In $$\Delta ABC,BC$$ is the greatest side. Then,

$$\angle A$$ must be greater than $${75^0}$$ but less than $${90^0}$$

$$\angle A$$ must be greater that $${60^o}$$

$$\angle A$$ must be greater than $${45^0}$$ but not necessarily greater than $${60^0}$$

MCQ Questions for Class 9 Maths Triangles Quiz 6

In the figure, ABC is a triangle in which AB = AC. X and Y are points on AB and AC such that AX =AY. Then

$\Delta ABY \cong \Delta ACX$

0%
True
0%
False

Explanation

$The\quad statement\quad is\quad true$

$Consider\triangle ABY\quad and\quad \triangle ACX$

$AB=AC\quad (given)$

$AX=AY\quad (given)$

$\angle A=\angle A\quad (common)$

$\therefore \triangle ABY\cong \triangle ACX\quad \{ By\quad SAS\quad congruency\}$

$\Delta ABC$ , E and D are the two points on BC such that CE = DB and

$\angle AEC= \angle ADB$ . Then the statement

$\Delta CAD \cong \Delta BAE$ is

0%
True
0%
False

Two line segments are congruent if they have the same length.

0%
True
0%
False

Explanation

We know, two shapes are said to be congruent when they exactly overlap each other when one is put over the other.

Similarly, when length of two line segments are equal, they overlap each other.

Therefore, we can say that, two line segments are congruent if they have the same length.

Hence they are congruent.

Therefore, the statement is true and and option

$A$ is correct.

In right triangle ABC, the right angle is 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 (see figure).

State whether the statement is true/false

$\Delta AMC \cong \Delta BMD$

0%
True
0%
False

Explanation

$In\triangle AMC\& \triangle DMB$

$\quad AM=MB=x$ (given)

$\quad DM=CN=y$ (given)

$\quad \angle DMB=\angle AMC$ (vertically opposite)

$\therefore \triangle AMC\cong \triangle DMB$ (by S.A.S)

1003231_1040288_ans_5f960e552d644887ad245eb047924f6c.png

In the above figure,

$\text AD$ =

$\text DC$ =

$\text CB$ and the measure of

$\angle DAC$ =

$40^o$ . What is the measure of

$\angle DCB$ ?

0%
$80^o$
0%
$40^o$
0%
$20^o$
0%
$300^o$

Explanation

$\because AD=DC$

$\implies$

$\angle DAC=\angle ACD$ (By isosceles triangle property, angles opposite equal sides are equal).

$\Delta ACD:$

${ 40 }^{ \circ }+{ 40 }^{ \circ }+\angle CDA={ 180 }^{ \circ }$ (By angle sum property of a triangle)

$\implies { 80 }^{ \circ }+\angle CDA={ 180 }^{ \circ }$

$\implies \angle CDA={ 180 }^{ \circ }-80^o$

$\implies \angle CDA={ 100 }^{ \circ }$ .

$\therefore \angle CDB={ 180 }^{ \circ }-{ 100 }^{ \circ }={ 80 }^{ \circ }$

$(\because$ linear pair, i.e. sum of angles on a straight line

$={ 180 }^{ \circ })$ .

$\Delta DCB$ :

$\angle BDC=\angle DBC$ (By isosceles triangle property, angles opposite equal sides are equal).

$\implies \angle DBC={ 80 }^{ \circ }$ .

$In\quad \Delta DBC$

$\implies$

${ 80 }^{ \circ }+{ 80 }^{ \circ }+\angle DCB={ 180 }^{ \circ }$ (By angle sum property of a triangle)

$\implies$

$\angle DCB={ 180 }^{ \circ }-{ 160 }^{ \circ }$

$={ 20 }^{ \circ }$ .

Therefore, option

$C$ is correct.

1001140_1054722_ans_e7ca9e7fd47941f3878debb7883b1164.png

In,

$\triangle DEF$ , if

$\angle$

$\text D$

$\equiv$

$\angle F$ , then:

0%
$\text DE$ $\equiv$ $\text EF$ $\equiv$ $\text DF$
0%
$\text DE$ $\equiv$ $\text DF$
0%
$\text DE$ $\equiv$ $\text EF$
0%
All the above

Explanation

By isosceles triangle property,

the angles opposite to equal sides in a triangle are equal.

Then, by converse of isosceles triangle property,

the sides opposite to equal angles in a triangle are equal.

That is, from figure,

$DE\equiv EF$ .

Therefore, option

$C$ is correct.

1001144_1054755_ans_d156a628ddad43ddbb1b6debaea9079a.png

$ABCD$ is a square and

$\Delta PAD$ is equilateral then

$PB = 2PC$ .

