BC.}}$$
$$\dfrac{1}{2}AB = \dfrac{1}{2}BC \Rightarrow PB = BQ \Rightarrow \angle 4 = \angle 1$$
$${\text{Now from}}$$ $$\Delta BPO$$ $$PB = BQ \Rightarrow \angle P = \angle Q$$
$${\text{In}}$$ $$\Delta APS$$ $${\text{and}}$$ $$\Delta CQR$$
$${\text{In line AB:-}}$$ $$\angle 3 + \angle SPQ + \angle 4 = {180^0}……(3)$$
$${\text{In line BC:-}}$$ $$\angle 1 + \angle PQR + \angle 2 = {180^0}$$
$${\text{Can be written as}}$$ $$\angle 4 + \angle PQR + \angle 3 = {180^0}…….(4)$$
$${\text{From (3) and (4)}}$$
$$\therefore $$$$\angle 3 + \angle SPQ + \angle 4 = \angle 3 + \angle PQR + \angle R$$
$$\therefore $$$$\angle SPQ = \angle PQR$$
$${\text{Now,}}$$ $$PS||QR [\because $$ $${\text{Opposite sides are equal]}}$$
$${\text{PQ is transversal.}}$$
$$\angle SPQ + \angle PQR = {180^0} \Rightarrow \angle SPQ + \angle SPQ = {180^0}$$
$$2\angle SPQ = {180^0} \Rightarrow \angle SPQ = {90^0}$$
$$\therefore $$ $${\text{PQRS is parallelogram and an angle}}$$ $$\angle SPQ = {90^0}$$
$${\text{So PQRS is a rectangle.}}$$
$$\therefore $$ $${\text{Hence proved.}}$$
$${\textbf{Thus, PQRS is a rectangle. Hence proved.}}$$
In the given figure, ABCD is a trapezium, Diagonals AC and BD interset at Q Prove that the area of $$\Delta $$AOD is the same as $$\Delta $$BOC
$$ABCD$$ is a rhombus. If $$\angle BCA=35^{o}$$, find $$\angle ADC$$.
Here in rhombus ABCD $$\angle BCA=35^o$$
in $$\Delta ABC$$
$$\angle BCA+\angle CBA +\angle BAC=100^o$$
$$35+\angle CBA+35=100$$
$$\angle CBA=100-70$$
$$\angle CBA=110^o$$
We know
$$\angle ABC+\angle ADC=100^o$$
$$110^o+\angle ADC=100$$
$$\angle ADC=100-110$$
$$\boxed{\angle ADC=70}$$.
State the angle sum property of a quadrilateral.
The area of a rectangular plot is 540 $${ m }^{ 2 }$$. If its length is 27 m, find its breadth and perimeter.
$$ABCD$$ is a parallelogram. The circle through $$A, B$$ and $$C$$ intersects $$CD$$ produced at $$E$$. If $$AB=10\ cm, BC=8\ cm, CE=14\ cm$$ . Find $$AE$$
Given : $$ABCD$$ is a parallelogram and a circle is drawn through $$A,B$$ and $$C$$ which intersects $$CD$$ produced, at $$E$$
$$AB=10cm$$ $$BC=8cm$$ $$CE=14cm$$
Now,
$$\angle AED+\angle ABC={ 180 }^{ 0 }\quad \rightarrow \left( 1 \right) $$ (Sum of opposite angles of a cyclic quadrilateral is supplementary)
$$\angle EDA+\angle ADC={ 180 }^{ 0 }\quad \rightarrow \left( 2 \right) $$ (linear pair)
$$\angle EDA+\angle ABC={ 180 }^{ 0 }\quad \rightarrow \left( 3 \right) $$ ($$\because$$ $$\angle ADC=\angle ABC$$ as they are the opposite angles of a parallelogram)
From $$(1)$$ & $$(3)$$ we have,
$$\angle AED=\angle EDA\quad \rightarrow \left( 4 \right) $$
$$\Rightarrow$$ in $$\Delta AED$$ two angles are equal, hence the opposite sides corresponding to these angles must also be equal
i.e. $$AE=AD$$
and $$AD=BC$$ (opposite sides of a parallelogram are equal)
$$\therefore$$ $$AE=8cm$$.
Parallelogram $$ABCD$$ and rectangle $$ABEF$$ are on the same base and have equal areas. Show that perimeter of the parallelogram is greater than that of rectangle.
