MCQ Questions for Class 9 Maths Triangles Quiz 9

$\Delta ABC$ and

$\Delta DEF$ ,

$AB=FD$ ,

$\angle A= \angle D$ . The two triangles will be congruent by SAS axiom if :

0%
$BC=DE$
0%
$AC=EF$
0%
$BC=EF$
0%
$AC=DE$

Explanation

For the triangles to be congruent by

$SAS$ axiom we need two sides and included angle of given triangles to be equal.

Given:

$AB=FD$ and

$\angle A=\angle D$

Now from the figures, we can say that we need the second pair of sides which is

$AC$ and

$DE$ according to the given definition.

So for given triangles to be congruent by

$SAS$ criteria

we need

$AC=DE$

Option

$D$ is correct.

$\Delta ABC$ , if

$\angle A = 35^{\circ}$ and

$\angle B = 65^{\circ}$ , then the longest side of the triangle is :

0%
AC
0%
AB
0%
BC
0%
None of these

Explanation

Using angle sum property of triangle

$\angle A+\angle B+\angle C={ 180 }^{ \circ }\\ \Rightarrow { 35 }^{ \circ }+{ 65 }^{ \circ }+\angle C={ 180 }^{ \circ }\\ \Rightarrow \angle C={ 180 }^{ \circ }-{ 100 }^{ \circ }={ 80 }^{ \circ }$

Side opposite to largest angle is the longest side of any triangle.

Here

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

$AB$

$AB$ is the longest side.

In the given 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)

$\angle AOD=\angle BOC$ (vertically opposite angle)

$\triangle AOD\cong \triangle BOC$ by

$SAS$ congruence rule.

$\triangle ABC$ , if

$AB >BC$ then:

0%
$\angle C < \angle A$
0%
$\angle C = \angle A$
0%
$\angle C > \angle A$
0%
$\angle A = \angle B$

Explanation

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

$AB$ is

$\angle C$ and angle opposite to

$BC$ is

$A$

Here,

$AB>AC$

$\Rightarrow \angle C>\angle A$

So, option

$C$ is correct.

$\Delta ABC, AB = 76$ cm,

$BC=69$ cm and

$CA=61$ cm

the greatest angle ?

0%
$\angle C$
0%
$\angle A$
0%
$\angle B$
0%
All are equal

Explanation

In a triangle, the greatest angle is opposite the largest side. Here,

$AB$ is the largest side, hence,

$\angle C$ is the greatest.

In the given figure, if

$AB=DC$ ,

$\angle ABD= \angle CDB$ , which congruence rule would you apply to prove

$\Delta ABD$

$\cong \Delta CDB$ ?

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

Explanation

$\triangle ABD$ and

$\triangle CDB$

$AB=DC$ (given)

$\angle ABD=\angle CDB$ (given)

$BD=DB$ (common side)

$\therefore \triangle ABD\cong \triangle CBD$ by

$SAS$ congruence rule.

Hence option (A) is correct

Two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of the third side of the triangle cannot be

0%
$\displaystyle3.6\,cm$
0%
$\displaystyle4.1\, cm$
0%
$\displaystyle3.8\, cm$
0%
$\displaystyle3.4\, cm$

Explanation

In a triangle, the difference between two sides should be less than the third side.
Hence,option D is correct

$3.4\ cm$

By which congruency are the following pairs of triangles congruent:

In $\Delta\,ABC$ and $\Delta PQR$ , $AB=PQ$ , $BC=QR$ and $AC=PR$

0%
$ASA$
0%
$SSS$
0%
$AAA$
0%
$ASS$

Explanation

In triangles

$ABC$ and

$PQR$ , we have

$AB = PQ$ ....(Given)

$AC = PR$ ....(Given)

$BC = QR$ ....(Given)
Thus,

$\triangle ABC \cong \triangle PQR$ ..... (SSS Postulate)

For any triangle

$ABC$ , the true statement is

0%
${ AC }^{ 2 }={ AB }^{ 2 }+{ BC }^{ 2 }$
0%
$AC=AB+BC$
0%
$AC>AB+BC$
0%
$AC<\,AB+BC$

Explanation

For any

$\triangle ABC$ , the sum of two sides must be greater than the third side.

Hence,

$AB + BC > AC$ .

State true/false:

If in $\Delta\, ABC$ and $\Delta \,DEF$ , AB = DE, BC = DE, BC = EF and $\angle\,B\,=\, \angle\,E$ , then both the triangles are congruent.

