If $$a, b, c$$ are sides of a triangle, then

$$\left| {\sqrt a - \sqrt b } \right| > \sqrt c $$

MCQ Questions for Class 9 Maths Triangles Quiz 8

$a, b, c$ are sides of a triangle, then

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$\sqrt a + \sqrt b > \sqrt c$
0%
$\left| {\sqrt a - \sqrt b } \right| > \sqrt c$
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$\sqrt a + \sqrt b < \sqrt c$
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None of these

Explanation

Given: Sides of the triangle

$a, b$ and

$c$ .

Triangle inequality theorem: According to this theorem,

[The sum if lengths of any two sides of triangle is greater than the length of its third side.]

i.e.

$a + b > c$

$b + c > a$

$a + c > b$

For example,

$a = 3cm, b = 4 cm, c = 5 cm$ then

$3 + 4 > 5 ; \, 7 > 5$

$4 + 5 > 3 ; \, 9 > 5$

$3 + 5 > 4 ; \, 6 > 4$

$\sqrt{a} + \sqrt{b} > \sqrt{c}$

$\sqrt{3} + \sqrt{4} > \sqrt{5}$

$1.73 + 2 > 2.24$

$3.73 > 2.24$

It means the correct option is

$(A)$ . i.e.

$\sqrt{a} + \sqrt{b} > \sqrt{c}$

2118781_1437961_ans_082f1d6ce05d4c199234ebff4a746cca.png

Two sides of a triangle are 7 cm and 25 cm, then the number of all possible integral values of third side of the triangle is

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11
0%
14
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12
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13

Which of the following pairs of triangle are congruent ?

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

Take any pointy

$O$ in the interior of a triangle

$PQR$ . Is

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$OP+OQ>PQ?$
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$OQ+OR>QR?$
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$OR+OP>RP?$
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All of the above.

If the sides of triangle are 4 cm and 6 cm then the third side cannot be ... cm.

0%
8
0%
10
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5
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8.5

$\Delta ABC\cong \Delta DEF$ write the part of

$\Delta ABC$ that correspond to

0%
DE
0%
$\angle E$
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DF
0%
EF
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$\angle F$

AM is a median of a triangle ABC. Is

$AB+BC+CA>2 AM$ ?
(Consider the sides of triangles

$\triangle ABM$ and

$\triangle AMC$ .)

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True
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False

$\Delta ABC$

$AB<AC$ Then...... holds good.

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$\angle A<\angle B$
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$\angle B<\angle C$
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$\angle C<\angle A$
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$\angle C<\angle B$

In a triangle ABC, AB=AC, BA is extended upto D, in such a manner that AC=AD is a circular measure of <BCD:

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$\dfrac{\pi}{6}$
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$\dfrac{\pi}{3}$
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$\dfrac{2 \pi}{3}$
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$\dfrac{\pi}{2}$

if ABC and DEF are congruent triangles such that

$\angle A={ 47 }^{ \circ }\quad and\quad \angle E={ 83 }^{ \circ },\quad then\quad \angle C=$

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${ 100 }^{ \circ }$
0%
${ 50 }^{ \circ }$
0%
${ 90 }^{ \circ }$
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NONE OF THESE

$\triangle ABC$ ,

$P$ is a point on the side

$BC$ then which of the following is correct.

0%
$AP<AB+BP$
0%
$AP<AC+PC$
0%
$AP<AB+BC$
0%
$AP> AB+BC$

The sum of any two sides of a triangle is always

0%
equal to the third side
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less than
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grater than or equal to the 3rd side
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grater than

Given two right angled triangle ABC and PQR, such that

$<A=20^o, <Q=20^o$ and

$AC=QP$ . Write the correspondence if triangles are congruent.

0%
$\Delta ABC\cong \Delta PQR$
0%
$\Delta ABC\cong \Delta PRO$
0%
$\Delta ABC\cong \Delta ROP$
0%
$\Delta ABC\cong \Delta QRP$

If one of the two equal sides of a triangle is 5 cm long. Then what can be the measure of the third side?

