MCQ Questions for Class 9 Maths Triangles Quiz 4

In a rectangle

$PQRS$ ,

$QS$ is a diagonal. Then

$\displaystyle \Delta PQS......\Delta RSQ$

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is similar to
0%
is not equal to
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is congruent to
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is not congruent to

Explanation

In a rectangle

$PQRS$ ,

$QS$ is a diagonal.
We know that , in a rectangle opposite sides are equal.

$\displaystyle \therefore PQ=SR$
Also,

$\displaystyle \angle R=\angle P={ 90 }^{ o }$
Ans

$QS$ is the hypotenuse for both right angled triangles

$PQS$ and

$RSQ$ .

$\displaystyle \therefore$ according to R.H.S condition for congruence,

$\displaystyle \Delta PQS\cong \Delta RSQ$ .

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF,\angle A={ 47 }^{ o },\angle E={ 53 }^{ o }$ and

$\displaystyle AB=DE=16 \ cm\\$ .

$\displaystyle \angle A=\angle D$ and

$\displaystyle \angle B=\angle E$ , then

$\displaystyle \angle C=$ ..............

0%
$\displaystyle { 47 }^{ o }$
0%
$\displaystyle { 53 }^{ o }$
0%
$\displaystyle { 80 }^{ o }$
0%
$\displaystyle { 100 }^{ o }$

Explanation

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$

$\displaystyle \angle A=\angle D,\angle B=\angle E$ and

$\displaystyle AB=DE$

$\displaystyle \therefore$ According to A.S.A condition for congruence,

$\displaystyle \Delta ABC\cong \Delta DEF$

$\displaystyle \angle A=\angle D={ 47 }^{ o }$

$\displaystyle \angle B=\angle E={ 53 }^{ o }$

$\displaystyle \angle C=\angle F$

$\displaystyle \therefore \quad \angle D+\angle E+\angle F=\angle A+\angle B+\angle C={ 180 }^{ o }\\$

$\displaystyle \therefore \quad \angle C={ 180 }^{ o }-{ 100 }^{ o }$

$\displaystyle \therefore \quad \angle C={ 80 }^{ o }$

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta PQR$ ,

$AB=PQ, AC=PR,$ and

$\displaystyle \angle A=\angle P,$ then:

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$\displaystyle \Delta BAC\cong \Delta PRQ$
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$\displaystyle \Delta BAC\cong \Delta PQR$
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$\displaystyle \Delta ABC\cong \Delta PRQ$
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$\displaystyle \Delta ABC\cong \Delta PQR$

Which condition for congruence is used to prove that the following pairs of triangles are congruent

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

Explanation

In the given

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ ,

we can clearly see that sides

$AB = DE, BC = EF$ and

$AC = DF.$

Thus, by

$SSS$ condition of congruence, the given

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ are congruent.

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF,$

$AB=FE, BC=ED$ and

$\displaystyle \angle B=\angle E$ . Therefore ............

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$\displaystyle \Delta ABC\cong \Delta DEF$
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$\displaystyle \Delta ABC\cong \Delta EDF$
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$\displaystyle \Delta BCA\cong \Delta DEF$
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$\displaystyle \Delta BCA\cong \Delta EDF$

Explanation

For,

$\displaystyle \Delta BCA$ and

$\displaystyle \Delta EDF,$

$\displaystyle AB=FE\quad \therefore \angle C=\angle D$

$\displaystyle BC=ED\quad \therefore \angle A=\angle F$

$\displaystyle \angle B=\angle E$

$\displaystyle \Delta BCA\cong \Delta DEF$
(By

$SAS$ condition for congruence).

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ ,

$\displaystyle \angle B=\angle D,\angle C=\angle F$ and

$BC=DF$ . Therefore which of the following is correct?

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$\displaystyle \Delta ABC\cong \Delta DEF$
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$\displaystyle \Delta BCA\cong \Delta DFE$
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$\displaystyle \Delta BCA\cong \Delta DEF$
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$\displaystyle \Delta ABC\cong \Delta DFE$

Explanation

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ , we have

$\displaystyle \angle B=\angle D,\angle C=\angle F$ .....Given

$BC=DF$ .....Given
Therefore, according to A.S.A condition for congruence,

$\displaystyle \Delta BCA\cong \Delta DFE$

Hence, option B is correct.

