MCQ Questions for Class 9 Maths Triangles Quiz 5

The distance from town A to town B is five miles. C is six miles from B. Which of the following could be the distance from A to C?
I.

$11$
II.

$1$
III.

$7$

0%
I only
0%
II only
0%
I and II only
0%
II and III only
0%
I, II or III

Explanation

The distance from two A nad B is five miles. C is six miles from B To find distance between A to C there are three possibilities.

Case:

$1$ In triangle

$ABC,AB=5$ and

$BC=6$

Here , the length of AC must be less than the sum of other two sides and greater than the difference of the other sides

$6.5<6+5$

$\Rightarrow 1<Ac<11$

Thus

$AC=7$ is a possible

$\therefore$ distance from

$A$ to

$C$ could be

$11,1$ or

$7$ miles.

2114881_535657_ans_11d7a9a00cde41c8846a0574c566c664.png

Two sides of a triangle have lengths

$7$ and

$9$ . Which of the following could not be the length of the third side?

0%
$4$
0%
$5$
0%
$7$
0%
$11$
0%
$16$

Explanation

An important rule to remember about triangles is called the third side rule:

The length of the third side of a triangle is less than the sum of the lengths of the other two sides and greater than the (positive) difference of the lengths of the other two sides.

For this triangle, the length of the third side must be greater than

$9-7=2$

97=2 and less than

$9+7=16$ 9+7=16. All the answers are possible except for answer E, which is equal to

$16$ 16 but not less than

$16$ 16.

Which of the following sets of measurements can be used to construct a triangle?

0%
$4\text{ cm}, 5\text{ cm}, 6\text{ cm}$
0%
$4\text{ cm}, 3\text{ cm}, 8\text{ cm}$
0%
$5\text{ cm}, 6\text{ cm}, 12\text{ cm}$
0%
$6\text{ cm}, 3\text{ cm}, 10\text{ cm}$

Explanation

Because the sum of the length of any

$2$ sides of the triangle should be greater than the third side, which is only satisfied by option

$A.$

$(4+5)>6$

$(4+6)>5$

$(5+6)>4$

In the figure above,

$ABCD$ is a rectangle. If

$AD = 6$ , which of the following could be the length of

$\overline {AC}$ ?

0%
$2$
0%
$4$
0%
$5$
0%
$6$
0%
$7$

Explanation

In triangle

$ACD,$ angle

$ADC$ is greater than the other two angles

By sine rule , we get

$\dfrac{AC}{AD}=\dfrac{\angle(ADC)}{\angle(DCA)}$

We know that

$\angle(ADC) > \angle(DCA)$

Therefore

$AC>AD=6$

Therefore option

$E$ is correct

In a triangle with sides of

$7$ and

$9$ , the third side must be

0%
more than $16$
0%
between $7$ and $9$
0%
between $2$ and $16$
0%
between $7$ and $16$
0%
between $9$ and $16$

Explanation

Given that two sides of triangle are

$7,9$
Let the third side be

$x$
We have

$7+9>x$ and

$x+9>7$ and

$x+7>9$ and

$x>0$
We get

$x<16$ and

$x>2$
Therefore the third side will lie between

$2$ and

$16$ .

Mark the triplet that can be the lengths of the sides of a triangle.

0%
$2, 3, 5$
0%
$1, 4, 2$
0%
$7, 4, 4$
0%
$5, 6, 12$
0%
$9, 20, 8$

Explanation

for triplet to be the length of triangle, it must satisfy triangle property.

sum of any two sides must be greater than third side
So,

$7,4,4$ is correct answer.
If we consider remaining options, observe that sum of two sides is less than third side, so they cannot form triangle.

$\triangle PQR$ is right angled at

$Q$ ,

$PR=5cm$ and

$QR=4cm$ . If the lengths of sides of another triangle

$ABC$ are

$3cm$ ,

$4cm$ ,

$5cm$ , then which one of the following is correct?

