MCQ Questions for Class 9 Maths Triangles Quiz 3

In the figure,

$\displaystyle \Delta ABC\cong \Delta ADC$ by the congruence postulate

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

In the figure,

$\displaystyle AD\parallel BC$ and

$\displaystyle AD=BC$ . Then

$\displaystyle \Delta ABD\cong \Delta CDB$ by congruence postulate.

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

Explanation

$\displaystyle \Delta ABD$ and

$\displaystyle \Delta CDB$

$\displaystyle AD=BC$

$\displaystyle \angle BDA=\angle DBC$ (Alternate angles)

$\displaystyle BD=BD$

$\displaystyle \therefore \quad \Delta ABD\cong \Delta CDB$ (By SAS postulate)

In the above figure,

$\angle{B}=\angle{C}=65^\circ$ and

$\angle{D}=30^\circ$ . Then

0%
$BC < CA < CD$
0%
$BC>CA>CD$
0%
$BC < CA$ , $CA>CD$
0%
$BC>CA$ , $CA < CD$

Explanation

Here

$\angle{B}=\angle{C}=65^\circ$ and

$\angle{D}=30^\circ$

$\therefore$ From the triangle ABD

$\angle{A}=85^\circ$
In triangle ABC

$\angle{A}=50^\circ$

$\because BC<CA$
In triangle ACD

$\angle{A}=85^\circ-50^\circ=35^\circ$

$\therefore CD>CA$

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%
$QS < PQ > PR$
0%
None of these

Explanation

Given:

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

$\angle QPR =180^o - (70^o + 30^o ) =80^o$ {angle sum property of a triangle}

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

$\angle PRQ < \angle PQR$

$\therefore PQ <PR$ ....... (i) {Side opposite to the larger angle is larger}

$\angle QPS < PQS$

$\therefore QS < PR$ ...... (ii) {Side opposite to the larger angle is larger}

$\Delta PQS$ ,

$\angle PSQ = 180^o -(70^o + 40^o) =70^o$ {angle sum property of a triangle}

$\angle QPS < \angle PSQ$

$\therefore QS < PQ$ ........ (iii) {Side opposite to the larger angle is larger}
From (i), (ii) & (iii)

$QS < PQ < PR$ .

1159892_332138_ans_e910c3a26f934d05b619cdb4eed0bcf7.png

$\overline{AC}$ and

$\overline{BD}$ intersect at

$O$ such that

$AO = CO$ and

$BO = DO$ , then:

0%
$AB = CD$
0%
$AB \parallel CD$ and $AB = CD$
0%
$AB \parallel CD$
0%
None of these

Explanation

$\triangle AOB$ and

$\triangle DOC$ ,

side

$OA =$ side

$OC$ ... given

side

$OB =$ side

$OD$ ... given

$\angle AOB \cong \angle COD$ ...Vertically opposite angles

$\therefore \triangle AOB \cong \triangle COD$ ...SAS test of congruence.

$\therefore$ side

$AB$

$\cong$ side

$CD$ ...c.s.c.t

$\therefore \angle BAO \cong \angle DCO$ ....c.a.c.t

i.e.

$\angle BAC \cong \angle DCA$ ....

$A-O-C$ and

$B-O-D$

$\therefore AB \parallel CD$ ....by converse of alternate angles test

So, option B is correct.

In a triangle ABC. The relation which is true for its sides is-

0%
AB + BC < AC
0%
AB + BC > AC
0%
AB +BC = CA
0%
AB = BC + AC

Explanation

$ABC$ ,
Sum of the two sides is always greater than third side.
So,

$AB + BC > AC$

In the figure,

$\displaystyle PS=QR$ and

$PQ=SR$ , then choose the correct option.

