If $$AB=QR$$, $$BC=PQ$$ and $$CA=PR$$, then:

$$\triangle ABC\cong \triangle PQR$$

$$\triangle BAC\cong \triangle RPQ$$

$$\triangle PQR\cong \triangle BCA$$

MCQ Questions for Class 9 Maths Triangles Quiz 2

In $\triangle ABC,$ $\angle\,B\,=\,30^{\circ}$ and $\angle\,C\,=\,70^{\circ}$ . The greatest side of the triangle is:

0%
$AC$
0%
$AB$
0%
$BC$
0%
Incomplete data

Explanation

Since the sum of angles in a triangle is equal to

${180}^{o}.$

$\angle A = {180}^{o} - {30}^{o} - {70}^{o}$

$= {80}^{o}$

Out of the three angles,

$\angle A$ is the largest.

This means, the side opposite to it, which is

$BC$ will be the greatest side of the triangle.

State true or false:

In the given figure, the diagonals

$AC$ and

$BD$ intersect at point

$O$ . If

$OB =OD$ and

$AB//DC$ , then

$Area \left ( \bigtriangleup \: DOC \right )=Area \left ( \bigtriangleup \: AOB \right )$ .

0%
True
0%
False

Explanation

Given, the diagonals AC and BD intersect at point O such that OB

$=$ OD and AB

$\parallel$ DC.
Now, In

$\triangle DOC$ and

$\triangle AOB$

$\angle BDC=\angle DBA$ (Alternate angles, As AB

$\parallel$ CD and BD intersects it.)

$OD= OB$ (Given)

$\angle DOC=\angle AOB$ ( Vertically opposite angles)
So,

$\triangle DOC\quad \cong \quad \triangle AOB$ (By ASA concurrence Property)
Since, Concurrent triangle have equal areas.

$\therefore$ Area of

$\triangle DOC =$ Area of

$\triangle AOB$

In triangle

$ABC, D$ is any point on side

$BC$ , then:

$AB + BC + AC > 2AD$

0%
True
0%
False

Explanation

In triangle ABD,

$AB + BD > AD$ [because, the sum of any two sides of a triangle is always greater than the third side]
In triangle

$ADC$ ,

$AC + DC > AD$ [because, the sum of any two sides of a triangle is always greater than the third side]
Adding these we get,

$AB + BD + AC + DC > AD + AD$

$\Rightarrow AB + (BD + DC) + AC > 2AD$

$\Rightarrow AB + BC + AC > 2AD$
Hence proved

In the same figure, if $y > x > z;$ arrange sides $AB, BC$ and $AC$ in descending order according their lengths.

0%
$BC > AB > AC$
0%
$AB > BC > AC$
0%
$BC > AB < AC$
0%
None of these

Explanation

$\text {Since, } y>x>z;\\ \\ \angle A\quad =\quad 180\quad -\quad x\quad \\ \angle B\quad =\quad 180\quad -\quad y\quad (\quad y\quad is\quad the\quad largest,\quad so\quad \angle B\quad \quad is\quad the\quad smallest)\\ \angle C\quad =\quad 180\quad -\quad z\quad (\quad z\quad is\quad the\quad smallest,\quad So\quad \angle C\quad is\quad the\quad largest)\\ \\ so\quad \angle C\quad >\quad \angle A\quad >\quad \angle B\\ So\quad AB\quad >\quad BC\quad >\quad AC\quad (\quad Angle\quad side\quad relationship)\\ \\ \quad$

D is a point on the side BC of a

$\triangle ABC$ such that AD bisects

$\angle BAC$ . Then

0%
$BD=CD$
0%
$BA>BD$
0%
$BD>BA$
0%
$CD>CA$

Explanation

Given-

AD is the angular bisector of a triangle ABC.

AD meets BC at D.

To find out-

which of the given options is true.

Solution-

AD is the bisector of

$\quad \angle BAC.$

$\therefore \quad \angle BAD=\angle CAD.\quad$

Again , in

$\quad \Delta ACD,\quad \quad \angle ADC\quad$ is the external angle .