0%
Area of $\triangle PQR$ is double that of $\triangle ABC$ .
0%
Area of $\triangle ABC$ is double that of $\triangle PQR$ .
0%
$\sqrt { B } =\cfrac { \sqrt { Q } }{ 2 }$
0%
Both triangles are congruent.

Explanation

Given:

$\Delta PQR$ is right angled at

$Q$ ,

$PR = 5 cm, QR = 4 cm$ .

Length of sides of another triangle are

$3 cm, 5 cm, 4 cm$ .

According to question,

Ref. image

Applying Pythagoras theorem in

$\Delta PQR$ ,

$H^2 = P^2 + B^2$

$5^2 = P^2 + 4^2$

$25 - 16 = P^2$

$9 = P^2$

$P = 3 cm, PQ = 3cm$

Now we can see that both the triangles have equal sides.

i.e.

$PQ = AB = 3cm$

$PR = AC = 5 cm$

$QR = BC = 4 cm$

It means by SSS congruence property both triangles are congruent.

{SSS congruence property: Two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of other triangle.}

i.e.

$\Delta PQR \cong \Delta ABC$

The correct option is (D). i.e. both triangles are congruent.

2106402_551937_ans_f3067abf757246458fd5df7a2d336307.png

In quadrilateral

$ACBD, AC = AD$ and

$AB$ bisects

$\angle CAD.$ Show that

$\triangle ABC \cong \triangle ABD$ . Can you say that

$BC$ =

$BD$ ?

0%
True
0%
False

Explanation

$\triangle ABC$ and

$\triangle ABD,$

$\angle CAB=\angle BAD$ [Given]

$AC=AD$ [Given]

$AB=AB$ [Common]

So, by

$SAS$ rule of congruence,

$\triangle ABC \cong \triangle ABD$

As corresponding sides in congruent triangles are equal so,

$BC=BD.$

$\triangle ABC$ , the bisector of

$\angle {A}$ intersects

$\overline { BC }$ at a point

$D$ . Then:

0%
$BD\times AC=BC\times AB$
0%
$BD\times AB=DC\times AC$
0%
$AC\times AB=DC\times BC$
0%
$BD\times AC=DC\times AB$

Explanation

$\triangle ABC$ , the bisector of

$\angle A$ intersects

$\overline BC$ at

$D$ .

$\therefore$

$AD$ bisects

$BC$

$\therefore BD=DC$ ........

$(i)$

In triangles,

$ABD$ and

$ADC$

$BD=CD$ ......... From

$(i)$

$\angle BAD=\angle DAC$ ....... (Given)

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

$\therefore$

$\triangle ABD\cong \triangle ADC$ ........ [By S.A.S criterion]

$\implies AB=AC$

Thus,

$BD\times AC=DC\times AB$

Which of the following set of measurements will form a triangle

0%
$11cm,4cm,6cm$
0%
$13cm,14cm,25cm$
0%
$8cm,4cm,3cm$
0%
$5cm.16cm.5cm$

Explanation

Triangle Inequality Theorem states that the sum of two side lengths of a triangle is always greater than the third side.

If this is true for all three combinations of added side lengths.

i.e.

$a+b> c,$

$b+c> a$ and

$c+a> b$ then the lengths form a triangle

(A)

$11+4=15> 6$

And

$4+6 =10\ngtr 11$

So, it does not form a triangle

(B)

$13+14=27> 25$

And

$14+25=39> 13$

And

$25+13=38> 14$

So, it forms a triangle

(C)

$8+4=12> 3$

And

$8+3=11> 4$

And

$4+3=7\ngtr 8$

So, it does not form a triangle

(D)

$5+16=21> 5$

And

$5+5=10\ngtr 16$

So, it does not form a triangle.

Hence option B is the correct answer

The two triangles in the figure are congruent by the congruence theorem. Here, it is given

$OQ=OR$ . Which of the following condition, along with the given condition, is sufficient to prove that the two triangles are congruent to each other?