0%
$\displaystyle \Delta QRP\cong \Delta PSR$
0%
$\displaystyle \Delta PQR\cong \Delta PRS$
0%
$\displaystyle \Delta PQR\cong \Delta RSP$
0%
$\displaystyle \Delta RQP\cong \Delta RSP$

Explanation

$\displaystyle \Delta PQR$ and

$\displaystyle \Delta PSR$

$\displaystyle PS=QR,PQ=SR,PR=PR$

$\displaystyle \therefore \quad \Delta RQP\cong \Delta RSP$ (By SSS postulate)

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DBC,AC=BD$ and

$\displaystyle \angle ABC=\angle BCD={ 90 }^{ o }$ . Then

$\displaystyle \Delta ABC\cong \Delta DCB$ by the postulate

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

Explanation

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DBC$ ,

$\displaystyle \angle ABC=\angle BCD$

$\displaystyle AC=BD$ (Givenn)

$\displaystyle BC=BC$ (Common Side)

$\displaystyle \Delta ABC\cong \Delta DCB$ WCB (By RHS postulate)

$\displaystyle \Delta PQR,\angle P={ 67 }^{ o }$ and

$\displaystyle \angle Q={ 76 }^{ o }$ . Arrange the sides in descending order according to the lengths we get,

0%
$PR> QR > PQ$
0%
$PQ> QR> PR$
0%
$PQ> QR>PR$
0%
$QR>PQ> PR$

Explanation

$\displaystyle \Delta PQR,\angle P={ 67 }^{ o }$ and

$\displaystyle \angle Q={ 76 }^{ o }$

$\displaystyle \therefore \quad \angle R={ 37 }^{ o }$

$\displaystyle \left( \because \angle P+\angle Q+\angle R={ 180 }^{ o } \right).$

Now arranging the angles in descending order, we get

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

Sides opposite to greater angle are greater in length and the angles and the opposite sides of a triangle have same relation.

Therefore, the sides arranged in descending order, we get

$PR > QR > PQ.$

$\displaystyle \Delta XYZ,\angle X={ 60 }^{ o },\angle Y={ 75 }^{ o }$ . The greatest side is ......... and the smallest side is ..............

0%
$XY, YZ$
0%
$YZ, XZ$
0%
$XZ, XY$
0%
$XZ, YZ$

Explanation

$\displaystyle \Delta XYZ,\angle X={ 60 }^{ o },\angle Y={ 75 }^{ o }$
Also,

$\displaystyle \angle X+\angle Y+\angle Z={ 180 }^{ o }$

$\displaystyle \therefore \quad \angle Z={ 45 }^{ o }$

$\displaystyle \therefore \quad \angle Y$ is the greatest angle and

$\displaystyle \angle Z$ is the smallest angle. Thus

$XZ$ is the greatest side and

$XY$ is the smallest side.

$\displaystyle \Delta PQR,\angle Q={ 67 }^{ o },\angle R={ 48 }^{ o }$ . The smallest side is ........ and the greatest side is ........

0%
$PQ, PR$
0%
$PQ, QR$
0%
$QR, PR$
0%
$PR, PQ$

Explanation

$\displaystyle \Delta PQR,\quad \angle Q={ 67 }^{ o },\angle R={ 48 }^{ o }$ .
Also,

$\displaystyle \angle P+\angle Q+\angle R={ 180 }^{ o }$ . Thus

$\displaystyle \angle P={ 65 }^{ o }$ .

The greatest angle is

$\displaystyle \angle Q$ and the smallest angle is

$\displaystyle \angle R$ .

Side opposite to greater angle is greater in length and side opposite to the smaller angle is smallest.

Thus, the smallest side is

$PQ$ and the greatest side is

$PR.$

If two sides of a triangle are unequal, the angle opposite to the greater side is ___.

0%
smaller
0%
equal
0%
greater
0%
does not exist

Explanation

Let we have triangle

$\Delta ABC$ and we consider a point

$D$ on

$AC$ such that

$AB=AD$

$\Delta ABD$

$AB=AD$

$\to \angle ABD=\angle ADB~~~~-(1)$ (Angle opposite to equal sides are equal)

$\Delta BDC$

$\angle BDA>\angle BCD$ (Exterior angle is greater than interior angle)

$\angle ABD>\angle BCA~~~~~~-(2)$ (from eqs.(1))

Also, as

$\angle ABC>\angle ABD\\\to \angle ABC>\angle BCA$

Hence, angle opposite to greater side is greater.

1912406_435372_ans_13e77758db4a44f5b243ca7aba56c679.png

$\displaystyle \Delta XYZ, XY=6 \ cm, YZ=8 \ cm$ and

$XZ=9 \ cm.$ The greatest angle is ___ and the smallest angle is ____.