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8 cm
0%
14 cm
0%
12 cm
0%
16 cm

The sides of a triangle are three consecutive natural numbers and its largest the smallest one then the sides of triangle are

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3, 4, 5
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5, 6, 7
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4, 5, 6
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6, 7, 8

$\triangle APQ$ is such that perpendicular bisector of side

$PQ$ bisect it at

$M$ and passes from vertex

$A$ . Then by using which of the following congruence rule it can be proved that

$\triangle \mathrm{AMP} \cong \triangle \mathrm{AMQ}$ .

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

Given that

$\Delta ABC = \Delta FDE$ and

$AB = 5 cm, \angle B = 40^{0}$ and

$\angle A = 80^{0}$ then which of the following relation is true?

0%
$DF=5cm,\angle F=60^{0}$
0%
$DF=5cm,\angle E=60^{0}$
0%
$DE=5cm,\angle E=60^{0}$
0%
$DE=5cm,\angle D=40^{0}$

$\triangle PQR$ , if

$\angle R\displaystyle>\angle Q$ , then

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

Explanation

$\triangle PQR$ ,

$\angle R > \angle Q$
Hence,

$PQ > PR$ (Sides opposite larger angles is large)

option B is correct

In a triangle, the difference of any two sides is ____ than the third side.

0%
Smaller
0%
Equal
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Greater
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None of these

ABC is an isosceles triangle with AB

$=$ AC and D is a point on BC such that

$AD \perp BC$ (Fig. 7.13). To prove that

$\angle BAD = \angle CAD,$ a student proceeded as follows:

$\Delta ABD$ and

$\Delta ACD,$
AB

$=$ AC (Given)

$\angle B = \angle C$ (because AB

$=$ AC)
and

$\angle ADB = \angle ADC$
Therefore,

$\Delta ABD \cong \Delta ACD (AAS)$
So,

$\angle BAD = \angle CAD (CPCT)$
What is the defect in the above arguments?

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It is defective to use $\angle ABD = \angle ACD$ for proving this result.
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It is defective to use $\angle ADB = \angle ADC$ for proving this result.
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It is defective to use $\angle BAD = \angle DCA$ for proving this result.
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Cannot be determined

Explanation

$\bigtriangleup ABD$ and

$\bigtriangleup ACD$

AB=AC (given )

Then

$\angle ABD=\angle ACD$ ( because AB=AC )

and

$\angle ADB=\angle ADC=90$ ( because AD⊥BC )

$\therefore \bigtriangleup ABD=\bigtriangleup ACD$

$\angle BAD=\angle CAD$

It is defective to use

$\angle ABD=\angle ACD$ for proving this result

Given

$\Delta OAP \cong \Delta OBP$ in figure, the criteria by which the triangles are congruent is:

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

Explanation

$\triangle OAP$ and

$\triangle OBP$

$OA=OB$ (given)

$\angle AOP=\angle BOP$ (given)

$OP=OP$ (common side)

$\therefore \triangle OAP\cong \triangle OBP$ by

$SAS$ congruent rule as two corresponding sides and the included angles of the triangles are equal.

$\Delta ABC$ , if

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

$\angle B = 60^{\circ}$ , then the greatest side is :

0%
AB
0%
BC
0%
AC
0%
Cannot say

Explanation

Using angle sum property of triangle

$\angle A+\angle B+\angle C={ 180 }^{ \circ }\\ \Rightarrow { 50 }^{ \circ }+{ 60 }^{ \circ }+\angle C={ 180 }^{ \circ }\\ \Rightarrow \angle C={ 180 }^{ \circ }-{ 110 }^{ \circ }={ 70 }^{ \circ }$

Side opposite to the largest angle is the longest side.

Here

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

$AB$

$AB$ is the longest side.

