MCQ Questions for Class 9 Maths Triangles Quiz 1

Two sides of a triangle are of lengths

$4$ cm and

$1.5$ cm. The length of the third side of the triangle cannot be

0%
$3.6$ cm
0%
$4.1$ cm
0%
$3.8$ cm
0%
$5.8$ cm

Explanation

Sum of given sides of the triangle is

$5.5$ cm.

The length of the third side cannot be greater than the sum of the two given sides.
Here, one option gives length as

$5.8$ cm is not possible.

The construction of a triangle

$ABC$ , given that

$BC =$

$6$ cm,

$B =$

$45 ^{\circ}$ is not possible when difference of

$AB$ and

$AC$ is equal to:

0%
$6.9$ cm
0%
$5.2$ cm
0%
$5.0$ cm
0%
$4.0$ cm

Explanation

According to the theorem of inequalities, the sum of any two sides of the triangle is greater than the third side.

Therefore,

$AC+BC>AB$

$\Rightarrow BC>AB-AC$

Therefore, only the first option that is

$6.9$ cm does not satisfy the above equation. Rest all the options satisfy the equation.

In the above figure, if OA

$=$ OB, OD

$=$ OC then

$\Delta AOD \cong \Delta BOC$ by congruence rule :

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

Explanation

$\triangle AOD$ and

$\triangle BOC$

$OA=OB$ -------(given)

$OD=OC$ -------(given)

$\angle AOD=\angle BOC$

$\therefore \triangle AOD\cong \triangle BOC$ by

$SAS$ criteria.

$\Delta ABC\, \angle A=50^{\circ}$ and

$\, \angle B=60^{\circ}$ . By arranging the sides of the triangle in ascending order, we get :

0%
$AB < BC < CA$
0%
$CA < AB < BC$
0%
$BC < CA < AB$
0%
$BC < AB < CA$

Explanation

Using angle sum property in

$\triangle ABC$

$\angle A+\angle B+\angle C={ 180 }^{ \circ }\\ \Rightarrow { 50 }^{ \circ }+{ 60 }^{ \circ }+\angle C={ 180 }^{ \circ }\\ \Rightarrow \angle C={ 70 }^{ \circ }$

In any triangle side opposite to the largest angle is the longest side.

Here

$\angle C$ is largest and side opposite to it is

$AB$

$\therefore AB$ is the longest side.

Then comes

$\angle B$ and side opposite to it is

$AC$

$\therefore AC$ is the second longest side.

Then comes

$\angle A$ and side opposite to it is

$AC$

$\therefore BC$ is the shortest side.

So the increasing order of sides is

$BC<AC<AB$

$\Delta ABC, \angle A=100^{\circ}, \angle B=30^{\circ}$ and

$\angle C= 50^{\circ}$ ,then

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

Explanation

In any triangle side opposite to the largest angle is the longest side.

Here

$\angle A$ is largest and side opposite to it is

$BC$

$\therefore BC$ is the longest side.

Then comes

$\angle C$ and side opposite to it is

$AB$

$\therefore AB$ is the second longest side.

Then comes

$\angle B$ and side opposite to it is

$AC$

$\therefore AC$ is the shortest side.

So the increasing order of sides is

$AC<AB<BC$

$\Rightarrow AB>AC$

So option

$A$ is correct.

In figure, if

$AB=AC$ and

$AP=AQ$ , then by which congruence criterion

$\Delta PBC \cong \Delta QCB$ ?

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

Explanation

In the given triangle:

$AB=AC$

$\Rightarrow \angle B=\angle C$ (As angles opposite to equal to opposite sides are equal.

$AB=AC$

$AP+PB=AQ+QC$

Now

$AP=AQ$

$\Rightarrow PB=QC$

Now in

$\triangle PBC$ and

$\triangle QCB$

$PB=QC$ (Proved above)

$\angle B=\angle C$ (Proved above)

$BC=CB$ (Common side)

$\therefore \triangle PBC\cong \triangle QCB$ by

$SAS$ criteria.

$Q$ is a point on side RS of

$\Delta PSR$ such that

$PQ =PR$ then which of the following is correct:

0%
$PS = PQ$
0%
$PS > PQ$
0%
$PS < PQ$
0%
Cannot be determined

Explanation

$\Delta PQR$ ,

side

$PQ =$ side

$PR$ ....given

$\angle PQR = \angle PRQ$ ....isosceles triangle theorem
Now,

$\angle PQS + \angle PQR = 180^o$ ....angles in linear pair

Since,

$\angle PQR$ is an acute angle,

$\angle PQS$ is obtuse.

i.e.