0%
True
0%
False

Explanation

$\Delta PAB$ &

$\Delta PDC$

$PA = PD$ [ Side of equilateral

$\Delta$ ]

$AB = CD$ [Side of square]

$\angle BAP = \angle CDP = 150^{\circ}$

$\Delta PAB \cong \Delta PDC$

$PB = PC$

1323457_1063214_ans_441f8a70c2784b7a9c1cc1af3bddf5bb.png

PQR is a right angled triangle in which

$\angle R=90^{\circ}$ . If RS

$\perp$ PQ, PR=3 cm, and RQ=4 cm, then what is the value of RS(in cm)?

0%
$\dfrac {12} 5$
0%
$\dfrac {36} 5$
0%
$5$
0%
$2.5$

In the given figure,

$ABC$ is an isosceles triangle in which

$AB = AD$ . Also,

$D$ is a point such that

$BD = CD$ . then "

$AD$ bisects

$\angle A$ and

$\angle D$ ." is ?

0%
True
0%
False

Explanation

Given:

$\triangle ABC$ is an isosceles triangle.

Consider

$\triangle ABD$ and

$\triangle ACD$

$\Rightarrow$

$AB=AC$ [ Given ]

$\Rightarrow$

$BD=CD$ [ Given ]

$\Rightarrow$

$AD=AD$ [ Common side ]

$\therefore$

$\triangle ABD\cong\triangle ACD$ [ By

$SSS$ congruence property ]

$\Rightarrow$

$\angle BAD=\angle CAD$ [ C.P.C.T ]

$\Rightarrow$

$\angle BDA=\angle CDA$ [ C.P.CT. ]

Hence, we have proved that

$AD$ bisects

$\angle A$ and

$\angle D.$

State the whether given statement is true or false

In the Fig. below,

$\angle B < \angle A$ and

$\angle C < \angle D$ . Show that

$A D < B C$ .

0%
True
0%
False

Explanation

$AO < BO$ (side opposite to smaller angle)

$DO < CO$ (side opposite to smaller angle)

So,

$AO + DO < BO + CO$

Or,

$AD < BC$

1085934_1174154_ans_4d7da755c85b46ccb8e3add4c6dca7e8.png

In the given figure if

$AB = AC = DC$ , then

$\angle BAD = \angle CAD$

0%
True
0%
False

Explanation

Given that,

$AB=AC=DC$

$\triangle ABD$ and

$\triangle ADC$ ,

$AB=AC$ (given)

$AD=AD$ (common side)

Here, we do not have any extra corresponding equal side or angle to prove the congruency of

$\triangle ABD$ and

$\triangle ADC$ ,

So,

$\angle BAD\neq\angle CAD$

The answer is

$\textbf{False}$

State true or false.

In the given figure,

$AB=AC$ , then

$\begin{array} { l } { \text {} \mathrm { DP } = \mathrm { DQ } } \end{array}.$

0%
True
0%
False

Explanation

$AB$ &

$AC$ are equal in length
So, in

$\triangle ABC$ ,

$\angle B=\angle C$

From

$\triangle BPD$ &

$\triangle DQC$

$i) \angle BPD=\angle CQD=90^0$

$ii) BD=DC$

$iii) \angle B=\angle C$

$\triangle BPD \cong \triangle CQD (by \ ASA \ criteria)$

by CPCT

$DP=DQ$

Which of the following will form the sides of a triangle?

0%
$23\,cm,\,17\,cm,\,8\,cm$
0%
$12\,cm,\,10\,cm,\,25\,cm$
0%
$6\,cm,\,7\,cm,\,16\,cm$
0%
$8\,cm,\,7\,cm,\,16\,cm$

Explanation

Triangle inequality theorem states that the sum of two sides of a triangle is always greater than the third side.

$(a)\ 17+8>23$ is true

Hence, the sides with the measure

$23\ cm,17\ cm,8\ cm$ will form the sides of a triangle.

$(b)\ 10+12>25$ is not true

Hence, the sides with the measure

$10\ cm,12\ cm,25\ cm$ will not form the sides of a triangle.

$(c)\ 6+7>16$ is not true

Hence, the sides with the measure

$6\ cm,7\ cm,16\ cm$ will not form the sides of a triangle.

$(d)\ 8+7>16$ is not true

Hence, the sides with the measure

$8\ cm,7\ cm,16\ cm$ will not form the sides of a triangle.