0%
True
0%
False

Explanation

$\triangle ABC$ and

$DEF$ ,

$AB = DE$ .......... (Given)

$\angle B = \angle E$ ......... (Given)

$BC = EF$ ............. (given)
Thus,

$\triangle ABC \cong \triangle DEF$ (SAS rule)

In a triangle

$ABC$ and

$DEF$ ,

$AB=FD$ and

$\angle A=\angle D$ . The two triangles will be congruent by SAS axiom if:

0%
$BC=EF$
0%
$AC=DE$
0%
$AC=EF$
0%
$BC=DE$

Explanation

$\triangle ABC$ and

$\triangle DEF$ ,

$\angle A = \angle D$ ....(Given)

$AB = FD$ ....(Given)
and

$\triangle ABC \cong \triangle DFE$ ....(SAS rule)
It is possible, if corresponding sides are equal,

$AC = ED$

State true or false:

If two sides and an angle of a triangle are respectively equal to two sides and an angle of another triangle,then the two triangles are always congruent.

0%
True
0%
False

Explanation

If two sides and one angle of one triangle is equal to the two sides and one angle of another triangle, then the two triangles may or may not be congruent.
For the two triangles to be congruent, the angle must be contained by the two sides. Thus, by

$SAS$ rule the two triangles will be congruent.

So, the triangles are not always congruent.

Hence,

$False$

Angles opposite to the equal sides of a triangle are equal.

0%
True
0%
False
0%
Incomplete information
0%
None

Explanation

Consider

$\triangle ABC$ ,

$\angle B = \angle C$
Construction: Draw a median AD on BC.

Hence,

$BD = DC$
Now, In

$\triangle ABD$ and

$\triangle ACD$

$\angle ADB = \angle ADC$ .... Common angle

$AD = AD$ ..... (Common)

$BD = DC$
Thus,

$\triangle ABD \cong \triangle ACD$ .... (SAS rule)
Thus,

$\angle B = \angle C$ .... (By CPCT)

In triangles

$ABC$ and

$PQR$ ,

$AB=AC$ ,

$\angle C=\angle P$ and

$\angle B=\angle Q$ . Then the two triangles are :

0%
Isosceles but not congruent
0%
Isosceles and congruent
0%
Congruent but not isosceles
0%
Neither congruent nor isosceles

Explanation

$\triangle ABC$ and

$\triangle PQR$

$\angle C = \angle P$ (Given)

$\angle B = \angle Q$ (Given)

$\angle A = \angle R$ (Third angle of the triangle)

Thus,

$\triangle ABC \sim \triangle PQR$
Also, given,

$AB = AC$
Thus,

$\angle B = \angle C$ (Isosceles triangle Property)

But,

$\angle B = \angle Q$ and

$\angle C = \angle P$
Hence,

$\angle Q = \angle P$
or

$PR = QR$

Thus, both the triangles are Isosceles but not congruent.

If any two sides of a triangle are unequal, then the larger side has the

$.........$ angle opposite to it.

0%
Greater
0%
Lesser
0%
Equal
0%
May be greater or lesser

Explanation

Given:

$\triangle ABC$ ,

$AB > AC$

To Prove:

$\angle ACB > \angle ABC$

Construction: Make a point D on

$AB$ such that

$AD = AC$ and join

$DC$

$\triangle ACD$ ,

$AD = AC$

$\angle ACD = \angle ADC$ (Angles opposite to equal sides)

But

$\angle ADC > \angle ABC$ (Exterior angle of the triangle is greater than the opposite interior angle)

Again,

$\angle ACB > \angle ACD$ (D lies in interior of

$\angle ACB$ )
Thus,

$\angle ACB > \angle ABC$

If in two triangles

$ABC$ and

$PQR$ ,

$AB=\, QR$ ,

$\angle A=\angle Q$ and

$\angle B=\angle R$ , then

$\triangle ABC\cong$

$\triangle$

0%
$QRP$
0%
$PQR$
0%
$PRQ$
0%
None of the above.

Explanation

Given:

$AB=\, QR$ ,

$\angle A=\angle Q$ and

$\angle B=\angle R$

Now, In

$\triangle ABC$ and

$\triangle PQR$

$AB = QR$

$\angle A = \angle Q$

$\angle B = \angle R$
Thus, by

$ASA$ congruence rule,

$\triangle ABC \cong \triangle QRP$

Hence,

$Op-A$ is correct.

