MCQ Questions for Class 9 Maths Triangles Quiz 11

Which of the following is congruent to the above figure?

0%
0%
0%
0%
None of these

Explanation

Two figures are said to be congruent if they have exactly same shape and size or one can exactly overlap the other.

Here, if we rotate the figure in option

$A$ , and place it over each other, we get the same figure given, i.e. they will overlap.

Therefore, option

$A$ is correct.

Which of the following sides are equal if

$\angle RPQ$ and

$\angle RQP$ are similar?

0%
$RP=RQ$
0%
$RP=PQ$
0%
$RQ=PQ$
0%
None of these

Explanation

Here, since two angles of the triangle are equal, the

$\triangle RPQ$ is an isosceles triangle.

Then, from the converse of isosceles angle property,

sides opposite to equal angles of a triangle are equal.

Here,

$RP$ is the side opposite to

$\angle RQP$

and

$RQ$ is the side opposite to

$\angle RPQ$ .

Thus, if

$\angle RPQ=\angle RQP$ , then

$RP=RQ$ .

Therefore, option

$A$ is correct.

Converse of isosceles triangle theorem states that ...........

0%
The two angles that lie opposite to the two congruent sides of the isosceles triangle would also be congruent
0%
If two angles of a triangle are congruent then the sides opposite to these angles are congruent.
0%
If two angles of a triangle are congruent then the sides opposite to these angles are notcongruent.
0%
None of these

Explanation

Converse of isosceles triangle theorm states that if two angles of a triangle are congruent then the sides opposite to these angles are congruent.
If

$\angle A =\angle C$ then

$AB=AC$ .

Therefore, option

$B$ is correct.

State if True or False:
If two angles of a triangle are congruent, then the sides opposite to these angles are congruent.

0%
True
0%
False

Explanation

If two angles of a triangle are congruent, then these angles are equal.

Then, by the converse of isosceles triangle property, the sides opposite to these angles are also equal.

Since the two sides of the triangle are equal, then these sides are congruent.

Hence, if two angles of a triangle are congruent then the sides opposite to these angles are congruent.

Therefore, the statement is true, and option

$A$ is correct.

A triangle with two congruent sides and congruent base angles is termed as?

0%
Equilateral triangle
0%
Isosceles triangle
0%
Scalene triangle
0%
None of these

Explanation

A triangle with two congruent sides and congruent base angles is termed as an isosceles triangle.

If two sides of a triangle are congruent, then by isosceles triangle property, the angles opposite to these sides are congruent.

That is, base angles of the triangle are congruent.

Then, such a triangle is known as Isosceles triangle.

Therefore, option

$B$ is correct.

Which of the following is congruent to the above figure?

0%
0%
0%
0%
None of these

Explanation

Two figures are said to be congruent if they have exactly same shape and size or one can exactly overlap the other.

Here, if we rotate the figure in option

$B$ , and place it over each other, we get the same figure given, i.e. they will overlap.

Therefore, option

$B$ is correct.

Consider isosceles triangle

$ABC$ , in which

$AB=AC$ and

$\angle ABC =50^o$ . Find the

$\angle BAC$ .

0%
$50^o$
0%
$65^o$
0%
$80^o$
0%
$130^o$

Explanation

Here,

$\triangle ABC$ is an isosceles triangle.

Given,

$AB=AC$ and

$\angle ABC=50^o$ .

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

Thus,

$\angle ABC=\angle ACB=50^o$ .

Therefore, by angle sum property of triangle,

$\angle BAC+\angle ABC + \angle ACB=180^o$

$\implies$

$\angle BAC+50^o + 50^o=180^o$

$\implies$

$\angle BAC=180^o-(50^o+50^o)=80^o$ .

Hence,

$\angle BAC=80^o$ .