0%
$\displaystyle \angle X,\angle Y$
0%
$\displaystyle \angle Z,\angle Y$
0%
$\displaystyle \angle Y,\angle X$
0%
$\displaystyle \angle Y,\angle Z$

Explanation

$\displaystyle \Delta XYZ$ , side

$XZ$ is the greatest and side

$XY$ is the smallest.

Angle opposite to greater side is greater and smaller side is smaller.

Thus,

$\displaystyle \angle XYZ$ i.e.

$\angle Y$ is the greatest and

$\displaystyle \angle YZX$ i.e.

$\angle Z$ is the smallest.

$\displaystyle \Delta XYZ,\angle Y={ 90 }^{ o }$ and

$\angle Z={ 40 }^{ o }$ . Arranging sides (according to length) in ascending order, we get:

0%
$XY< YZ<YZ$
0%
$XY< YZ< ZX$
0%
$YZ< ZX< XY$
0%
$YZ< XY< ZX$

Explanation

$\displaystyle \Delta XYZ,\angle Y={ 90 }^{ o }$ and

$\angle Z={ 40 }^{ o }$

$\displaystyle \therefore \quad \angle X={ 50 }^{ o }\quad \left( \because \angle X+\angle Y+\angle Z={ 180 }^{ o } \right)$
Now, arranging the angles in ascending order, we get

$\displaystyle \angle Z,\angle X$ and

$\angle Y$

Sides opposite to greater angle are greater in length and the angles and the opposite sides of a triangle have same relation.
Thus, arranging the sides in ascending order

$XY< YZ < ZX.$

Which is the greatest side in the following triangle?

0%
$AB$
0%
$BC$
0%
$AC$
0%
cannot be determined

Explanation

$\displaystyle \angle A={ 69 }^{ o },\angle C={ 32 }^{ o }$
Now,

$\displaystyle \angle A+\angle B+\angle C={ 180 }^{ o }$

$\displaystyle \therefore \quad \angle B={ 79 }^{ o }$
Thus,

$\displaystyle \angle B$ is the greatest angle and hence

$AC$ is the greatest side.

$\displaystyle \Delta ABC,\angle B={ 55 }^{ o }$ and

$\displaystyle \angle A={ 45 }^{ o }$ . Arranging the sides in descending order we get:

0%
$AB > BC> AC$
0%
$BC>AB> AC$
0%
$AC>AB> BC$
0%
$AB> AC> BC$

Explanation

$\displaystyle \Delta ABC,\angle B={ 55 }^{ o }$ and

$\displaystyle \angle A={ 45 }^{ o }$

$\displaystyle \therefore \quad \angle C={ 80 }^{ o }\left( \because \angle A+\angle B+\angle C={ 180 }^{ o } \right)$

$\displaystyle \therefore$ Arranging the angles in descending order, we get

$\displaystyle \angle C,\angle B$ and

$\angle A$

Since, the angles and the opposite sides of a triangle have same relation

Therefore, arranging the sides in descending order we get,

$AB, AC$ and

$BC.$

The two quadrilaterals are congruent.
Which angle in quadrilateral WXYZ corresponds to

$\displaystyle \angle BCD$ in quadrilateral ABCD?

0%
$\displaystyle \angle XYZ$
0%
$\displaystyle \angle XWZ$
0%
$\displaystyle \angle WZY$
0%
$\displaystyle \angle WXY$

Explanation

Clearly,

$\angle WXY$ corresponds to

$\angle BCD$ because

$BC$ faces the angles marked with one arc and three arc and

$XY$ also faces the angles marked with one arc and three arcs.

So,

$\text{D}$ is the correct option.

$\displaystyle \bigtriangleup ABC,AD\perp BC,\angle B=\angle C$ and AB = AC State by which property

$\displaystyle \Delta ADB\cong \Delta ADC$ ?

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

$\displaystyle \Delta ABC,\angle A={ 12 }^{ o }$ and

$\angle C={ 97 }^{ o }$ . Arranging the sides (according to lengths) in ascending order, we get:

0%
$BC<AB< AC$
0%
$AB< AC< BC$
0%
$BC< AC< AB$
0%
$AB< BC < AC$

Explanation

$\displaystyle \Delta ABC,\angle A={ 12 }^{ o }$ and

$\angle C={ 97 }^{ o }$

$\displaystyle \therefore \quad \angle B={ 71 }^{ o }\quad \left( \because \angle A+\angle B+\angle C={ 180 }^{ o } \right)$
Now arranging the angles in ascending order

$\displaystyle \angle A<\angle B <\angle C$ .