The construction of a triangle ABC, given that BC = 3 cm is possible when difference of AB and AC is equal to :

0%
3.2 cm
0%
3.1 cm
0%
3 cm
0%
2.8 cm

Explanation

Let the length of

$AB$ be

$x$ and

$AC$ be

$y$

A triangle can be formed if the sum of any two sides is greater then the third

$\Rightarrow BC+AC>AB\\ \Rightarrow 3+AC>AB\\ \Rightarrow 3>AB-AC\\ \Rightarrow AB-AC<3$

So only option

$D$ is correct.

$\Delta PQR,$ if

$\angle R > \angle Q,$ then:

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

Explanation

We know that the side opposite to the greater angle is greater.

Given:

$\angle R> \angle Q$

The side opposite to

$\angle R=PQ$ and the side opposite to

$\angle Q=PR$ .

Thus,

$PQ>PR$ .

Hence, option

$B$ is correct.

$\Delta ABC$ and

$\Delta DEF$ , AB = DF and

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

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

Explanation

For triangle to be congruent

$SAS$ (Side angle Side) axiom we need to have two equal sides and the included angle same.

$\Delta ABC$ and

$\Delta DFE$

We have

$AB=DF$

and

$\angle A=\angle D$

But

$\angle A$ is formed by the two sides AB and AC of

$\Delta ABC$

But

$\angle D$ is formed by the two sides DE and DF of

$\Delta DFE$

So we need

$AC=DE$ for the two triangles to be congruent.

In triangles 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

For triangle to be congruent by

$SAS$ axiom two of the sides and the included angle of the triangles must be equal.

Given

$AB=DF$ and

$\angle A=\angle D$

No for

$\triangle ABC\cong \triangle DFE$ by

$SAS$ axiom we need

$AC=DE$

So option

$B$ is correct.

$\Delta ABC, \angle B = 30^{\circ}, \angle C = 80^{\circ}$ and

$\angle A = 70^{\circ}$ then,

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

Explanation

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

$\angle A$ and side opposite to it is

$BC$ .

$\therefore BC$ is second longest.

Then comes

$\angle B$ and side opposite to it is

$AC$

So it is the smallest side.

So the decreasing order of sides is

$AB>BC>AC$

$\Delta ABC \cong \Delta DEF$ by SSS congruence rule then :

0%
$AB=EF, BC=FD, CA=DE$
0%
$AB=FD, BC=DE, CA=EF$
0%
$AB=DE, BC=EF, CA=FD$
0%
$AB=DE, BC=EF, \angle C=\angle F$

Explanation

If triangles are congruent by

$SSS$ congruence rule then their corresponding sides are equal.

Here

$\triangle ABC\cong \triangle DEF$

$\therefore AB=DE,BC=EF,AC=DF$ or

$(CA=FD)$

So option

$C$ is correct.

In the given figure , which of the following statement is true ?

0%
$\angle B=\angle C$
0%
$\angle B$ is the greatest angle in triangle
0%
$\angle B$ is the smallest angle in triangle
0%
$\angle A$ is the smallest angle in triangle

Explanation

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

Here

$BC$ is the longest side and angle opposite to it is

$\angle A$ so

$\angle A$ is the largest angle.

Smallest side is

$AC$ and angle opposite to it is

$\angle B$ so

$\angle B$ is the smallest angle.

So option

$C$ is correct.

$\Delta ABC \cong \Delta DEF$ by SSS congruence rule then

0%
$AB=EF,BC=FD,CA=DE$
0%
$AB=FD,BC=DE,CA=EF$
0%
$AB=DE,BC=EF,CA=FD$
0%
$AB=DE,BC=EF,\angle C = \angle F$

Explanation

If triangles are congruent by

$SSS$ then their corresponding sides are equal , therefore

$AB=DE$

$BC=EF$

$AC=DF$ or

$CA=FD$

So option

$C$ is correct.

CBSE Questions for Class 9 Maths Triangles Quiz 8 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Triangles Quiz 8 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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