0%
$\angle P=\angle S$
0%
$\angle Q=\angle R$
0%
$OP=OS$
0%
$PQ=SR$

Explanation

Given

$OQ=OR$ and

$\triangle POQ\cong \triangle ROS$
We know that, congruent parts of congruent triangles are congruent

$\angle POQ\cong \angle ROS$ (vertically opposite angles)
If

$OP=OS$ then by using the (SAS) congruent, we can conclude the congruency of two triangles.

Hence, option C is sufficient to prove the congruency.

$\triangle FGC$ is an isosceles triangle, which of the following method will prove

$\triangle ABC$ congruent to

$\triangle DEF$ ?

0%
SSS
0%
RHS
0%
AAS
0%
SAS

Explanation

$\triangle ABC$ and

$\triangle DEF$

Given,

$AB=DE$

$\angle ABC=\angle DEF=90^\circ$

$\triangle FCG$ is isoceles,

$GF=GC$

$\implies FD=CA$ ...... (as

$AB||DE$ ,

$GD$ should be equal to

$GA$ )

$AC=FD$

Hence,

$\triangle ABC \cong \triangle DEF$ by

$RHS \ postulate$

Triangles with sides

$3$ cm,

$4$ cm and

$5$ cm is possible.

0%
True
0%
False

Explanation

Yes, as

$3^2+4^2=5^2$
1309449_703062_ans_488cf80961b744e9b14afe2e019e3a52.jpg

Consider isosceles triangle

$ABC$ , in which

$\angle ABC=\angle ACB$ ,then which of the two sides are similar?

0%
$AB=AC$
0%
$AB=BC$
0%
$AC=BC$
0%
All of the above

Explanation

Here,

$\triangle ABC$ is an isosceles triangle.

Given,

$\angle ABC=\angle ACB$ .

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

Thus, if

$\angle ABC=\angle ACB$ , then

$AB=AC$ (as

$\angle ACB$ is opposite to side

$AB$ and

$\angle ABC$ is opposite to side

$AC$ ).

Hence, option

$A$ is correct.

To prove that

$\Delta DEF$ and

$\Delta ABC$ are congruent by SAS, what additional information is needed?

0%
$EF \cong BC$
0%
$\angle DFE \cong \angle ABC$
0%
$DE \cong AB$
0%
$\angle DFE \cong \angle ACB$

Explanation

Given

$FD=CA$ and

$\angle D=\angle A$

Now for triangles to be congruent by

$SAS$ we need a third side that include the given angles with the given equal sides.

$DE=AB$

$\implies DE\cong AB$

Hence, option

$C$ is correct.

Consider isosceles triangle

$ABC$ , in which

$\angle ABC=\angle ACB$ ,

$AB=2BC$ and

$AB=8cm$ . What is the perimeter of the

$\triangle ABC$ ?

0%
$24cm$
0%
$32cm$
0%
$20cm$
0%
$18cm$

Explanation

$\triangle ABC$ is an isosceles triangle.

Gi en that

$\angle ABC=\angle ACB$ ,

$AB=2BC$ and

$AB=8$ cm$$

So,

$AB=AC=8$ cm

$\Rightarrow BC=\dfrac {AB}{2}=4$ cm

Perimeter of

$\triangle ABC= 8+8+4=20$ cm.

Here,

$\triangle ABC$ is an isosceles triangle.

Given,

$\angle ABC=\angle ACB$ ,

$AB=2BC$ and

$AB=8 cm$ .

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

Then, So,

$AB=AC=8 cm$ .

Since

$AB=2BC$ ,

$\Rightarrow BC=\dfrac {AB}{2}=4 cm$ .

Thus, perimeter of

$\triangle ABC= 8+8+4=20 cm$ .

Hence, option

$C$ is correct.

State Whether the following statement is True or False:
In the figure,

$C$ is the mid-point of

$AB$ ,

$\angle BAD = \angle CBE, \angle ECA= \angle DCB$ then

$DA = EB$ .

0%
True
0%
False

Explanation

Given

$\angle P=\angle Q$

$\implies$

$\angle ECA=\angle DCB$ .

Now adding

$\angle X$ both side we get:

$\angle P+\angle X=\angle Q+\angle X$ ........