Referring the figure given below, for

$\displaystyle \Delta PQR$ and

$\displaystyle \Delta XYZ$ ,

$PQ=YZ, QR=XZ$ and

$\displaystyle \angle Q=\angle Z$ . The value of

$\displaystyle \angle P$ is ___.

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$\displaystyle { 42 }^{ \circ }$
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$\displaystyle { 75 }^{ \circ }$
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$\displaystyle { 63 }^{ \circ }$
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Cannot be determined

Explanation

$\displaystyle \Delta PQR$ and

$\displaystyle \Delta XYZ,$

$\displaystyle PQ=YZ$

$\displaystyle QR=XZ$

$\displaystyle \angle Q=\angle Z$

$\displaystyle \therefore \Delta PQR\cong \Delta XYZ$ (

$S.A.S.$ condition for congruence)

$\displaystyle \therefore \angle R=\angle X={ 63 }^{ \circ }$

$\displaystyle \Rightarrow \angle P=\angle Y$

$\displaystyle \Rightarrow \angle Q=\angle Z={ 75 }^{ \circ }$

Now,

$\displaystyle \angle P+\angle Q+\angle R={ 180 }^{ \circ }$

$\displaystyle \therefore \angle P+{ 75 }^{ \circ }+{ 63 }^{ \circ }={ 180 }^{ \circ }$

$\displaystyle \Rightarrow \angle P={ 42 }^{ \circ }$

In the given figures, which of the following satisfies A.S.A. condition for congruence?

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Figure II
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Figure I and II
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Figure I
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None of these

Explanation

Figure 1 satisfy the condition of A.S.A

Here

$\angle ACB=\angle DCE$ {vertical angle|

$DC=BC$ and

$\angle EDC=\angle ABC$ [Alternate angles]

$\therefore$ It satisfy A.S.A Congruency.

For

$\displaystyle \Delta ABC \cong \displaystyle \Delta DEF$ . Then according to the given figure,

$\angle D = 40^o$ ,

$\displaystyle \angle C=$ .....

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$\displaystyle { 40 }^{ o }$
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$\displaystyle { 90 }^{ o }$
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$\displaystyle { 60 }^{ o }$
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$\displaystyle { 50 }^{ o }$

Explanation

According to the given figure,

$\displaystyle \angle B=\angle E={ 90 }^{ o }$

$\Rightarrow AC=DE=5$ cm (hypotenuse) and

$BC=EF=2$ cm

$\displaystyle \therefore \Delta ABC\cong \Delta DEF$ ..... (R.H.S condition for congruence)

$\displaystyle \Rightarrow \angle A=\angle D={ 40 }^{ o }$

$\displaystyle \Rightarrow \angle B={ 90 }^{ o }$

$\displaystyle \angle C=?$
We know

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

$\displaystyle \therefore \angle C={ 180 }^{ o }-{ 90 }^{ o }-{ 40 }^{ o }$

$\displaystyle \therefore \angle C={ 50 }^{ o }$

Given are two congruent triangles

$ABC$ and

$XYZ.$ Which of the following statements is incorrect?

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$\displaystyle \angle A=\angle X$
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$\displaystyle AB=XY$
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$\displaystyle \Delta ABC\cong \Delta XYZ$
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$\displaystyle AC=YZ$

Explanation

Given,

$\triangle ABC$ and

$\triangle XYZ$ are congruent

From figure, we have

$AC=XZ, BC=YZ$ and

$\angle B=\angle Y=90^o$

$\displaystyle \therefore AC=YZ$ is an incorrect statement.

Which condition of congruency proves that the following pair of triangles are congruent

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$A.A.S$
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$S.S.S$
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$S.A.S$
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$A.S.A$

Explanation

From the given figure, we have

$DE=EG, DF=GP$ and

$FE=PE.$

Thus, from

$S.S.S$ condition of congruency,

$\displaystyle \Delta DEF\cong \Delta GEP$

Complete the correspondence statement:

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

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$\displaystyle \Delta DEF$
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$\displaystyle \Delta DFE$
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$\displaystyle \Delta FDE$
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$\displaystyle \Delta EFD$

Explanation

Here,

$\displaystyle \angle A=\angle D,\angle B=\angle F$ and

$\displaystyle \angle C=\angle E$ .