In the given figure,

$PA$

$\perp$

$AB$ ,

$QB$

$\perp$

$AB$ and

$\Delta$

$OAP$

$\cong \Delta$

$OBQ$ , then:

0%
$PA$ $=$ $OB$
0%
$AP$ $=$ $QB$
0%
$OP$ $=$ $BQ$
0%
$OA$ $=$ $OQ$

Explanation

It is given that

$\Delta OAP$

$\cong$

$\Delta OBQ$ both triangles are congruent.

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

Then,

$OA = OQ , AP = QB , OP = OQ$ ,

$\angle O=\angle O$ ,

$\angle A=\angle B$ and

$\angle P=\angle Q$ .

So, option

$B$ is correct.

If for

$\triangle$ ABC and

$\triangle$ DEF, the correspondence CAB

$\leftrightarrow$ EDF gives a congruence, then which of the following is NOT true?

0%
AC = DE
0%
AB = EF
0%
$\angle$ A = $\angle$ D
0%
$\angle$ C = $\angle$ E

Explanation

$\triangle ABC$ and

$\triangle DEF$

$\Rightarrow$ The correspondence

$CAB\leftrightarrow EDF$ gives a congruence.

$\Rightarrow$ So,

$CA=ED,\,AB=DF,\,CB=EF$

$\Rightarrow$

$\angle CAB=\angle EDF,\,\angle ACB=\angle DEF,\,\angle ABC=\angle DFE$

$\therefore$

$AB=EF$ is not true.

The ________ criterion is used to construct a triangle congruent to another triangle whose length of three sides are given.

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

Explanation

When the length of all three sides of a triangle are given, then by Side-Side-Side i.e.

$SSS$ criterion we can say that the sides of the other triangle will be equal to that by

$CPCT$ .

Hence, the answer is

$SSS$ .

$\triangle ABC$ , if

$AB = 7$ cm,

$\angle A= 40^o$ and

$\angle B = 70^o$ , which criterion can be used to construct this triangle?

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

Explanation

Given : In a triangle

$ABC, AB = 7 cm, \angle A=40^{o}$ and

$\angle B=70^{o}$

So, here we know two of the angles and a side including these angles.

Hence, Angle Side Angle i.e ASA criterion can be used to construct this triangle.

706o

Which congruence criterion can be used to conclude

$\triangle$

$XYZ$

$\cong$

$\triangle$

$QPR$ ?

0%
SAS
0%
SSS
0%
RHS
0%
None of these

Explanation

$\triangle$

$XYZ$ and

$\triangle$

$PQR$ ,

$\angle$

$YXZ$

$=$

$\angle$

$QPR$ (given)

$XY = PQ$ (given)

$\angle$

$XYZ =$

$\angle$

$PQR$ (given).
Therefore,

$\triangle$

$XYZ$

$\cong$

$\triangle$

$PQR$ (By

$ASA$ congruency criterion).

Hence, option

$D$ is correct.

$a, b$ and

$c$ are the sides of a triangle, then __________.

0%
$a-b > c$
0%
$c > a +b$
0%
$c =a+b$
0%
$b< c+a$

Explanation

This problem is totally based on the theorem which says any triangle with sides let say

$a , b ,c$ , then

$a+b > c$ and

$c+b > a$ and

$c+a > b$

Therefore, option D is correct.

Which of the following pair of triangles are congruent by RHS criterion?

0%
(i) and (ii)
0%
(iii) and (iv)
0%
(i) and (iii)
0%
(ii) and (iv)

Explanation

$\triangle ABC$ and

$\triangle QRP$

$\Rightarrow$

$AB=QR=4\,cm$

$\Rightarrow$

$BC=RP=5\,cm$ [Hypotenuse]

$\Rightarrow$

$\angle A=\angle Q=90^o$

$\therefore$

$\triangle ABC\cong\triangle QRP$ [By RHS criteria]

Direction (14 - 15) : Study the figure and information given below carefully and answer the following questions.
CF and AE are equal perpendiculars on BD, BF = FE = ED.