$[ 1]$ .

Now in triangles

$\Delta ACD$ and

$\Delta BCE$ ,

$\angle A=\angle B$ ........[given]

$\implies$

$\angle BAD=\angle ABE$

$AC=BC$ ...........[

$C$ is midpoint of

$AB$ ]

$\angle ACD=\angle BCE$ .......

$[ 1]$

So,

$\Delta ACD\cong \Delta BCE$ ......[by

$ASA$ congruency criterion]

Hence by

$CPCT$ , the corresponding parts are also equal.

Then,

$AD=BE$ .

Therefore, the given statement is true.

Hence, option

$A$ is correct.

1370647_712201_ans_3d08563ebd484d7eb12d9c7294a0d7d3.png

In the figure,

$\angle BCD = \angle ADC$ and

$\angle ACB = \angle BDA$ .

0%
$\angle B = \angle B$
0%
$\angle A = \angle D$
0%
$\angle A = \angle B$
0%
$\angle C = \angle D$

Explanation

$\bigtriangleup ACD$ and

$\bigtriangleup BDC$

$\angle BCD=\angle ADC$ [Given].

Given,

$\angle ACB=\angle BDA$ .

Adding these two above equation , we get,

$\angle BCD+\angle ACB=\angle ADC+\angle BDA$

$\implies$

$\angle ACD=\angle BDC$ .

Also,

$CD=CD$ [Common]

$\bigtriangleup ACD \cong \bigtriangleup BDC$ .....(By

$ASA$ congruency criterion).

Therefore, by

$CPCT$ rule, the corresponding parts are equal.

Then,

$AD=BC$ and

$\angle A=\angle B$ .

Hence, option

$C$ is correct.

In a quadrilateral

$ACBD, AC = AD$ and

$AB$ bisect

$\angle A$ , then:

0%
$BC \cong AD$
0%
$BD \cong AC$
0%
$BC \cong BD$
0%
$AB \cong AC$

Explanation

$\bigtriangleup ABC$ and

$\bigtriangleup ABD$ ,

$AB=AB$ (Common)

$\angle CAB=\angle DAB$ (

$AB$ bisect

$\angle A$ )

$AC=AD$ (Given)

By using

$SAS$ rule of congruence,

$\triangle ABC\cong \triangle ABD$

$\Rightarrow BC\cong BD$ [by

$CPCT$ ]

Therefore, option

$C$ is correct.

State whether the following statement is True or False?
Two sides

$AB, BC$ and median

$AM$ of one triangle

$ABC$ are respectively equal to sides

$PQ$ and

$QR$ and median

$PN$ of

$\triangle PQR$ . then

$\triangle ABM\cong \triangle PQN$

0%
True
0%
False

Explanation

$\bigtriangleup ABM$ and

$\bigtriangleup PQN$

$AB=PQ$

$BC=QR$

$AM=PN$

$AM$ is the median of

$\bigtriangleup ABC$

So,

$BM=CM=\dfrac{1}{2}BC$

$PN$ is the median of

$\bigtriangleup PQR$

So,

$QN=RN=\dfrac{1}{2}QR$

Therefore,

$BM=QN$

Hence,

$\bigtriangleup ABM\cong \bigtriangleup PQN$ {By SSS congruence}

In given figure

$\triangle PQR \cong \triangle XYZ$ by _______ congruency rule.

0%
$SSS$
0%
$AAA$
0%
$SAS$
0%
$ASA$

Explanation

$\triangle PQR$ and

$\triangle XYZ$ ,

$\angle$

$QPR$

$=$

$\angle$

$YXZ$ (given)

$PQ = XY$ (given)

$\angle$

$PQR$

$=$

$\angle$

$XYZ$ (given)

Therefore,

$\triangle PQR \cong \triangle XYZ$ by

$ASA$ rule of congruency.

State the following statement is True or False:
In an isosceles triangle any two angles of a triangle are same.

0%
True
0%
False

Explanation

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

Then, in an isosceles triangle only the angles opposite to equal side are equal.