Also,

$AB=DF$ ,

$BC=FE$ and

$AC=DE$ .

Since, all the corresponding parts are equal,

$\displaystyle \Delta ABC\cong \Delta DFE$ .

Therefore, option

$B$ is correct.

In the figure,

$\displaystyle \Delta ACB\cong \Delta DCB$ , then the value of

$x$ and

$y$ are:

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$14$ cm, $\displaystyle { 20 }^{ o }$
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$28$ cm, $\displaystyle { 20 }^{ o }$
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$\displaystyle { 14 }^{ o }$ , $20$ cm
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$\displaystyle { 28 }^{ o }$ , $20$ cm

Explanation

From the figure, we have,

$\angle D= 20^o, BD=28$ cm.

We need to find value of

$x$ and

$y$ .

$\displaystyle \Delta ACB\cong \Delta DCB$ , we have,

Then the corresponding parts will also be equal.

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

Then,

$AC=DC$ ,

$CB=CB$ ,

$AB=DB$ ,

$\angle A=\angle D$ ,

$\angle C=\angle C$ and

$\angle B=\angle B$ .

$\displaystyle \therefore AB=BD$ cm

$\displaystyle \therefore 2x=28$ cm

$\displaystyle \therefore x=14$ cm.

Also,

$\displaystyle \angle A=\angle D$

$\displaystyle \therefore y={ 20 }^{ o }$ .

Therefore, option

$A$ is correct.

Which condition of congruence proves that the following pair of triangles are congruent

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$A.A.S$
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$S.S.S$
0%
$S.A.S$
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$A.S.A$

Explanation

From the given figure, we have

$AC=DF, BC=ED$ and

$\displaystyle \angle C=\angle D$

$\displaystyle \therefore$ According to

$S.A.S$ condition of congruence,

$\displaystyle \Delta ABC\cong \Delta FED$

Which condition of congruency proves that the following pair of triangles are congruent

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

Explanation

According to the given figure, we have

$\displaystyle \angle B=\angle F,\angle C=\angle G$ and

$BC=FG$

$\displaystyle \therefore \Delta ABC\cong \Delta EFG$ ....

$(ASA$ condition for congruence

$)$

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ ,

$\displaystyle \angle B=\angle D={ 90 }^{ o }$ ,

$\displaystyle AC=EF$ and

$\displaystyle AB=ED$ . Then

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

0%
$\displaystyle \Delta EDF$
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$\displaystyle \Delta DEF$
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$\displaystyle \Delta EFD$
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$\displaystyle \Delta FED$

Explanation

For

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DEF$ ,

$\displaystyle \angle B=\angle D={ 90 }^{ o }$ ,

$\displaystyle AC=EF$ and

$\displaystyle AB=ED$

$\displaystyle \therefore$ According to R.H.S condition for congruence, both triangles are

$\displaystyle \therefore \Delta ABC\cong \Delta EDF$

In a

$\triangle ABC$ ,

$D$ is the midpoint of the side

$BC. DE$ is perpendicular to

$AB$ and

$DF$ is perpendicular to

$AC$ . Also,

$DE=DF$ , Then

$\displaystyle \angle B$ =

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

Explanation

Given:

$D$ is the midpoint of

$BC$

$\displaystyle \therefore \quad BD=DC$

Also,

$DE = DF$

$DE$ is perpendicular to

$AB$

$\displaystyle \therefore \angle E={ 90 }^{ o }$

$DF$ is perpendicular to

$AC$

$\displaystyle \therefore \angle F={ 90 }^{ o }$

If we consider two triangles so formed,

$\displaystyle \Delta DEB$ and

$\displaystyle \Delta CFD$ , then their hypotenuse

$BD$ and

$CD$ are equal.

$\displaystyle \angle E=\angle F={ 90 }^{ o }$ . Also

$DE=CF$ .

Thus according to R.H.S condition of congruence

$\displaystyle \Delta DEB\cong \Delta CFD$

Thus, the corresponding angles

$\displaystyle \angle B$ and

$\angle C$ are equal, i.e

$\displaystyle \angle B=\angle C$

Complete the correspondence statement:

$\displaystyle \Delta GHK\cong$ ____.