In the figure, $$\Box ABCD$$ is a trapezium, $$AB\parallel DC$$. Point $$P$$ and $$Q$$ are midpoints of $$seg\ AD$$ and $$seg\ BC$$ respectively. Then prove that, $$PQ\parallel AB$$ and $$PQ=\dfrac{1}{2}(AB+DC)$$.
Given,
$$ABCD$$ is a trapezium in which $$AB\parallel DC$$ and $$P$$, $$Q$$ are the midpoints of $$AD$$ & $$BC$$ respectively
Construction : Join $$CP$$ and produce it to meet $$AB$$ produced to $$R$$.
In $$\Delta PDC$$ & $$\Delta PAR$$
$$PD=PA$$ ($$\because$$ $$P$$ is the midpoint of $$AD$$)
$$\angle CPD=\angle RPA$$ (Vertically opposite angles)
$$\angle PCD=\angle PRA$$ (alternate angle)
$$\therefore$$ $$\Delta PDC\cong \Delta PAR$$ using $$A$$ as criterion
$$\Rightarrow CP=CR$$ $$(CPCTC)$$
Also,
In $$\Delta CRB$$,
$$P$$ is the midpoint of $$CR$$ (proved above)
Also, $$Q$$ is the midpoint of $$BC$$
$$\Rightarrow$$ By midpoint theorem $$PQ\parallel AB$$ and $$PQ=\dfrac { 1 }{ 2 } \left( RB \right) $$
But $$RB=RA+AB$$
$$=CD+AB$$ ($$\therefore$$ $$AR=CD$$ by $$CPCTC$$)
$$\therefore$$ $$PQ=\dfrac { 1 }{ 2 } \left( CD+AB \right) $$
Hence proved.
If the tangents $$P A$$ and $$P B$$ from a point $$P$$ to the circle with cetre $$O$$ inclined to each other at the angle of $${ 110 }^{ },$$ then find $$\angle POA.$$
Given $$\angle BPA=110°$$
Since the angle subtended by the tangent at the boundary of circle is $$90°$$
$$\Rightarrow \angle A =\angle B=90°$$
In $$PBAO$$
$$\Rightarrow \angle P +\angle B +\angle O +\angle A=360°$$
Sum of angles of a quadrilateral is $$360°$$
$$\Rightarrow 110° +90+90° +\angle O=360°$$
$$\Rightarrow 290+\angle O=360°$$
$$\Rightarrow \angle O =360° -290°$$
$$\Rightarrow \angle AOB=70°$$
Hence, the answer is $$70°.$$
In a parallelogram, the bisector of any two consecutive angles intersect at right angle, Prove it.
Let ABCD is a parallelogram
as we know
$$\begin{array}{l} \angle A+\angle B={ 180^{ 0 } } \\ OA\, \, bi\sec ts\, \, \angle DAB\, \, \& \, \, OB\, \, bi\sec ts\, \, \angle CBA \\ to\, \, prove\, \, \angle AOB={ 90^{ 0 } } \\ now\, \, in\, \, \Delta AOB \\ \angle OAB+\angle CBA+\angle AOB={ 180^{ 0 } } \\ \frac { 1 }{ 2 } \angle DAB+\frac { 1 }{ 2 } \angle CBA+\angle AOB={ 180^{ 0 } } \\ \frac { 1 }{ 2 } \left( { \angle DAB+CBA } \right) +\angle AOB={ 180^{ 0 } } \\ \frac { 1 }{ 2 } \times { 180^{ 0 } }+\angle AOB={ 180^{ 0 } } \\ { 90^{ 0 } }+\angle AOB={ 180^{ 0 } } \\ \angle AOB={ 180^{ 0 } }-{ 90^{ 0 } } \\ \angle AOB={ 90^{ 0 } } \end{array}$$
The diagonals of a rectangle ABCD intersect at o. if $$\angle $$ BOC =$$70^{\circ}$$. Find $$\angle $$ODA.
Since in a rec diagonal are equal and bisect each other
$$\therefore OD=OA$$
and $$\angle S$$ opp to equal sides are equal
$$\therefore \angle 1=x$$
Now, In $$\Delta ODA$$
$$\angle DOA=70^0$$ (vertically opp. $$\angle S$$)
$$x+\angle 1+70=180^0$$ ($$\angle$$ sum prop)
and $$\angle 1=x$$
$$\therefore x+x+70=180$$
$$2x=110$$
$$x=\dfrac{110}{2}$$
$$x=55^0$$
$$\Rightarrow \boxed{\angle ODA=55^0}$$
PQRS is a parallelogram. PM is the height from P to $$\overline {SR} $$ and PN is the height from P to $$\overline {QR} $$. If SR=12 cm and PM=7.6 cm. Find the area of the parallelogram PQRS.