We know that the external angle of a triangle is greater than each intrenal opposite angle of the same triangle.

$\quad \therefore \quad \angle ADB=\angle DAC+DCA.$

$=\angle BAD+\angle DCA\quad \left( \because \quad \angle DAC=\angle BAD \right) \\ i.e\quad \angle ADC>\angle BAD.$

$\therefore \quad BA>BD.\quad$

So option B is true.

Option A is true only when the triangle is equlateral or isosceles.

Option C is false since Option B is true.

option D is false since

$\angle ADC>\angle BAD\quad or\ \angle CAD\quad$

$\therefore \quad CD\ngtr CA\quad$ \

Ans- Option B.

If $AB > AC > BC,$ arrange the angles $x, y$ and $z$ in decending order of their values.

0%
$y< z < x$
0%
$x>y>z$
0%
$z>y>x$
0%
Incomplete question

Explanation

Given

$AB > AC > BC,$

This means that angles opposite these sides are in the same order.

$,$

$\angle C > \angle B > \angle A$

$\angle z = 180 - \angle C\\ \angle y = 180 - \angle B \\ \angle x = 180 - \angle A$

From the above statements it is clear that

$x > y > z.$

Hence

$,$

$B$ is the answer.

$\Delta ABC$ , if

$\angle A = 50^0$ and

$\angle B = 60^0$ , then the shortest and largest sides of the triangle, respectively are:

0%
$AC$ and $AB$
0%
$BC$ and $AB$
0%
$AC$ and $BC$
0%
$BC$ and $AC$

Explanation

We have,

$\angle A = 50$ and

$\angle B = 60$

$\therefore \angle A + \angle B + \angle C = 180$

$[$ Angle sum property of a

$\Delta$

$]$

$\Rightarrow 50 + 60 + \angle C = 180$

$\Rightarrow 110 + \angle C = 180$

$\Rightarrow \angle C = 180 - 110$

$\Rightarrow \angle C = 70$
Since

$\angle A$ and

$\angle C$ are the smallest and largest angles respectively, therefore corresponding sides

$BC$ and

$AB$ are the smallest and largest sides of the triangle respectively.

Hence, option

$B$ is the correct answer.

In the adjoining figure,

$AB=AC$ and

$AP\bot BC$ . Then:

0%
$AB=AP$
0%
$AB < AP$
0%
$AB>AP$
0%
$AB\le AP$

Explanation

Given:

$AB = AC$ ,

$AP \perp BC$
In

$\triangle ABP$ , we have

$\angle APB = 90^{\circ}$

$\angle ABP = 90 - \angle BAP$ ....(Angle sum property)
Thus,

$\angle APB > \angle ABP$

$AB > AP$ .... (Side opposite to larger angle is large)

Which pair of triangles is not congruent by

$SAS$ congruence criterion

0%
0%
0%
0%
None of these

Explanation

In option (3), we have

$\angle BAC = \angle DFE = 120^{\circ}$

$BC = DE = 5$

$AC = DF = 3$

For SAS to hold true corresponding angle included by two congruent sides must be congruent

here, angle A and F have not included angles by the congruent sides.
Hence,

$\triangle ABC \not\cong \triangle DEF$

$AB=QR$ ,

$BC=PQ$ and

$CA=PR$ , then:

0%
$\triangle ABC\cong \triangle PQR$
0%
$\triangle CBA\cong \triangle PQR$
0%
$\triangle BAC\cong \triangle RPQ$
0%
$\triangle PQR\cong \triangle BCA$

Explanation

$\triangle ABC$ and

$\triangle PQR$

$AB=QR$

$BC=PQ$

$CA=PR$
Thus,

$\triangle CBA \cong \triangle PQR$ (SSS rule)

State the property by which

$\Delta ADB\, \cong\, \Delta ADC$ in the following figure.,

0%
SAS property
0%
SSS property
0%
RHS property
0%
ASA property

Explanation

From figure, we have

$AB = AC$ ....Given

$\angle BAD\, =\, \angle CAD$

$AD = AD$ ....Common sides

$\therefore$ By SAS property,

$\Delta ADB \cong \Delta ADC$ .