0%
$\displaystyle \Delta GTK$
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$\displaystyle \Delta KTG$
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$\displaystyle \Delta GKT$
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$\displaystyle \Delta TGK$

Explanation

Here,

$\displaystyle \angle K={ 90 }^{ o }$ (common angle)

$GH=GT$ and

$KH=KT$

$\displaystyle \therefore$ According to

$R.H.S$ condition of congruency of triangles,

$\displaystyle \Delta GHK\cong \Delta GTK$ .

Therefore, option

$A$ is correct.

Identify the largest angle of the triangle.

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A
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B
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C
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D

Explanation

In a triangle inequality theorem, the largest angle is across from the longest side.
So,

$9$ is the longest side in the triangle,

$< A$ , across from it, is the largest angle.
Therefore,

$A$ is the largest angle.

$AD$ is the bisector of BC.
Which property of triangle applies here?

0%
$SAS$
0%
$SSS$
0%
$AAS$
0%
none

Explanation

$\triangle ABD$ &

$\triangle ADC$

$AD = AD$

$BD = DC$

$\angle ADB = \angle ADC$

$\triangle ABD \cong \triangle ADC$

$\therefore$ Both triangles are congruent by

$SAS$ congruence condition.

$\Delta ABC \cong \Delta DEF$ by ASA congruence condition. Find the value of

$x$ .

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$x = 20$
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$x = 30$
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$x = 40$
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$x = 50$

Explanation

We know

$\Delta ABC \cong \Delta DEF$ by ASA congruence.
Therefore,

$\angle B = \angle F$

$\Rightarrow 60 = x + 30$

$\Rightarrow x = 60 - 30$

$\Rightarrow x = 30$

What is the largest angle of the triangle?

0%
X
0%
Y
0%
Z
0%
P

Explanation

In a triangle inequality theorem, the largest angle is across from the longest side.
So,

$20$ is the longest side in the triangle,

$\angle Z$ , across from it is the largest angle.
Therefore,

$Z$ is the largest angle.

In triangle

$PQR$ , an exterior angle at

$A$ measures

$160^o$ , and

$\angle$

$Q = 70 ^o$ . Which is the longest side of the triangle?

0%
$\overline{PR}$
0%
$\overline{PA}$
0%
$\overline{PQ}$
0%
$\overline{QR}$

Explanation

By using exterior angle property,

$\angle P+\angle Q = \angle A$

$\Rightarrow \angle P +70^o=160^o$

$\Rightarrow \angle P = 160^o-70^o$

$\Rightarrow \angle P = 90^o$

According to the triangle inequality theorem, the side that is opposite to the largest angle is largest.
So, in a right angle triangle,

$90^o$ is the longest angle in any,

$\overline{QR}$ is the side opposite to it.

Therefore,

$\overline{QR}$ is the largest side.

$\displaystyle \Delta BAC\cong \Delta DEC,AB\parallel DE$ , if

$\displaystyle \angle B$ is

$\displaystyle { 90 }^{ o }$ and

$\displaystyle \angle C$ is

$\displaystyle { 50 }^{ o }$ ,
Find

$\displaystyle \angle A$ and

$\displaystyle \angle D$ =.......

0%
$\displaystyle { 35 }^{ o }$
0%
$\displaystyle { 50 }^{ o }$
0%
$\displaystyle { 40 }^{ o }$
0%
$\displaystyle { 55 }^{ o }$

Explanation

Given $\triangle ABC \cong \triangle DEC$ .

Then the corresponding parts will also be equal.

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

Then,

$AB=DE$ ,

$BC=EC$ ,

$AC=DC$ ,

$\angle A=\angle D$ ,

$\angle B=\angle E$ and

$\angle C=\angle C$ .

Also given,

$\angle B=90^o$ and

$\angle C=50^o$ .

$\triangle ABC$ ,

by angle sum property,

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

$\implies$

$\angle A+90^o + 50^o=180^o$

$\implies$

$\angle A+140^o=180^o$

$\implies$

$\angle A=180^o-140^o$

$\implies$

$\angle A=40^o$ .