Sides opposite to greater angle are greater in length and the angles and the opposite sides of a triangle have same relation.

Thus, arranging the sides in ascending order, we get

$BC< AC <AB.$

For a

$\displaystyle \Delta XYZ, ZY = 12\ m, YX = 8\ m$ and

$XZ = 10\ m$ . If

$\displaystyle \Delta ZYX\cong \Delta ABC$ , then

$AC =$ _____

$m$ .

0%
$10$
0%
$12$
0%
$8$
0%
$1$

Explanation

Given :

$\displaystyle \Delta XYZ, ZY = 12\ m, YX = 8\ m$ and

$XZ = 10\ m$ .

Also,

$\displaystyle \Delta ZYX\cong \Delta ABC$ .

We know, congruent triangles have their sides congruent.

$\implies AB=ZY, BC=YX$ and

$AC=ZX$ .

$\implies \displaystyle ZX=AC$ .

$\displaystyle \therefore AC=10\ m$ .

Therefore, option

$A$ is correct.

In the figure given below,

$\displaystyle \Delta ABC\cong \Delta MBC.$ The corresponding side to

$BA$ is ......

0%
$BM$
0%
$AB$
0%
$AC$
0%
$MC$

Explanation

Given $\triangle ABC \cong \triangle MBC$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=MB$ , $BC=BC$ , $AC=MC$ , $\angle A=\angle M$ , $\angle B=\angle B$ and $\angle C=\angle C$ .

Thus, $AB=MB$ $\implies$ $BA=BM$ .

Hence, option $A$ is correct.

In the figure given below,

$\displaystyle \Delta ABC\cong \Delta XYZ$ . The corresponding side to

$XZ$ is .......

0%
$AB$
0%
$BC$
0%
$AC$
0%
$YZ$

Explanation

Given $\triangle ABC \cong \triangle XYZ$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=XY$ , $BC=YZ$ , $AC=XZ$ , $\angle A=\angle X$ , $\angle B=\angle Y$ and $\angle C=\angle Z$ .

Thus, $AC=XZ$ .

Hence, option $C$ is correct.

In the figure given below,

$\displaystyle \Delta BCA\cong \Delta BCD$ . Corresponding angle to

$\displaystyle \angle D$ is ...........

0%
$\displaystyle \angle B$
0%
$\displaystyle \angle C$
0%
$\displaystyle \angle D$
0%
$\displaystyle \angle A$

Explanation

Given $\triangle BCA \cong \triangle BCD$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $BC=BC$ , $CA=CD$ , $BA=BD$ , $\angle B=\angle B$ , $\angle C=\angle C$ and $\angle A=\angle D$ .

Thus, $\angle A=\angle D$ .

Hence, option

$D$ is correct.

In the figure given below, both the triangles are congruent. The corresponding side to

$BC$ is .......

0%
$EF$
0%
$AC$
0%
$DF$
0%
$DE$

Explanation

Given $\triangle ABC \cong \triangle DEF$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=DE$ , $BC=EF$ , $AC=DF$ , $\angle A=\angle D$ , $\angle B=\angle E$ and $\angle C=\angle F$ .

Thus, $BC=EF$ .

Hence, option

$A$ is correct.

In the figure given below, both the triangles are congruent. The corresponding side to

$AB$ is ___.

0%
$AB$
0%
$DE$
0%
$BC$
0%
$DF$

Explanation

Given $\triangle ABC \cong \triangle DEF$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=DE$ ,

$BC=EF$ ,

$AC=DF$ ,

$\angle A=\angle D$ ,

$\angle B=\angle E$

$\angle C=\angle F$ .

Thus, $AB=DE$ .

Hence, option $B$ is correct.

Which of the following sets of side lengths form a triangle?

0%
4 m, 3 m, 11 m
0%
7 mm, 4 mm, 4 mm
0%
3 cm, 1.1 cm, 5 cm
0%
3 m, 4 m, 8 m

Explanation

A triangle can be formed only if sum of any two sides is greater than the third side.