$m \angle PQS > 90^o$
Now in

$\Delta PQS$ ,

$m \angle PQS > m \angle PSQ$

$\therefore PS > PQ$

$\therefore PS > PR$ ...(

$\because PQ=PR$ )

If length of the largest side of a triangle is 12 cm then other two sides of triangle can be

0%
4.8 cm, 8.2 cm
0%
3.2 cm, 7.8 cm
0%
6.4 cm, 2.8 cm
0%
7.6 cm, 3.4 cm

Explanation

Sum of any two sides of a triangle is greater than the third side.

Here the sum must be greater than

$12\ \ cm$

In option

$A$

$4.8\ \ cm+8.2\ \ cm=13\ \ cm$

$\Rightarrow 13\ \ cm>12 \ \ cm$

In rest of the options sum is less than

$12\ \ cm$

So option

$A$ is correct.

If $AB = 1$ , then radius of the inscribed circle

is:

0%
$\frac{1}{3}$
0%
$\frac{1}{2}$
0%
$\frac{1}{\sqrt{3} + 1}$
0%
$\frac{1}{\sqrt{3} - 1}$

$\Delta ABC$ is congruent of

$\Delta DEF$ , if

0%
$AB \neq DE$ , $AC = 6 = DF$ , $\angle A = \angle D$
0%
$AB = DE$ , $AC = DF$ , $\angle B =\angle E$
0%
$AB = DE$ , $AC \neq DF$ , $\angle C =\angle F$
0%
$AB = DE$ , $AC = DF$ , $\angle C \neq \angle F$

Explanation

We say that triangle

$\triangle ABC$ is congruent to

$\triangle DEF$ if -

AB = DF
BC = EF
CA = FD
$\angle A = \angle D$
$\angle B = \angle E$
$\angle C = \angle F$

For this equation all options satisfies the above conditions.

Therefore for this question,

$\triangle ABC$ is congruent of

$\triangle DEF$ if all the options are correct.

$E$ is a point on side

$QR$ of

$\Delta PQR$ such that

$PE$ bisects

$\angle QPR$ , then :

0%
$QE = ER$
0%
$QP > QE$
0%
$QE > QP$
0%
$ER > RP$

Explanation

According to the given condition

$\angle QPE = \angle RPE$

And from external angle theorem,

$\angle PEQ= \angle RPE + \angle PRE$

$\Rightarrow \angle PEQ = \angle QPE + \angle PRE$

$\Rightarrow \angle PEQ > \angle QPE$

In triangle, side opposite angle is greater

$\Rightarrow QP > QE$

$\Delta PQR, \angle P = 60^{\circ}$ and

$\angle Q = 50^{\circ}$ . Which side of the triangle is the longest ?

0%
PQ
0%
QR
0%
PR
0%
None

Explanation

Using angle sum property of triangle

$\angle P+\angle Q+\angle R={ 180 }^{ \circ }\\ \Rightarrow { 60 }^{ \circ }+{ 50 }^{ \circ }+\angle R={ 180 }^{ \circ }\\ \Rightarrow \angle R={ 180 }^{ \circ }-{ 110 }^{ \circ }={ 70 }^{ \circ }$

$\angle R$ is the largest angle and side opposite to largest angle is the longest side.

$\therefore PQ$ is the longest side.

T is a point on side QR of

$\Delta$ PQR. "S" is a point such that

$RT = ST.$ then

0%
$PQ + QR > QS$
0%
$PQ + PR > QS$
0%
$QR + PR > QS$
0%
$PQ + PR < QS$

Explanation

$\Delta PQR$ ,

$PQ + PR > RQ...........(1)$

$RQ = TR + TQ$

$Also, in$

$\Delta STQ$ ,

$= S T + TQ > QS...........(2)$

From (1) and (2) [since ST+TQ=RQ]

$PQ + PR > QS$

State true/false:

If in

$\bigtriangleup ABC and \bigtriangleup QRP, AB =QR,\angle B=\angle R\:and \: \angle C=\angle P.$

Then, by A.S.A criteria,

$\triangle ABC$ and

$\triangle QRP$ are congruent.