Since,

$\angle A=\angle D$

Therefore,

$\angle A=\angle D$

$=40^o$ .

Hence, option

$C$ is correct.

In figure,

$BD$ is a bisector of

$\angle B, CD = DA$
In order to show the

$\triangle BDA$ and

$\triangle BDC$ are congruent, which property of triangle applies here?

0%
$SAS$
0%
$SSS$
0%
$AAS$
0%
None

Explanation

$\triangle ABD$ and

$\triangle CBD$

$BD$ is a bisector of

$\angle B$

$BD$ is common side.

$\angle ABD=\angle CBD$

$CD=DA$

Two triangles are congruent from

$SAS$ postulate

Hence option A is correct.

Identify the largest side of the triangle. If the angles are given as:

$\angle A = 50 ^o$ ,

$\angle B = 10^o$ and

$\angle C = 22^o$

0%
$\overline{AB}$
0%
$\overline{BC}$
0%
$\overline{CA}$
0%
$\overline{AC}$

Explanation

In a triangle inequality theorem, the largest side is opposite to the longest angle.
So,

$50$ degree is the longest angle in the triangle,

$\overline{BC}$ , opposite to it, is the largest side.
Therefore,

$\overline{BC}$ is the largest side.

Which condition will prove these triangles as congruent?

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

Explanation

$\triangle$

$HJG$ and

$\triangle$

$MNK$ ,

$\angle$

$HJG$

$=$

$\angle$

$MNK$ (given)

$JG = NK$ (given)

$\angle$

$JGH =$

$\angle$

$NKM$ (given).

Therefore,

$\triangle$

$HJG$

$\cong$

$\triangle$

$MNK$ (By

$ASA$ congruency criterion).

Hence, option

$D$ is correct.

$\Delta ABC$ and

$\Delta DEF$ are congruent, then find the value of

$x$ and

$y$ .

0%
$x = 40, y = 60$
0%
$x = 30, y = 60$
0%
$x = 60, y = 60$
0%
$x = 60, y = 40$

Explanation

We know

$\Delta ABC \cong \Delta DEF$ by ASA congruence.
Therefore,

$\angle A = \angle D=40$

$\therefore x = 40$

$\angle B = \angle F=60$

$\therefore y=60$
So,

$x = 40, y = 60$

$\displaystyle \Delta XYZ$ , side

$\displaystyle \overline { XZ }$ is longer than side

$\displaystyle \overline { YZ }$ , which of the following statements is true?

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The measure of $\displaystyle \angle X$ is always less than the measure of $\displaystyle \angle Y$ .
0%
The measure of $\displaystyle \angle X$ is always equal than the measure of $\displaystyle \angle Y$ .
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The measure of $\displaystyle \angle X$ is always greater than the measure of $\displaystyle \angle Y$ .
0%
The measure of $\displaystyle \angle X$ is sometimes less than the measure of $\displaystyle \angle Y$ and sometimes equal to the measure of $\displaystyle \angle Y$ .
0%
The measure of $\displaystyle \angle X$ is sometimes greater than the measure of $\displaystyle \angle Y$ and sometimes equal to the measure of $\displaystyle \angle Y$ .

Explanation

Angles opposite to longer sides are larger.

Here

$XZ>YZ$

Angle opposite to

$XZ$ is

$\angle Y$ and to

$YZ$ is

$X$

$\therefore \angle Y>\angle X$

So option

$A$ is correct.

$\triangle ABC$ , if

$\angle A = 45^{\circ}$ and

$\angle B = 70^{\circ}$ , find the shortest and the largest sides of the triangle respectively

0%
$AB, BC$
0%
$BC, AC$
0%
$AB, AC$
0%
Either (A) or (C)

Explanation

Given,

$\angle A = 45^{0}$ and

$\angle B = 70^{0}$

By angle sum property,

$A+B+C = 180^{0}$

$\implies 45+70+C=180$

$\implies C=65^{0}$

Angle opposite to longest side is larger and the side opposite to

$\angle B$ is AC .

Also, shortest side is opposite to the shorter angle,

$\angle A$ is BC.

CBSE Questions for Class 9 Maths Triangles Quiz 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Triangles Quiz 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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