$\triangle$ ABE is congruent to

0%
$\triangle$ AED
0%
$\triangle$ BFC
0%
$\triangle$ CDF
0%
$\triangle$ BCD

Explanation

$\triangle ABE$ and

$\triangle CDF$

$\Rightarrow$

$AE = CF$ [given]

$\Rightarrow$

$\angle AEB=\angle CFD$ [given]

$\Rightarrow$

$BF + FE = DE + EF$

$\Rightarrow$

$BE = DF$

$\therefore$

$\triangle ABE \cong \triangle CDF$ [By SAS criteria]

By which congruency criterion,

$\triangle$ PQR

$\cong$

$\triangle$ PQS?

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

Explanation

$\triangle PQR$ and

$\triangle PQS$

$\Rightarrow$

$PR = PS = a \,cm$ [given]

$\Rightarrow$

$RQ = SQ = b\,cm$ [given]

$\Rightarrow$

$PQ = PQ$ [common side]

$\Rightarrow$

$\triangle PQR \cong \triangle PQS$ (By SSS)

In the given figure, triangles

$ABC$ and

$DCB$ are right angled at

$A$ and

$D$ respectively and

$AC$

$=$

$DB$ , then

$\Delta$

$ABC$

$\cong\Delta$

$DCB$ of from.

0%
$AAA$
0%
$SAS$
0%
$RHS$
0%
None of these

Explanation

$\Delta ABC$ and

$\Delta DCB,$

$AC = DB$ [Given]

$\angle BAC = \angle CDB = 90^{\circ}$

$BC=BC$ [Common Hypotenuse ]

So, by

$RHS$ rule of congruence,

$\Delta ABC$

$\cong$

$\Delta DCB$

Hence, option

$C$ is correct.

State True or False. If false, give reasons for that:
If two triangles are congruent, their corresponding angle are equal.

0%
True
0%
False

Explanation

If two triangles are congruent, then we know, their corresponding parts are equal. This rule is known as CPCT rule. That is, their corresponding angles and corresponding sides are equal.

Therefore, by CPCT rule, we can say that if two triangle are congruent then their corresponding angles are equal.

Hence the given statement is true.

Therefore, option

$A$ is correct.

State True or False. If false, give reasons:
A 1-rupee and a 5-rupee coins are congruent.

0%
True
0%
False

Explanation

We know, two shapes are congruent, if they have the same shape and size.

Consider a

$1-rupee$ coin and a

$5-rupee$ coin.

Both the coins are in the shape of a circle.

But considering the size,

$5-rupee$ coin is thicker than

$1-rupee$ coin.

Hence, the given statement is false.

Therefore, option B is correct.

$\triangle ABC$ and

$\triangle DBC$ are on the same base BC,AB=DC and AC=DB,then which of the following gives a CORRECT congruence relationship?

0%
$\triangle ABC=\triangle DBC$
0%
$\triangle ABC=\triangle CBD$
0%
$\triangle ABC=\triangle DCB$
0%
$\triangle ABC=\triangle BCD$

Explanation

In triangle

$\Delta ABC$ and

$\Delta DCB$ we have

$AB=DC$ ,

$AC=DB$ and

$BC=BC$

then from

$SSS$ congruence ,

$\Delta ABC=\Delta DCB$

Which of the following triangles is congruent to the given triangle?

Two triangles are congruent, if two angles and the side included between them in one triangle is equal to the two angles and the side included between them of the other triangle.This is known as

0%
RHS congruence criterion
0%
ASA congruence criterion
0%
SAS congruence criterion
0%
SSS congruence criterion

Explanation

When two angles and side included in between these angles of one triangle are same as another one, both these triangle are Congruent.

This criteria of congruency is knows as ASA criterion as a side included between two angles are all same.

$\textbf{Hence the answer is option (b)}$

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

0%
True
0%
False

D is a point on side BC of

$\Delta$ ABC such that AD

$=$ AC then AB

$>$ AD.