Hence, in an isosceles triangle, any two angles of the triangle are not same.

Therefore, the statement is false and option

$B$ is correct.

In the given figure,

$AL\parallel DC, E$ is mid point of

$BC$ , then

$\triangle EBL\cong \triangle ECD$ .

0%
True
0%
False

Explanation

$\bigtriangleup EBL$ and

$\bigtriangleup ECD$ , we have,

$\angle BEL=\angle CED$ ....{vertical oppsoite angle}

$BE=CE$ .....{

$E$ is the mid-point of

$BC$ }

$\angle EBL=\angle ECD$ ....{alternate interior angle}.

Therefore,

$\bigtriangleup EBL\cong \bigtriangleup ECD$ ...{By

$ASA$ Congruency criterion}.

Hence, the given statement is true.

Therefore, option

$A$ is correct.

Which congruence criteria can be used to state that

$\triangle$ XOY

$\cong$

$\triangle$ POQ?

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

Explanation

$\triangle XOY$ and

$\triangle POQ$

$\Rightarrow$

$\angle YXO=\angle QPO=65^o$ [Given]

$\Rightarrow$

$OX=OP$ [Given]

$\Rightarrow$

$\angle XOY=\angle POQ$ [Vertically opposite angles]

$\therefore$

$\triangle XOY\cong\triangle POQ$ [By ASA criteria]

In Fig if

$\angle C >\angle A,\angle D> \angle E$ then

0%
$AE>CD$
0%
$AE < CD$
0%
$AB < BC$
0%
$BE < BD$

Explanation

Given,

$\angle C > \angle A\implies AB>BC$ ------ Side opposite to largest angle is longest

Similarly,

$\angle D > \angle E\implies EB>BD$

From above,

$(AB+EB)>(BC+BD)$

$\implies AE>CD$ Option A

In the given figure, which of the following is correct?

0%
$\Delta PQR \cong \Delta RSP$
0%
$\Delta PQR \cong \Delta SPR$
0%
$\Delta PQR \cong \Delta RPS$
0%
$\Delta PQR \cong \Delta PSR$

Explanation

$\triangle PQR$ and

$\triangle RSP$

$\angle QPR = \angle SRP = 45^o$

$PQ = RS = 5.5 cm$

$PR = RP$ (common)

$\triangle PQR \cong \triangle RSP$ [By SAS congruency]

Hence option (A) is correct

In given figure,
CF and AE are equal perpendiculars on BD, BF = FE = ED.

$\angle$ BAE = ______.

0%
$\angle$ BCD
0%
$\angle$ CBA
0%
$\angle$ ADC
0%
$\angle$ DCF

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]

$\therefore$

$\angle BAE=\angle DCF$ [By CPCT]

In the given figure, the line segment

$AB$ is parallel to another line segment

$RS$ and

$O$ is the midpoint of

$AS$ , Then

$\Delta AOB \cong \Delta SOR \ ?$

0%
True
0%
False

Explanation

$AO=OS$

$\\\angle ROS=\angle AOB(\ Vertically\ Opposite\ Angle)$

$\\\angle RSA=\angle BAS\ (\ Alternating\ Angle )$

Hence, According to ASA Congruence Criterion

$\triangle AOB\cong\triangle SOR$

In the figure given below,

$BA\parallel DF$ and

$CA\parallel EG$ and

$BD=EC$ , then

$BG=DF$ .

0%
True
0%
False

In the figure ,

$\angle A = \angle C$ and

$AB = BC$ . Then

$\Delta ABD \cong \Delta CBE$

0%
True
0%
False

Explanation

Since,

$\angle B$ is common to both Triangles

Hence,

$\angle B=\angle B\\\angle A=\angle C\\AB=AC$

From ASA Congruency Criterion

$\triangle ABD\cong\triangle CBE$

CBSE Questions for Class 9 Maths Triangles Quiz 5 - MCQExams.com

0:0:2

Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Triangles Quiz 5 - MCQExams.com

0:0:2

Practice Class 9 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.