In an isosceles traingle ABC, with

$AB=\,AC$ , the bisectors of

$\angle B$ and

$\angle C$ intersect each other at the point

$O$ . Join

$A$ to

$O$ . Then:

0%
$\angle OAB = \angle OAC$
0%
$OB=\, OC$
0%
$AO$ bisects $\angle A$
0%
None of these

Explanation

$AB=AC$ .... (Given)

$\angle B= \angle C$ .... (angles opposite to equal sides of triangle are equal)

$\dfrac {1}{2}\angle B= \dfrac {1}{2} \angle C$ .... (

$BO$ and

$CO$ are bisectors of

$\angle B$ and

$\angle C$ , respectively)

$\therefore OBC=\angle OCB$

$\therefore OB=OC$ .... (Sides opposite to equal angles of triangle are equal).

$\triangle OAB$ and

$\triangle OSC$ ,

$AB=AC$ .... (Given)

$OB=OC$ .... (Proved)

$\angle B=\angle C$ ..... (Angles opposite to equal sides of triangle are equal)

$\dfrac {1}{2} \angle B=\dfrac {1}{2} \angle C$

$\therefore \angle ABO= \angle ACO$ .

$\therefore \triangle OAB \cong \triangle OAC$ .... (by

$SAS$ Rule)

$\therefore AO$ bisects

$\angle A$ .... (By

$CPCT$ ).

Hence, option

$A$ ,

$B$ and

$C$ are correct.

$AB$ is a line segment and

$P$ is its mid-point.

$D$ and

$E$ are points on the same side of

$AB$ such that

$\angle BAD=\angle ABE$ and

$\angle EPA=\angle DPB$ . Then:

0%
$AB = AD$
0%
$\triangle DAP\cong \triangle EBP$
0%
$AD=\,BE$
0%
$\angle DPA = \angle EPB$

Explanation

$\angle EPA=\angle DPB$ (Given)

$\angle EPA$

$+\angle DPE$

$=\angle DPB$

$+\angle DPE$

$\implies$

$\angle APD=\angle BPE$ ...........

$(1)$ .

Now, in

$\triangle DAP$ and

$\triangle EBP$ ,

$AP=\,PB$ .....(

$\because P$ is the mid point of

$AB$ )

$\angle PAD=\angle PBE$ .....(Given)

$\angle APD=\angle BPE$ .....(From

$1$ ).

Therefore

$\triangle DAP\cong \triangle EPB$ (By

$ASA$ congruency criterion)
Thus,

$AD = BE$ (By

$CPCT$ rule).

Hence, options

$B$ ,

$C$ and

$D$ are correct.

$l$ and

$m$ are two parallel lines intersected by another pair of parallel lines

$p$ and

$q$ . Then:

0%
$\triangle ABC\cong \triangle CDA$ .
0%
$\triangle ABC\cong \triangle ADC$ .
0%
$\triangle ABC\cong \triangle DCA$ .
0%
$\triangle ABC\cong \triangle CAD$ .

Explanation

$\triangle ABC$ and

$\triangle CDA$

$\angle CAB=\angle ACD$ (Pair of alternate angle)

$\angle BCA=\angle DAC$ (Pair of alternate angle)

$AC=AC$ (Common side)

$\triangle ABC\cong \triangle CDA$ (ASA criteria)

If in two triangles

$PQR$ and

$DEF$ ,

$PR=\,EF$ ,

$QR=\,DE$ and

$PQ=\,FD$ , then

$\triangle PQR\cong$

$\triangle$ ___.

0%
$FDE$
0%
$DEF$
0%
$FED$
0%
$DFE$

Explanation

Given:

$PR=\,EF$ ,

$QR=\,DE$ and

$PQ=\,FD$
Now, In

$\triangle PRQ$ and

$\triangle DEF$

$PR = EF$

$QR = DE$

$PQ = FD$
Thus,

$\triangle PQR \cong \triangle FDE$ (SSS rule).

