MCQ Questions for Class 9 Maths Triangles Quiz 12

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

0%
True
0%
False

If two triangles are congruent, then the corresponding angles are equal.

0%
True
0%
False

If two angles and a side of a triangle are equal to two angles and a side of another triangle, then the triangles are congruent.

0%
True
0%
False

If the hypotenuse of one right triangle is equal to the hypotenuse of another right triangle, then the triangles are congruent.

0%
True
0%
False

In Figure 6.28, two triangles are congruent by RHS.

0%
True
0%
False

$\Delta ABC, \angle A=100^0, \angle B=30^0$ and

$\angle C=50^0$ then

0%
$AB > AC$
0%
$AB < AC$
0%
$BC < AC$
0%
none of these

Explanation

We know that the side opposite to largest angle is longest side of triangle .

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

$BC$

$BC$ is the longest side.

Then comes

$\angle C$ and side opposite to it is

$AB$ .

$AB$ is second longest side.

Then comes

$\angle B$ and side opposite to it is

$AC$ .

So it is the smallest side.

So the decreasing order of side is

$BC>AB>AC$

So option

$A$ is correct.

If in two triangles

$\Delta ABC$ and

$\Delta PQR$ ,

$AB = QR, BC = PR$ and

$CA = PQ,$ then :

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$\Delta ABC\cong \Delta PQR$
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$\Delta CBA\cong \Delta PRQ$
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$\Delta BAC\cong \Delta RPQ$
0%
$\Delta PQR\cong \Delta BCA$

Explanation

$\triangle ABC$ and

$\triangle PQR$ , we are given

$AB=QR$ ,

$BC=PR$

and

$CA=PQ$

$\therefore \Delta CBA\cong \Delta PRQ$ .... SSS test

$\Delta PQR$ ,

$\angle P =$

$70^{\circ}$ and

$\angle R = 30^{\circ}.$ Which side of this triangle is the longest?

0%
$PR$
0%
$QR$
0%
$PQ$
0%
All sides are equal.

Explanation

Given,

$\triangle PQR$ ,

$\angle P = 70^{\circ}$ ,

$\angle R = 30^{\circ}$
Sum of angles of triangle

$= 180^{\circ}$

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

$\implies 70^{\circ} + 30^{\circ} + \angle Q = 180^{\circ}$

$\implies \angle Q = 80^{\circ}$
Since

$\angle Q$ is the largest

$\implies PR$ will be the longest side this is because when the two sides of a triangle are unequal, the angle opposite to the longer side is larger.

$D$ and

$E$ are the mid-points of the sides

$AB$ and

$AC$ respectively of

$\triangle ABC$ .

$DE$ is produced to

$F$ . To prove that

$CF$ is equal and parallel to

$DA$ , we need an additional information which is:

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$\triangle DAE$ = $\triangle EFC$
0%
$AE = EF$
0%
$DE = EF$
0%
$\triangle ADE = \triangle ECF$

Explanation

Given

$CF$ is equal and parallel to

$DA$

$D$ is the midpoint of

$AB$

So,

$AD$ =

$DB$ =

$CF$ =

$\cfrac { c }{ 2 }$

Similarly

$AE$ =

$EC$ =

$\cfrac { b }{ 2 }$

Let

$\angle AED=\alpha$

$ADE=\beta \quad$ and

$\quad DAE=\gamma$

Since

$AD$ is parallel to

$FC$

$\angle FEC = \alpha \angle EFC=\beta$ and

$\angle ECF=\gamma$

Applying

$S.A.S$ Congruency for triangle

$ADE$ and

$CFE$ ,

$AD= CF$ ,

$AE= AC$

$\angle DAE=\angle ECF$

${ \Delta }^{ k } DAE$ and

${ \Delta }^{ k } FCE$ are congruent

$DE = EF$

It is given that

$DE= EF$ ,then by

$S.S.S.$ congruency all angles will be equal and

$CF$ and parallel to

$DA$

$\therefore$ Additionsl information needed is

$DE= EF$ .

$AB$ and

$CD$ are the smallest and largest sides of a quadrilateral

$ABCD$ .

Out of

$\angle B$ and

$\angle D$ , decide which is greater.

0%
$\angle B$ will be greater.
0%
$\angle D$ will be greater.
0%
Both the angles are equal.
0%
Cannot be determined.

Explanation

In the fig.

${ABCD}$ is a quadrilateral with

${AB}$ as the smallest side, and

${CD}$ as the largest side.

We join the diagonal

${BD}$ .

We label the angles as shown in the figure as

$\theta 1$ ,

$\theta 2$ ,

$\theta 3$ ,

$\theta 4$ as shown in figure.

Now in

$\Delta {ABD}$ ,

${AB}<{AD}$

$\implies \theta 2 < \theta 1 \rightarrow {(1)}$

Similarly in

$\Delta {BCD}$ ,

${BC}<{CD}$

$\implies \theta 4 < \theta 3 \rightarrow {(2)}$

Adding

${(1)} and {(2)}$ ,

$\theta 2 + \theta 4 < \theta 1 + \theta 3$

which is nothing but,

$\angle{D} < \angle{B}$

Thus

$\angle(B)$ will be greater.

In the given figure,

$B < A$ and

$D > C$ , then:

0%
$AD > BC$
0%
$AD = BC$
0%
$AD < BC$
0%
$AD = 2BC$

Explanation

$\triangle AOB$ ,

$\angle {A}>\angle {B}$
Hence,

$OB>OA$ ..... (i)
In

$\triangle ODC$ ,

$\angle {D}>\angle {C}$
Hence,

$OC>OD$ .... (ii)
Adding inequalities (i) and (ii), we get

$OB+OC>OA+OD$

$BC>AD$

$\triangle ABC$ and

$\triangle DEF$ ,

$AB = FD$ 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

$SAS=$ If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent.

Here,

$AB = FD$ ,

$\angle A = \angle D$

So,

$AC = DE$

Hence, option B is correct.

CBSE Questions for Class 9 Maths Triangles Quiz 12 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Triangles Quiz 12 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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