In option

$B$ sum of any two sides taken at a time is greater than the third side.

$7+4 = 11>4$

$4+4= 8>7$

$4+7= 11>4$
In all other options, sum of first two sides is less than the third side:
A)

$4+3<11$
C)

$3+1.1<5$
D)

$3+4<8$

So option

$B$ is correct.

It is given that

$\Delta ABC\cong \Delta DEF$ . In the given figure the corresponding angle to

$\displaystyle \angle C$ is ........

0%
$\displaystyle \angle B$
0%
$\displaystyle \angle C$
0%
$\displaystyle \angle F$
0%
$\displaystyle \angle A$

Explanation

Given $\triangle ABC \cong \triangle DEF$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=DE$ , $BC=EF$ , $AC=DF$ , $\angle A=\angle D$ , $\angle B=\angle E$ and $\angle C=\angle F$ .

Thus, $\angle C=\angle F$ .

Hence, option

$C$ is correct

It is given that

$\Delta ABC\cong \Delta DEF$ . In the given figure the corresponding angle to

$\displaystyle \angle B$ is___.

0%
$\displaystyle \angle A$
0%
$\displaystyle \angle C$
0%
$\displaystyle \angle E$
0%
$\displaystyle \angle F$

Explanation

Given $\triangle ABC \cong \triangle DEF$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=DE$ , $BC=EF$ , $AC=DF$ , $\angle A=\angle D$ , $\angle B=\angle E$ and $\angle C=\angle F$ .

Thus,

$\angle B=\angle E$ .

Hence, option

$C$ is correct.

Referring the figure below, for

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ ,

$AB=DE, BC=EF$ and

$\displaystyle \angle B=\angle E.$ The value of

$\displaystyle \angle E$ is .........

0%
$\displaystyle { 30 }^{ o }$
0%
$\displaystyle { 60 }^{ o }$
0%
$\displaystyle { 90 }^{ o }$
0%
$\displaystyle { 45 }^{ o }$

Explanation

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$

$\displaystyle AB=DE,BC=EF$ and

$\displaystyle \angle B=\angle E$

$\displaystyle \therefore \quad \Delta ABC\cong \Delta DEF$ (S.A.S condition for congruence)
As,

$\displaystyle AB=DE\Rightarrow \angle C=\angle F={ 60 }^{ o }$

$\displaystyle BC=EF\Rightarrow \angle A=\angle D={ 60 }^{ o }$

$\displaystyle AC=DF\Rightarrow \angle B=\angle E$
Now,

$\displaystyle \angle D+\angle E+\angle F={ 180 }^{ o }$

$\displaystyle \therefore \quad { 120 }^{ o }+\angle E={ 180 }^{ o }$

$\displaystyle \therefore \quad \angle E={ 60 }^{ o }$

In the triangles

$ABC$ and

$PQR,$

$AC=QP, BC=RP$ and

$AB=QR.$ The measure of

$\angle A$ is ___.

0%
$\displaystyle { 30 }^{ \circ }$
0%
$\displaystyle { 120 }^{ \circ }$
0%
$\displaystyle { 20 }^{ \circ }$
0%
$\displaystyle { 40 }^{ \circ }$

Explanation

Given: In

$\triangle ABC$ and

$\triangle PQR,$

$\angle C=120^\circ$ and

$\angle R=40^\circ$

Now,

$AC=QP$ ...(1)

and

$BC=RP$ ...(2)

and

$AB=QR$ ...(3)

Hence

$ABC \cong PQR$

$($ by

$SSS$ congruency rule

$)$

By CPCT,

$\angle B=\angle R=40^\circ$

and

$\angle C=\angle P={ 120 }^{ \circ }$

and

$\angle A=\angle Q$

Now,

$\displaystyle \angle A+\angle B+\angle C={ 180 }^{ o }$

$\displaystyle \Rightarrow \angle A+{ 40 }^{ \circ }+{ 120 }^{ \circ }={ 180 }^{ \circ }$

$\displaystyle \Rightarrow \angle A={ 20 }^{ \circ}$

CBSE Questions for Class 9 Maths Triangles Quiz 3 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Triangles Quiz 3 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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