By which congruency property, the two triangles connected by the following figure are congruent.

0%
SAS property
0%
SSS property
0%
RHS property
0%
ASA property

Explanation

Separate the given figure into two triangles.

$\Delta ABC$ and

$\Delta ABD$

We can observe that,

$AC = AD$ {Both are

$x \,\,cm$ }

$BC = BD$ {Both are

$y\,\,cm$ }

$AB = AB$ {Common Side}

$\Delta ABC\, \cong\, \Delta ABD$ by SSS property.

$4$ cm and

$3$ cm are the lengths of two sides of a triangle then the length of the third side may be_____.

0%
$11$ cm
0%
$3$ cm
0%
$5$ cm
0%
$12$ cm

Explanation

In a triangle, sum of two sides

$>$ third side.

So, if two sides are

$4 cm$ and

$7 cm$ , then considering the given options,

A.

$11 cm$

$4 + 7 \ngtr 11$

Hence, this is not the correct option.

B.

$6 cm$
We see that

$4 + 7 > 6$

$4 + 6 > 7$

$7 + 6 > 4$
Hence,

$6 cm$ is the correct option.

In the given fig. if AD = BC and AD || BC, then:

0%
AB = AD
0%
AB = DC
0%
BC = CD
0%
None

Explanation

In the give fig. AD = BC, AC = AC,

$\angle ADC\, =\, \angle ABC$ (

$\because$ AD || BC)
By SAS theorem

$\Delta ABC\, \cong\, \Delta CDA$
So AB = DC.

ABC is an isosceles triangle in which the altitudes

$BE$ and

$CF$ are drawn to the equal sides

$AC$ and

$AB$ respectively. Then

0%
$BE = CF$
0%
$BE = AB$
0%
$AB = BC$
0%
$AC = BC$

Explanation

$\triangle ABE$ and

$\triangle ACF$ ,

we have

$\angle BEA=\angle CFA$ (Each

${ 90 }^{ 0 }$ )

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

$AB=\,AC$ (Given)

$\therefore \triangle ABE \cong \triangle ACF$ (By SAS congruence criteria)

$\therefore BF=\,CF$ [C.S.C.T]

In the following fig. if AB = AC and BD = DC then

$\angle ADC\, =$

0%
$60^{\circ}$
0%
$120^{\circ}$
0%
$90^{\circ}$
0%
None

Explanation

Given that, in

$\triangle ABC$ ,

$AB = AC, \ BD = DC$

To find out:

$\angle ADC$

$\triangle ADB$ and

$\triangle ADC$ ,

$AB = AC\quad \quad [given]$

$BD = DC\quad \quad [given]$

Also,

$AD = AD\quad \quad [common]$

$\therefore \$ By

$SSS$ congruence criterion,

$\Delta ADB\, \cong\, \Delta ADC$ .

Hence,

$\angle ADB=\angle ADB$
Also,

$\angle ADB\, +\, \angle ADC\, =\, 180^{\circ}\quad [linear\ pair]$

$\therefore \ \angle ADB\, =\, \angle ADC\, =\, 90^{\circ}$

Hence,

$\angle ADC=90^\circ$

In the given figure,

$OA=OB, OC=OD, \angle AOB=\angle COD.$ Which of the following statements is true

$\: ?$

0%
$AC=CD$
0%
$OA=OD$
0%
$AC=BD$
0%
$\angle OCA=\angle ODC$

Explanation

$\triangle$ s

$AOC$ and

$BOD$

$OA=OB$ (Given)

$OC=OD$ (Given)

$\angle AOB-\angle COB=\angle COD-\angle COB,$
i.e.,

$\angle AOC=\angle BOD$

$\therefore$

$\triangle AOC\cong \triangle BOD$ (SAS)

$\Rightarrow$

$AC=BD$ (cpct)

In a

$\triangle PQR$ , PS is bisector of

$\angle P$ and

$\angle Q =70^o, \angle R = 30^o$ , then:

0%
QS > PQ > PR
0%
QS < PQ < PR
0%
PQ > QS > SR
0%
PQ < QS < SR

Explanation

According to question,

$\angle QPR=180^o-(70^o+30^o)=80^o$

So,

$\angle QPS=\angle SPR=40^o$

$\angle PSQ=180^o-(40^o+70^o)=70^o$

As,

$\angle PRQ<\angle PQR$

So,

$PQ<PR$ ......(1)

Similarly,

$\angle QPS<\angle PQS$

So,

$QS<PR$ ........(2)

And,

$\angle QPS<\angle PSQ$

So,

$QS<PQ$ .......(3)

Therefore, from eq. (1), (2), (3)

$QS<PQ<PR$ .

1023877_331736_ans_d279521ba7814fc780584d7cab70fa73.png

Ankita wants to prove

$\Delta ABC\cong \Delta DEF$ using

$SAS$ . She knows

$AB=DE$ and

$AC=DF$ . What additional piece of information does she need?

0%
$\;\angle A=\angle D$
0%
$\;\angle C=\angle F$
0%
$\;\angle B=\angle E$
0%
$\;\angle A=\angle B$

Explanation

By SAS, two corresponding sides and their included angles should be congruent.
Since

$AB=DE$ and

$AC=DF$ ,

The included angles of

$AB$ and

$AC = \angle A$

and included angle of

$DE$ and

$DF = \angle D.$

$\Rightarrow \angle A = \angle D$ is needed.

If a

$\triangle PQR$ is constructed taking

$QR = 5\text{ cm},$

$PQ = 3\text{ cm},$ and

$PR = 4\text{ cm}$ then the correct order of the angles of the triangle is:

0%
$\angle P < \angle Q < \angle R$
0%
$\angle P>\angle Q<\angle R$
0%
$\angle P > \angle Q >\angle R$
0%
$\angle P < \angle Q>\angle R$

Explanation

In a triangle, the angle is determined by its sides if it is given. The largest side will have the largest angle opposite to it. The smallest side will have the smallest angle opposite to it.

So,

$QR=5\text{ cm},$ is the largest side. Hence the angle opposite to it will also be largest that is

$\angle P.$

Then the side

$PR=4 \text{ cm},$ , smaller than

$QR$ . Hence the

$\angle Q$ will be smaller than

$\angle P$

Finally, the smallest side is

$PQ=3\text{ cm}$ with its corresponding angle

$\angle R$ is smallest.

Hence the option

$C$ is correct.

Which of the following statement is false?

0%
The sum of two sides of a $\Delta$ is greater than the third side
0%
In a right angled $\Delta$ hypotenuse is the longest side
0%
A, B, C are collinear if AB + BC = AC
0%
None of these

Explanation

${\textbf{Step 1: Consider, option A}}{\textbf{.}}$

${\text{The sum of two sides of a}}$ $\vartriangle$ ${\text{is greater than the third side}}{\text{.}}$

${\text{From triangle inequality theorem we have,}}$

${\text{The sum of the lengths of any two sides of a triangle is greater than the length of the third side}}{\text{.}}$

${\text{Thus, option A}}{\text{. is true}}{\text{.}}$

${\textbf{Step 2: Consider, option B}}{\textbf{.}}$

${\text{In a right angled}}$ $\vartriangle$ ${\text{hypotenuse is the longest side}}{\text{.}}$

${\text{By Pythagoras theorem we have,}}$

${\text{In a right angled triangle, the square of the hypotenuse side is equal to the sum of}}$

${\text{square of the two sides}}{\text{.}}$

${\text{It implies that hypotenuse is the longest side}}{\text{.}}$

${\text{Thus, option B}}{\text{. is true}}{\text{.}}$

${\textbf{Step 3: Consider, option C}}{\textbf{.}}$

${\text{A, B, C are collinear if }}AB + BC = AC.$

${\text{As we know that, collinear points are the points that lie on the same line}}{\text{.}}$

${\text{These points lie on a line close to or far from each other}}{\text{.}}$

${\text{Thus, option C}}{\text{. is true}}{\text{.}}$

${\text{Hence, all of the above statements are true}}{\text{.}}$

${\text{So, none of these is false}}{\text{.}}$

${\textbf{Hence, option D is correct answer.}}$

Two sides of a triangle are

$7$ and

$10$ units, which of the following length can be the length of the third side?