$\Delta PQR$ and

$\Delta SQR$ ,

$\overline{PQ}$ =

$\overline{SR}$ and

$\angle PQR = \angle QRS$ then:

0%
$\Delta PQR \sim \Delta SRQ$
0%
$\Delta PQR \equiv \Delta SRQ$
0%
$\Delta PQR = \Delta SRQ$
0%
None of these

Explanation

$\Delta PQR$ and

$\Delta SQR$ ,

$\overline{PQ} =\overline{SR}$ (Given)

And

$\angle PQR = \angle QRS$

$QR=QR$ [common side]

$\Delta PQR \sim \Delta SRQ$ (SAS property)

The bisectors of the base angle of an isosceles triangle are equal.

0%
True
0%
False

Take any point O in the interior of a triangle XYZ. Which option is/are correct?

0%
$OX+OY>XY?$
0%
$OY+OZ>YZ?$
0%
$OZ+OX>XZ?$
0%
All of the above.

In the given figure,

$\triangle A B C$ is an isosceles triangle in which

$A B = A C$ . If

$A B$ and

$A C$ are produced to

$D$ and

$E$ respectively such that

$B D = C E$ .

$B E = C D$ ?

0%
True
0%
False

Explanation

Given:

$(1)$

$\Delta ABC$ is an isosceles triangle.

$\implies$

$AB=AC$ .

$(2)$

$BC=CE$ .

Since

$AB=AC$ ,

$\therefore \angle ABC=\angle ACB$ [Since angles opposite to equal sides of a triangle are equal]

$\therefore 180^o-\angle ABC=180^o-\angle ACB$

$\angle DBC=\angle ECB$ ………..

$(1)$

$[$ since

$ABD$ and

$ACE$ are straight lines

$]$ .

In triangles

$DBC$ and

$ECB$ ,

$\angle DBC=\angle ECB$ (from

$(1)$ )

$BD=CE$ (given)

$BC=BC$ (common side)

$\therefore \Delta DBC\sim \Delta ECB$ [by

$SAS$ criteria].

So ,since triangles are congruent,

$\therefore DC=EB$ .

Hence,

$BE=CD$ is true.

Therefore, the statement is true.

Hence, option

$A$ is correct.

In a triangle

$ABC, \angle ABC=50^{o}, \angle BAC=30^{o}$ , then the shortest sides is

0%
$AB$
0%
$BC$
0%
$CA$
0%
$None$

Explanation

The side opposite to the largest angle is the longest side of the triangle and the side opposite to the smallest angle is the shortest side of the triangle.

By angle sum property,

$\angle{A}+\angle{B}+\angle{C}={180}^{\circ}$

$\Rightarrow {30}^{\circ}+{50}^{\circ}+\angle{C}={180}^{\circ}$

$\Rightarrow \angle{C}={180}^{\circ}-{80}^{\circ}={100}^{\circ}$

The side opposite to the smallest angle

$\left({30}^{\circ}\right)$ is the side

$BC$

$\Delta ABC$ and

$\Delta XYZ$ are equilateral triangles and

$AB = XY,$ the condition under which

$\Delta ABC \cong\Delta XYZ$ is

0%
$ASA$
0%
$RHS$
0%
$SSS$
0%
$AAS$

Explanation

Given that

$\Delta ABC$ and

$\Delta XYZ$ are equilateral triangles.

i.e.,

$AB=BC=CA$

and

$XY=YZ=ZX$ …………….

$(1)$

and

here

$AB=XY$ ………………

$(2)$

By equation

$(1)$ &

$(2)$

$BC=YZ$ and

$CA=ZX$ …………….

$(3)$

We have to show that

$\Delta ABC \cong\Delta XYZ$

$\Delta ABC$ and

$\Delta XYZ$

$AB=XY$ (Given)

$BC=YZ$ (by equation

$(3)$ )

$CA=ZX$ (by equation

$(3)$ )

$\Delta ABC \cong\Delta XYZ$ (By

$SSS$ rules).

1221372_1450931_ans_0574deedf4ee4e5586fe20f3a179e7f1.png

$\Delta ABC,BC$ is the greatest side. Then,

0%
$\angle A$ must be greater that ${60^o}$
0%
$\angle A$ must be greater than ${75^0}$ but less than ${90^0}$
0%
$\angle A$ must be greater than ${45^0}$ but not necessarily greater than ${60^0}$
0%
none of these

Which of the following statements are true (T) and which are false (F)?
Sum of any two sides of a triangle is greater than the third side.