0%
True
0%
False

Explanation

$\triangle$ s, ABC and QRP,
Given,

$BC = RP$

$\angle B = \angle R$

$\angle C = \angle P$
Thus,

$\triangle ABC \cong \triangle QRP$ (By

$ASA$ criteria)

State whether the given statement is true or false:

$\displaystyle \bigtriangleup ABC$ and

$\bigtriangleup DEF, AB=DE,BC=EF$ and

$\angle B=\angle E$ . The triangles are congruent by

$SSS$ test.

0%
True
0%
False

Explanation

$\triangle$ ABC and

$\triangle$ DEF,

$AB = DE$ (Given)

$BC = EF$ (Given)

$\angle B = \angle E$ (Given)
Hence,

$\triangle ABC \cong \triangle DEF$ (By SAS postulate)

So, the given statement is False.

$AB = 7$ cm,

$BP = 4$ cm,

$AP = 5.4$ cm, then compare the segments.

0%
$AP > BP > AB$
0%
$BP < AP < AB$
0%
$AP > BP < AB$
0%
None of these

Explanation

$AB=7cm$ ,

$BP=4cm$ and

$AP=5.4cm$ . Therefore

$AB>AP>BP$ .

In triangle

$ABC$ ,

$\angle B=30^{o}$ and

$\angle$ C

$=70^{o}$ . The greatest side of the triangle is

0%
$AB$
0%
$BC$
0%
$AC$
0%
Data insufficient

Explanation

Since sum of angles in a triangle

$= {180}^{o}$

$\angle A = {180}^{o} - {30}^{o} - {70}^{o} = {80}^{o}$
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.

In the figure given above,

$AP$

$\perp l$ i.e.,

$AP$ is the shortest line segment that can be drawn from

$A$ to the line

$l$ . If

$PR > PQ$ , which of the following is true.

0%
$AQ>AR$
0%
$AR>AQ$
0%
$AQ=2AR$
0%
$AQ=\sqrt{2}AR$

Explanation

Cut off

$PS$

$=$

$PQ$ from

$PR$ join

$AS$
In

$\Delta APQ$ and

$\Delta APS$

$\angle APQ = \angle APS$ (each

$90^{\circ}$ )

$AP$

$=$

$AP$ .... (common side)

$PQ$

$=$

$PS$

$\therefore \Delta APQ \cong \Delta APS$

$AP$

$=$

$AS$ and

$\angle 1 = \angle 3$ ....(1)
In

$\Delta ARS$ ,

$\angle 3 > \angle 2$ ....(2)
[exterior angle property]
from (1) and (2)

$\angle 1 > \angle 2$
from

$\Delta AQR$ ,

$\angle 1 > \angle 2$

$\therefore AR > QR$

$\therefore AR > AQ$

In the figure, given below,

$AB = AC$ .
Hence,

$\angle BAC = \angle ACD$ .

0%
True
0%
False

Explanation

Given,

$AB = AC$ .
Hence,

$\angle ABC = \angle ACB$ (Isosceles triangle property).
Also,

$OB$ bisects

$\angle B$ and

$OC$ bisects

$\angle C$ .
Thus,

$\angle OBC = \angle ABO = \dfrac{1}{2} (\angle B)$
and

$\angle OCB = \angle ACO = \dfrac{1}{2} (\angle C)$ .

Since,

$\angle B = \angle C$ ,

hence,

$\angle OBC = \angle ABO = \angle ACO = \angle OCB$ ...

$(i)$ .

Now, in

$\triangle OBC$ ,

$\angle OBC + \angle OCB + \angle BOC = 180^o$ (Angles of a triangle)..

$(I)$ .

Also,

$\angle ACB + \angle ACD = 180^o$ (Angles on a straight line) ...

$(II)$ .

Equating

$(I)$ and

$(II)$ ,

$\angle OBC + \angle OCB + \angle BOC = \angle ACB + \angle ACD$

$\angle OBC + \angle OCB + \angle BOC = \angle ACO + \angle OCB + \angle ACD$ .

Hence,

$\angle BOC = \angle ACD$ (From

$(i)$ ).

Therefore, the statement is false and option

$B$ is correct.