If in two triangles

$ABC$ and

$DEF$ ,

$AB=\,DF$ ,

$BC=\,DE$ and

$\angle B=\angle D$ , then

$\triangle ABC\cong$

$\triangle$ ____.

0%
$FDE$
0%
$DEF$
0%
$EDF$
0%
$EFD$

Explanation

Given:

In

$\triangle ABC$ and

$\triangle DEF,$

$AB = DF$

$\angle B = \angle D$

$BC = DE$

Thus,

$\triangle ABC \cong \triangle FDE$ (SAS rule)

If the two angles of a triangle are unequal, then the smaller angle has the

$.......$ side opposite to it.

0%
Smaller
0%
Larger
0%
May be smaller may be larger
0%
The side will be equal to one of the opposite sides

Explanation

Let, In

$\triangle ABC$ ,

$\angle ACB > \angle ABC$
To Prove:

$AB > AC$
Construction: Make a point

$D$ on

$AB$ , such that

$\angle ACD = \angle ADC$ and join

$DC$
Now, in

$\triangle ACD$ ,

$\angle ACD = \angle ADC$

$\therefore AD = AC$ (Sides opposite to equal angles)

But

$AB = AD + DB$
thus,

$AB > AD$
Hence,

$AB > AC$ [as

$AD = AC$ ]

Thus, sides opposite smaller angles in a triangle are smaller.

If the two sides and the ____ angle of one triangle are respectively equal to two sides and the included angle of the other triangle, then the triangles are congruent.

0%
Included
0%
Excluded
0%
Adjacent
0%
Any

Explanation

Two sides and an included angle will fulfill the condition of the

$SAS$ rule of congruence so, the two triangles will be equal. Hence,

Two triangles are congruent if the two sides and the

$\underline{\text{included}}$ angle are equal to the two sides and the included angle of another triangle.

Which of the following pairs of triangles are congruent?

0%
$\triangle ABC$ and $\triangle DEF$ in which : $BC=\,EF$ , $AC=\,DF$ and $\angle C=\angle F$
0%
$\triangle ABC$ and $\triangle DEF$ in which : $AB=\,PQ$ , $BC=\,QR$ and $\angle C=\angle R$
0%
$\triangle ABC$ and $\triangle LMN$ in which : $\angle A=\angle L={ 90 }^{ 0 }$ , $AB=\,LM$ , $\angle C={ 40 }^{ 0 }$ and $\angle M={ 50 }^{ 0 }$
0%
$\triangle ABC$ and $\triangle DEF$ in which : and $\angle B=\angle E={ 50 }^{ 0 }$ and $AC=\,DF$

Explanation

From Option A,
Given:

$\triangle ABC$ and

$\triangle DEF$

$BC = EF$

$AC = DF$

$\angle C = \angle F$
Thus,

$\triangle ABC \cong \triangle DEF$ (SAS rule)

From Option C,
In

$\triangle ABC$ ,
Sum of angles = 180

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

$90 + \angle B + 40= 180$

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

Given,

$\triangle ABC$ and

$\triangle LMN$ ,

$\angle A = \angle L = 90^{\circ}$

$AB = LM$

$\angle B = \angle M = 50^{\circ}$
Thus,

$\triangle ABC \cong \triangle LMN$ (ASA rule)

If ____sides of a triangle are respectively equal to the ____ sides of the other triangle, then the triangles are congruent.

0%
Three
0%
Two
0%
One
0%
None

In a right triangle, the hypotenuse is the

$.......$ side

0%
Largest
0%
Shortest
0%
Can be largest or shortest
0%
None of these

Explanation

In a right triangle, using Pythagoras theorem,

$Hypotenuse^2 = perpendicular^2 + base^2$
Thus, Hypotenuse is the largest side.

In the

$\triangle ABC$ ,we have

$\angle A>\angle B>\angle C$ , then determine the shortest and the longest side of the triangle.

0%
Shortest side is $AB$ and the longest side is $BC$
0%
Shortest side is $BC$ and the longest side is $AB$
0%
Shortest side is $AB$ and the longest side is $AC$
0%
Shortest side is $AC$ and the longest side is $BC$

Explanation

Given,

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

Hence,

$BC > AC > AB$ (Side opposite largest angle is the longest and vice versa)

Thus,

$BC$ is the largest and

$AB$ is the shortest.