0%
True
0%
False

Explanation

Here,

$ABC$ is a triangle

Construction : Extend

$AB$ to

$D$ such that

$AD=AC$

Now,

$\triangle ACD,$

$AD=AC$ [ By construction ]

$\angle 1=\angle 2$ [ angles opposite to equal sides are equal ]

Now,

$\angle BCD=\angle BCA+\angle 1$

$\angle BCD=\angle BCA+\angle 2$ [ Since,

$\angle 1=\angle 2$ ]

$\therefore$

$\angle BCD>\angle 2$

Now, in

$\triangle BCD,$

$\Rightarrow$

$\angle BCD>\angle 2$

$\Rightarrow$

$BD>BC$ [ Side opposite to the larger angle is longer ]

$\Rightarrow$

$AB+AD>BC$ [ Since,

$BD=AB+AD$ ]

$\Rightarrow$

$AB+AC>BC$ [ Since,

$AD=AC$ by construction ]

Similarly, we can prove

$AB+BC>AC$

And

$AC+BC>AB$

$\therefore$ The given statement is true.

1522948_1669990_ans_a8691b643a77425eb35dd8eb841645af.jpeg

State the congruency of these triangles

0%
True
0%
False

Explanation

$\triangle ADB$ and

$\triangle ACB$ , we have

$\angle ADB=\angle ACB\quad$ [Each

$90^{\circ}$ ]

$AB=AB\quad$ [Common]

$AD=AC\quad[Each\space2\space\mathrm{cm}]$

$\therefore \triangle ADB\cong\triangle ACB\quad$ [By RHS congruency]

So,

$\text{A}$ is the correct option.

Are the triangles congruent?

0%
True
0%
False

Explanation

$\triangle PQS$ and

$\triangle PRS$ , we have

$\angle PSQ=\angle PSR\quad$ [Each

$90^{\circ}$ ]

$PQ=PR\quad[Each\space3\space\mathrm{cm}]$

$PS=PS\quad$ [Common]

$\therefore \triangle PQS\cong\triangle PRS\quad$ [By RHS congruency]

So,

$\text{A}$ is the correct option.

State the congruency of these triangles.

0%
True
0%
False

Explanation

$\triangle IJK$ and

$\triangle ONM$ , we have

$\angle IJK=\angle ONM\quad$ [Each

$90^{\circ}$ ]

$IK=OM\quad$ [Each

$5\space\mathrm{cm}$ ]

$IJ=ON\quad$ [Each

$4\space\mathrm{cm}$ ]

$\therefore \triangle IJKB\cong\triangle ONM\quad$ [By RHS congruency]

So,

$\text{A}$ is the correct option.

State the congruency for these triangles

0%
True
0%
False

Explanation

$\triangle ACD$ and

$\triangle ACB$ , we have

$\angle ADC=\angle ABC\quad$ [Each

$90^{\circ}$ ]

$AD=AB\quad[Each\space3.6\space\mathrm{cm}]$

$AC=AC\quad$ [Common]

$\therefore \triangle ACD\cong\triangle ACB\quad$ [By RHS congruency]

So,

$\text{A}$ is the correct option.

The difference between the length of any two sides of a triangle is smaller than the length of third side.

0%
True
0%
False

$\Delta ABC$ ,

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

Explanation

In a triangle ABC,

$\Rightarrow AB + BC > AC$

$\Rightarrow BC+ AC > AB$

$\Rightarrow AC + AB > BC$
Since the third side of the triangle is less than the sum of any two sides of its.

$\Delta PQR$ ,

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

Explanation

In a triangle PQR,

$\Rightarrow \,PQ + QR > PR$

$\Rightarrow \,QR + PR > PQ$

$\Rightarrow \,PR + PQ > QR$
Since the third side of the triangle is less then the sum of any two sides of its.
Now,

$\Rightarrow \,PQ - QR < PR$

$\Rightarrow \,QR - PR < PQ$

$\Rightarrow \,PR - PQ < QR$
Since the length of third side of the triangle is always greater than the difference between two sides.

$\Delta PQR$ in congruent to

$\Delta STU$ , then what is the length of

$TU$ ?

0%
$5 \,cm$
0%
$6 \,cm$
0%
$7 \,cm$
0%
cannot be determined

Explanation

Given,

$\Delta PQR \cong \Delta STU$ .

Then the corresponding parts will also be equal.

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

Then, $PQ=ST$ , $QR=TU$ , $PR=SU$ , $\angle P=\angle S$ , $\angle Q=\angle T$ and $\angle R=\angle U$ .

Thus, $QR=TU$ .

$\therefore TU = 6 \ cm$ .

Therefore, option

$B$ is correct.

Which of Which of the following statements is not correct?

0%
The sum of any two sides of a triangle is greater than the third side
0%
A triangle can have all its angles acute
0%
A right-angled triangle cannot be equilateral
0%
Difference of any to sides of a triangle is greater than the third side

CBSE Questions for Class 9 Maths Triangles Quiz 6 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Triangles Quiz 6 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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