0%
True
0%
False

Explanation

$AD=AC(given)$

$\therefore \{ADC\}!=\{ACD\}!$ [isosceles

$\Delta$ ]

$\Rightarrow \{ADC\}!$ is an acute angle

[two angles of a triangle cannot be obtuse]

$\{ADB\}!=180^0-ADC$

Which : obtuse

$\triangle ADB,$

$\{ADB\} ! >B!$

$\Rightarrow AB > AD$

[side opposite to greater angle is greater]

1330083_1073623_ans_753c9073305f4304a29ef858baf5c173.png

State True or False:

$\triangle ABC$ and

$\triangle PQR$ ,

$AB=PQ,BC=QR$ and

$CB$ and

$RQ$ are extended to

$X$ and

$Y$ respectively and

$\angle ABX=\angle PQY$ then

$\triangle ABC \cong \triangle PQR$ .

0%
True
0%
False

Explanation

From the above figure we can conclude that,

$\angle ABX+\angle ABC = 180^o\quad\quad\dots(1)$

$\angle PQY+\angle PQR = 180^o\quad\quad\dots(2)$

But it is given that,

$\angle ABX =\angle PQY\quad\quad\dots(3)$

From

$(1),\ (2),$ and

$(3)$ we can conclude that,

$\angle ABC = \angle PQR$

Now, in

$\triangle ABC$ and

$\triangle PQR,$

$AB=PQ$ [Given]

$\angle ABC = \angle PQR$

$BC=QR$ [Given]

So, by

$SAS$ rule of congruence,

$\triangle ABC\cong\triangle PQR$

Hence, the given statement is true.

1913942_1057825_ans_665e398531604d01be44f0cec6e7f580.png

In right triangle

$ABC$ , right-angled at

$C, M$ is the mid-point of hypotenuse

$AB.\ C$ is joined to

$M$ and produced to a point

$D$ such that

$DM = CM$ . Point

$D$ is joined to point

$B$ .Which of the following is correct.

0%
$\Delta \mathrm { AMC } \cong \Delta \mathrm { BMD }$
0%
$\angle D B C$ is a right angle.
0%
$\Delta D B C \equiv \Delta A BC$
0%
$\mathrm { CM } = \dfrac { 1 } { 2 } \mathrm { AB }$

Explanation

REF.Image

Given

$\triangle ABC$ is right-angled at

$C$

$M$ is the midpoint of side

$AB$

$\Rightarrow CM \perp AB$ (As M is the midpoint and

$\triangle ABC$ is right-angled)

$\Rightarrow \angle CMA = \angle DMB \rightarrow (1)$

$\triangle ABC$ and

$\triangle DBC$

$BC= BC$

$\angle DBC= \angle ACB$

$\angle DCB= \angle ABC$

$\Rightarrow \triangle ABC$ is concorent to

$\triangle DBC$

$\therefore$ In

$\triangle DBM$ and

$\triangle MAC$

$DB= AC$

$\angle CMA= \angle DMB$ and

$MB= AM$

$\therefore \triangle DBM$ is congruent to

$\triangle CMA$

In any triangle, the side opposite to the larger (greater) angle is longer

0%
True
0%
False

Explanation

When two line segments are inclined at an angle betweem

$0^o$ and

$180^o$ , the length of the line segment joining the other ends of initial lines increases with angle of inclination between them.

The construction forms a triangle.

Hence the side opposite to greater angle is longer.

In a triangle ABC , if AB , BC and AC are the three sides of the triangle , then which of the following statements is necessarily true ?

0%
$\displaystyle AB + BC < AC$
0%
$\displaystyle AB + BC > AC$
0%
$\displaystyle AB + BC = AC$
0%
$\displaystyle AB^2 + BC^2 = AC^2$

Explanation

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

$\triangle ABC, AB, BC$ and

$AC$ are the three sides ,

Now ,

$AB + BC > AC$

Two right angles are congruent.

0%
True
0%
False

If two legs of a right triangle are equal to two legs of another right triangle. then the right triangles are congruent.

0%
True
0%
False

CBSE Questions for Class 9 Maths Triangles Quiz 11 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Triangles Quiz 11 - MCQExams.com

0:0:1

Practice Class 9 Maths Quiz Questions and Answers

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