Both the pair of opposite angles of quadrilateral are equal and supplementary. Find the measure of each angle.
If non parallel sides of a trapezium are equal, prove that it is cyclic.
$${\textbf{Step -1:}}{\textbf{ Prove that }}\mathbf{ABCD}{\textbf{ as cyclic trapezium}}{\textbf{.}}$$
$${\text{In }}\Delta DAM{\text{ and }}\Delta CBN,$$
$$AD = BC{\text{ }}\left[ {{\text{Given}}} \right]$$
$$\angle AMD = \angle BNC{\text{ }}\left[ {{\text{Right angles}}} \right]$$
$$DM = CN{\text{ }}\left[ {{\text{Distance between the parallel lines}}} \right]$$
$$\therefore \Delta DAM \cong \Delta CBN{\text{ by }}RHS{\text{ congruence condition}}{\text{.}}$$
$${\textbf{Step -2: Prove the other angles}}{\textbf{.}}$$
$$\angle A = \angle B{\text{ }}\left[ {{\text{by }}CPCT} \right]$$
$$\angle B + \angle C = 180^\circ {\text{ }}\left[ {{\text{sum of the co - interior angles}}} \right]$$
$$\angle A + \angle C = 180^\circ $$
$${\textbf{}}{\textbf{Hence, }}\mathbf{ABCD}{\textbf{ is a cyclic quadrilateral as the sum of the pair of opposite angle is 180}}^\circ {\textbf{.}}$$
Line segment joining the mid - points M and N of parallel sides AB and DC respectively of a trapezium ABCD is perpendicular to both the sides AB and DC . Prove that AD = BC.
In triangle ABC, P is the mid-point of side BC. A line through P and parallel to CA meets AB at point Q; and a line through Q meets at $$S$$. QS parallel to BC meets median AP at point R. prove that : (i) AP = 2AR (ii) BC= 4QR
.png)
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.
To prove: 1. AP = 2 AR
2. BC = 4 QR
Proof:
In ∆ABC, P is the mid point of BC and PQ||AB.
∴ Q is the mid point of AB (Converse of mid-point theorem)
In ∴ ABP, Q is the mid point of AB and QR||BP.
∴ R is the mid point of AP. (Converse of mid point theorem)
⇒ AP = 2AR
In ∆ABP, Q is the mid point of AB and R is the mid point of AP.
.png)
In the above figure RISK and CLUE are parallelograms, Find the value of x.
In figure above, the bisectors of $$\angle P$$ and $$ \angle Q$$ meet at a point O. If $$\angle R={ 100 }^{ \circ }$$ and $$\angle S={ 50 }^{ \circ}$$, find $$\angle {POQ}$$.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementarily angles at the center of the circle.
In the adjoining figure, D, E and F are mid points of the side BC, CA and AB of $$\Delta ABC$$. If $$AB =4.3$$ cm , $$BC =5.6$$ cm and $$AC =3.5$$ cm. Find the perimeter of
(i) $$\Delta DEF$$ (ii) Quad. BDEF
$$ABCD$$ is a trapezium in which $$AB || DC, AB=16\ cm$$ and $$DC=24\ cm$$. If $$E$$ and $$F$$ are respectively the midpoints of $$AD$$ and $$BC$$, prove that $$ar(ABEF)=9/11\ ar (EFCD)$$.
Image
The length of the diagonals of a rhombus is in the ratio 4:3 .If its area is 384 $$cm^{2}$$, find its side.
Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.
$$\textbf{Step -1: Apply mid point theorem in triangles ABC and ADC.}$$
$$\text{Mid point theorem states that the line segment in a triangle joining the mid points of the two}$$
$$\text{sides of the triangle is parallel to its third side and is also half of the length of the third side.}$$
$$\text{Consider }\triangle ABC,$$
$$\text{We know that }P\text{ and }Q\text{ are the mid points of }AB\text{ and }BC.$$
$$\text{By using the midpoint theorem},$$
$$PQ \parallel AC\text{ and }PQ = \dfrac{1}{2}AC\ldots(1)$$
$$\text{Consider }\triangle ADC,$$
$$\text{We know that }R\text{ and }S\text{ are the mid points of }DC\text{ and }AD.$$
$$\text{By using the midpoint theorem},$$
$$RS \parallel AC\text{ and }RS=\dfrac{1}{2}AC\ldots(2)$$
$$\text{From (1) and (2),} $$
$$PQ \parallel RS\text{ and }PR=RS=\dfrac{1}{2}AC\ldots(3)$$
$$\textbf{Step
-2: Apply mid point theorem in triangles BAD and BCD.}$$
$$\text{Consider }\triangle BAD,$$
$$\text{We know that }P\text{ and }S\text{ are the mid points of }AB\text{ and }AD.$$
$$\text{By using the midpoint theorem,}$$
$$PS\parallel BD\text{ and }PS = \dfrac{1}{2} BD\ldots(4)$$
$$\text{Consider }\triangle BCD,$$
$$\text{We know that }Q\text{ and }R\text{ are the mid points of }BC\text{ and }CD.$$
$$\text{By using the midpoint theorem,}$$
$$RQ \parallel BD\text{ and }RQ = \dfrac{1}{2} BD\ldots(5)$$
$$\text{From (4) and (5), }$$
$$PS || RQ\text{ and }PS=RQ=\dfrac{1}{2} BD\ldots(6)$$
$$\textbf{Step -3: Prove PQRS is a rectangle.}$$
$$\because \text{The diagonals intersects at right angles in a rhombus.}$$
$$\therefore\angle EOF=90^\circ.$$
$$\text{As, we know that }RQ\parallel BD \text{ so, }ER\parallel OF.$$
$$\text{In the same way }SR \parallel AC\text{ so, }FR\parallel OE.$$
$$\Rightarrow OERF\text{ is a parallelogram.}$$
$$\because\text{Opposite angles are equal in a parallelogram.}$$
$$\text{So, } \angle FRE=\angle EOF=90^\circ.$$
$$\text{From equations (3) and (6), we get}$$
$$PQRS\text{ is a parallelogram.}$$
$$\text{and }\angle QPS=\angle QRS=90^\circ.$$
$$\Rightarrow PQRS \text{ is a rectangle.}$$
$$\textbf{Hence, the quadrilateral formed by joining the midpoints of the pairs of adjacent}$$$$\textbf{sides of a rhombus is a rectangle.}$$
Bisectors of $$\angle{B}$$ and $$\angle{D}$$ of a quadrilateral ABCD meet CD and AB produced at P and Q respectively. Prove that $$\angle{P} + \angle{Q} = 1/2(\angle{ABC} + \angle{ADC})$$
$$P,Q,R$$ and $$S$$ are respectively the midpoints of the sides $$AB,BC,CD$$ and $$DA$$ of a quadrilateral $$ABCD$$, Show that $$PQ\parallel AC$$ and $$PQ=\cfrac{1}{2}AC$$
$$ABCD$$ is a trapezium in which $$AB \parallel DC$$ and its diagonals intersect each other at $$O$$. Using Basic Proportionality theorem, prove that $$AO/BO = CO/DO$$.
The midpoints of the sides $$AB,BC,CD$$ and $$DA$$ of a quadrilateral $$ABCD$$ are joined to form a quadrilateral. If $$AC=BD$$ and $$AC\bot BD$$ then prove that the quadrilateral formed is a square.
If $$D,\ E$$ and $$F$$ mid-points of sides $$AB,\ BC$$ and $$CA$$ respectively of an isosceles triangle $$ABC$$, prove that $$\triangle DEF$$ is also isosceles.
The angles of quadrilateral are in the ratio 3 : 4 : 5 : 6 Show that the quadrilateral is a trapezium.
Class 9 Maths Extra Questions
- Areas Of Parallelograms And Triangles Extra Questions
- Circles Extra Questions
- Constructions Extra Questions
- Coordinate Geometry Extra Questions
- Heron'S Formula Extra Questions
- Introduction To Euclid'S Geometry Extra Questions
- Linear Equations In Two Variable Extra Questions
- Lines And Angles Extra Questions
- Number Systems Extra Questions
- Polynomials Extra Questions
- Probability Extra Questions
- Quadrilaterals Extra Questions
- Statistics Extra Questions
- Surface Areas And Volumes Extra Questions
- Triangles Extra Questions