In the figure,

$\displaystyle AB=CD$ and

$\displaystyle \angle A={ 90 }^{ o }=\angle D$ . Then

0%
$\displaystyle \Delta ABC\cong \Delta DBC$ by SAS postulate
0%
$\displaystyle \Delta ABC\cong \Delta DCB$ by RHS postulate
0%
$\displaystyle \Delta ABC\cong \Delta DBC$ by AAS postulate
0%
$\displaystyle \Delta ABC\cong \Delta DCB$ by SSS postulate

Explanation

$\displaystyle \Delta ABC$ and

$\displaystyle \Delta DBC$

$\displaystyle \angle A={ 90 }^{ o }=\angle D$ given

$\displaystyle BC=BC$ common side

$\displaystyle AB=CD$ given

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

Two sides of a triangle are

$7$ and

$10$ units. Which of the following length can be the length of the third side?

0%
$19$ cm
0%
$17$ cm
0%
$13$ cm
0%
$3$ cm

Explanation

In a triangle the sum of any two sides is always greater than the third side.

Consider each option :

$7+10 \ngtr 19$

$7+10 \ngtr 17$

$7+10> 13, 13+10>7, 7+13>10$

It satisfies the condition.

$7+3 \ngtr 10$

Hence : Option C is the correct answer.

$\Delta ABC$ , AB = AC and AD is perpendicular to BC. State the property by which

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

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

Explanation

$\triangle ADB$ and

$\triangle ADC$

$AB=AC$ (given)

$AD=AD$ (common side)

$\angle ADB=\angle ADC=90^{\circ}$ (As

$AD\perp BC)$

$\therefore \triangle ADB\cong \triangle ADC$ by

$RHS$ property.

$\triangle PQR$ ,

$\angle P={ 50 }^{ 0 }$ and

$\angle R={ 70 }^{ 0 }$ . Name
i) the shortest side.
ii)the longest side of the triangle

0%
i) $PQ$ ,

(ii) $QR$
0%
i) $PR$ ,

(ii) $PQ$
0%
i) $QR$ ,

(ii) $PQ$
0%
i) $PQ$ ,

(ii) $PR$

Explanation

$\triangle PQR$ ,

$\angle P = 50^{\circ}$ ,

$\angle R = 70^{\circ}$
Sum of angles of the triangle = 180

$\angle P + \angle Q + \angle R = 180$

$50 + \angle Q + 70 = 180$

$\angle Q = 60^{\circ}$
Side opposite smaller angle is shorter and side opposite to larger angle is larger.
Hence,

$QR$ is the smallest side,

$PQ$ is the largest side.

In a

$\Delta\, ABC$ , if

$\angle B$ is an obtuse angle, then the longest side is:

0%
$AB$
0%
$BC$
0%
$AC$
0%
none

Explanation

Side opposite to the largest angle is the longest side.

It is given that,

$\angle B$ is obtuse , so the other two angles must be acute angles.

So,

$\angle B$ is the largest angle and the side opposite to it is

$AC$

Hence,

$AC$ is the longest side.

2113561_271584_ans_c4a0eb23b18040378171e8a9d9834439.png

In the following fig. if

$AB=AC$ and

$BD= DC$ then

${\angle ADC}$ =

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

Explanation

$\triangle ABD$ and

$\triangle ACD$

$AB=AC$ (Given)

$BD=DC$ (Given)

$AD=AD$ (Common side)

$\therefore \triangle ABD\cong ACD$ by

$SSS$ criteria.

Now corresponding angles of congruent triangles are equal

$\Rightarrow \angle ADB=\angle ADC$

Now

$\angle ADB+\angle ADC=180^{\circ}$ (Adjacent angles on staraight line)

$\Rightarrow \angle ADC+\angle ADC={ 180 }^{ \circ }\\ \Rightarrow 2\angle ADC={ 180 }^{ \circ }\\ \Rightarrow \angle ADC={ 90 }^{ \circ }\\$

Can

$6$ cm,

$5$ cm and

$3$ cm form a triangle?

0%
Yes
0%
No
0%
Sometimes
0%
None of these

Explanation

Given,

$6$ cm,

$5$ cm and

$3$ cm are the sides of triangle.

Lets check if the triangle is possible or not.

$(6 + 5=11)$

$11 > 3$ Inequality property

$(5 + 3=8)$

$8 > 6$

$(3 + 6=9)$

$9 > 5$
They can form

$\Delta$

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

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Practice Class 9 Maths Quiz Questions and Answers

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

0:0:2

Practice Class 9 Maths Quiz Questions and Answers

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