In triangle

$ABC$ , bisector of angle

$BAC$ meets opposite side

$BC$ at point

$D$ . If

$AB=AC$ , then

$BD=CD.$

0%
True
0%
False

Explanation

$\triangle ABD$ and

$\triangle ACD$ ,

$AB = AC$ [Given]

$AD = AD$ [Common]

$\angle BAD = \angle CAD$ [

$BD$ is bisector of

$\angle A$ ]

So, by

$SAS$ rule of congruence,

$\triangle ABD \cong \triangle ACD$

$\Rightarrow BD = CD$

State true or false:

$ED= EC$ , then

$AB\, +\, AD> BC$ .(Refer given fig.)

0%
True
0%
False

Explanation

In $\triangle ABD$ , we have sum of two sides $>$ third side
$=> AB + AD > BD$ OR

$=> AB + AD > BE + ED -- (1)$

Also, in $\triangle BCE$ , we have sum of two sides $>$ third side

$=> EB + EC > BC$ ---- (2)

Adding equations $1, 2$ we get

$=> AB + AD + EB +EC > BE + ED +BC$

$=> AB + AD + EC > ED + BC$

But $EC = ED$

Hence, $AB + AD > BC$

In the figure

$\angle ABC=135^{\circ},\angle ABX=90^{\circ} ,\angle XCD=55^{\circ}, \angle BCD=100^{\circ}$ ,then determine whether

$\angle XBC$ and

$\angle XCB$ are congruent to each other.

0%
$\angle XBC = 45^{\circ}, \angle XCB = 46{\circ}$
0%
$\angle XBC = 45^{\circ}, \angle XCB = 45^{\circ}$
0%
$\angle XBC = 45^{\circ}, \angle XCB = 60^{\circ}$
0%
$\angle XBC = 45^{\circ}, \angle XCB = 25^{\circ}$

Explanation

Given,

$\angle ABC={ 135 }^{ o },\quad ABX={ 90 }^{ o },\quad XCD={ 55 }^{ o },\quad BCD={ 100 }^{ o }$ .

Here,

$\angle ABC=\angle ABX+\angle XBC\Rightarrow XBC=\angle ABC-ABX={ 135 }^{ o }-{ 90 }^{ o }\Rightarrow XBC=45^o.$

Again

$\angle BCD=\angle XCB+\angle XCD\Rightarrow \angle XCB=\angle BCD-\angle XCD\Rightarrow \angle XCB={ 100 }^{ o }-{ 55 }^{ o }$

$=45^o$ .

Since,

$\angle XBC=\angle XCB=45^o$ , they are congruent.

Hence, option

$B$ is correct.

From the following figure, we can say:

$AD= CE$

State TRUE or FALSE

0%
True
0%
False

Explanation

$\triangle$ ABC,

$AB = AC$
thus,

$\angle ABC = \angle ACB = x$ (Opposite angles of equal sides are equal)
Now,

$\angle ABC + \angle CBE = 180$ (Angles on a straight line)

$x + \angle CBE = 180$

$\angle CBE = 180 - x$ (I)
Also,

$\angle ACB + \angle ACD = 180$ (Angles on a straight line)

$x + \angle ACD = 180$

$\angle ACD = 180 - x$ (II)
Hence, from I and II

$\angle ACD = \angle CBE$
In triangles ACD and CBE

$\angle ACD = \angle CBE$

$CD = CB$ (Given)

$AC = BE$ (Given)
Thus,

$\triangle ACD \cong \triangle ECB$
Hence,

$AD = CE$

In a triangle

$PQR,\:QR= PR$ and

$\angle P= 36^{0} .$ Which is the largest side of the triangle?

0%
$QR$
0%
$PR$
0%
$PQ$
0%
None of these

Explanation

Since

$QR = PR$ , the angles opposite to them will be equal as well,

$=> \angle P = \angle Q = {36}^{0}$
Since sum of angles in a triangle

$= {180}^{o}$

$\angle R= {180}^{o} - {36}^{o} - {36}^{o} = {108}^{o}$
Out of the three angles,

$\angle R$ is the largest.
This means, the side opposite to it, which is

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

In the given figure,

$AB= AC$ and

$\angle DBC= \angle ECB= 90^{\circ}$ , then

$AD= AE$ .

0%
True
0%
False

Explanation

In triangle ABC,

$AB = AC$ thus,

$\angle ABC = \angle ACB$ (I) (Opposite angles of equal sides of triangle are equal)
Now, in

$\triangle$ s DBC and ECB,

$BC = BC$ (Common)

$\angle DBC = \angle ECB = 90$ (Given)

$\angle EBC = \angle DCB$ (From I)
Hence,

$\triangle DBC \cong \triangle ECB$ (ASA Congruency)
or

$DC = BE$ (equal sides of triangle of opposite angle are equal)
Also, Given

$AB = AC$

$DC - AC = BE - AB$

$AD = AE$

In the following figure;

$AC= CD$ ,

$AD= BD$ and

$\angle C= 58^{\circ}$ . Find angle

$CAB$ .