0%
$19$ units
0%
$17$ units
0%
$13$ units
0%
$3$ units

Explanation

The sum of any two sides is greater than the third side.

and difference between two sides should be lesser than the third side.

For option

$A,$

$7+10\ngtr 19$

For option

$B,$

$7+10\ngtr 17$

For option

$C,$

$7+10 > 13$

Also,

$|10-7| < 13$

For option

$D,$

$7+10 > 3$

but

$|7-10| \nless 3$

Only,

$C$ is correct.

If a triangle

$PQR$ has been constructed taking

$QR = 6$ cm,

$PQ = 3$ cm and

$PR = 4$ cm, then the correct order of the angle of triangle is

0%
$\displaystyle \angle P< \angle Q< \angle R$
0%
$\displaystyle \angle P> \angle Q< \angle R$
0%
$\displaystyle \angle P> \angle Q> \angle R$
0%
$\displaystyle \angle P< \angle Q> \angle R$

Explanation

Given, in

$\triangle PQR$ ,

$QR=6$ cm,

$PQ=3$ cm,

$PR=4$ cm

We know,

(i) the shortest side is always opposite the smallest interior angle.

(ii) the longest side is always opposite the largest interior angle.

Here,

$QR=6$ cm is the largest side, therefore

$\angle P$ is the greatest.

And

$PQ=3$ cm is the smallest side, therefore

$\angle R$ is the smallest angle.

Therefore, the correct order is

$\angle P>\angle Q>\angle R$ .

In the given figure,

$\triangle RTQ\cong \triangle PSQ$ by ASA congruency condition. Which of the following pairs does not satisfy the condition?

0%
$PQ=QR$
0%
$\angle P=\angle R$
0%
$\angle TQP=\angle SQR$
0%
None of these

Explanation

Given,

$\triangle RTQ \cong \triangle PSQ$
Thus,

$PQ = QR$ (by cpct)

$\angle P = \angle R$ (by cpct)

Also,

$\angle RQT = \angle PQS$ (By CPCT)

$\angle RQT - \angle TQS = \angle PQS - \angle TQS$
Thus,

$\angle TQP = \angle SQR$

In the given figure,

$AD=BC, AC=BD.$ Then

$\triangle PAB$ is

0%
equilateral
0%
right angled
0%
scalene
0%
isosceles

Explanation

$\triangle ADB$ and

$\triangle ACB$

$AD=BC$ (Given)

$AC=BD$

$AB=BA$ (Common)

$\therefore$

$\triangle ADB\cong \triangle ACB$ (SSS)

$\Rightarrow$

$\angle ABD=\angle CAB$ (cpct)

$\Rightarrow$

$\angle ABP=\angle PAB$

$\Rightarrow$

$PA=PB$ (Sides opp. equal angles are equal)

$\Rightarrow$

$\triangle PAB$ is isosceles.

It is given that

$AB=BC$ and

$AD=EC.$ The

$\triangle ABE\cong \triangle CBD$ by __________ congruency.

0%
$SSS$
0%
$ASA$
0%
$SAS$
0%
$AAS$

Explanation

Given,

$AD=EC$

$\Rightarrow$

$AD+DE=DE+EC$

$\Rightarrow$

$AE=DC$
Also,

$AB=BC$

$\Rightarrow$

$\angle BCA=\angle BAC$ (isos.

$\triangle$ property)

$\Rightarrow$

$\angle BCD=\angle BAE$

$\therefore$ In

$\triangle$ s

$ABE$ and

$CBD,$

$AB=CB$ (Given)

$AE=DC$ (proved above)

$\angle BAE=\angle BCD$ (Proved above)

$\therefore$

$\triangle ABE\cong \triangle CBD$ (SAS)

CBSE Questions for Class 9 Maths Triangles Quiz 2 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Triangles Quiz 2 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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