0%
$91.5^o$
0%
$82^o$
0%
$88.5^o$
0%
none of the above

Explanation

$\triangle ACD$ ,

by isosceles triangle property, angles opposite to equal sides are equal.

Then, let $\angle CAD = \angle CDA = x$ .

We know, by angle sum property, the sum of angles

$= 180^o$ .

$\angle CAD + \angle CDA + \angle ACD = 180^o$

$\implies$

$x + x + \angle ACD = 180^o$

$\implies$

$2x + 58^o = 180^o$ ......(as

$\angle C=\angle ACD=58^o$ )

$\implies$

$2x = 122^o$

$\implies$

$x = 61^{\circ}$ .
Therefore,

$\angle CAD = \angle CDA = 61^{\circ}$ .

Now, in

$\triangle ADB$ ,

$AD = BD$ (Given)

$\angle DAB = \angle ABD = y$ ......(Isosceles triangle property).
Also,

$\angle CDA = \angle DBA + \angle DAB$ ......(Exterior angle property)

$\implies$

$61^o = y + y$

$\implies$

$2y = 61^o$

$\implies$

$y = 30.5^0$

Then,

$\angle CAB = \angle CAD + \angle DAB$

$\implies$

$\angle CAB = 61^o + 30.5^o$

$\implies$

$\angle CAB = 91.5^{\circ}$ .

Therefore, option

$A$ is correct.

$ABC$ and

$DBC$ are two isosceles triangle on the same side of

$BC$ . Then,

$\angle BDA= \angle CDA$ .

0%
True
0%
False

Explanation

Join AD.
In

$\triangle$ ABD and

$\triangle$ ACD,

$AB = AC$ (Given)

$BD = CD$ (Given)

$AD = AD$ (Common side)

$\therefore$

$\triangle$

$ABD \cong \triangle ACD$ (By SSS congruence rule)
Hence,

$\angle BDA = \angle CDA$ (Corresponding angles of congruent triangles)

In the given figure,

$AB= AC$ and

$\angle DBC= \angle ECB= 90^{\circ}$ ,then

$BD= CE$ .

0%
True
0%
False

Explanation

In triangle ABC,

$AB = AC$ thus,

$\angle ABC = \angle ACB$ (I) (Opposite angles of equal sides of triangle are equal)
Now, in

$\triangle$ s DBC and ECB,

$BC = BC$ (Common)

$\angle DBC = \angle ECB = 90$ (Given)

$\angle EBC = \angle DCB$ (From I)
Hence,

$\triangle DBC \cong \triangle ECB$ (ASA Congruency)
or

$BD = CE$ (equal sides of triangle of opposite angle are equal)

In a $\triangle$ $PQR, QR = PR$ and $\angle\,P\,=\,36^{\circ}.$ The largest side of the triangle is:

0%
$PR$
0%
$QR$
0%
$PQ$
0%
Incomplete data

Explanation

Since, $QR = PR$ , their opposite angles are also equal.
Thus $\angle P = \angle Q = {36}^{0}$
Since sum of angles in a triangle PQR $= {180}^{o}$
$\Rightarrow \angle R = 180 - 2(36) = {108}^{o}$

Thus, $\angle R$ is the largest angle of the triangle, the side opposite to it, which is $PQ$ will be the greatest side of the triangle.

By which congruency are the following pair of triangles congruent:

In $\Delta\,ABC$ and $\Delta \,DEF$ , $\angle\,B = \angle\,E = 90\,^{\circ}, AC = DF$ and $BC = EF.$

0%
RHS
0%
SAS
0%
AAA
0%
SSS

Explanation

$\triangle ABC$ and

$DEF$ ,

$\angle B = \angle E = 90^{\circ}$ .......... (Given)

$AC = DF$ ........ (Given)

$BC = EF$ ......... (Given)

Thus,

$\triangle ABC \cong \triangle DEF$ (RHS postulate)

CBSE Questions for Class 9 Maths Triangles Quiz 1 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Triangles Quiz 1 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

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