If $$AB=QR,BC=PR$$ and $$CA=PQ$$ then

$$\triangle ABC\cong \triangle PQR$$

$$\triangle BAC\cong \triangle RPQ$$

$$\triangle PQR\cong \triangle BCA$$

MCQ Questions for Class 9 Maths Triangles Quiz 7

By which congruency criterion, the two triangles in Fig. are congruent?

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

Explanation

In triangle

$PQR$ and

$PQS$ ,

$PQ = PQ$ ...Common

$PR = PS = a \,cm$ ...Given

$QR = QS = b \,cm$ ...Given

$\therefore \Delta PSQ = \Delta PRQ$ ...

$SSS$ criterion

AS, all three sides are same so, it is congruent by

$SSS$ test.

So, option

$C$ is correct.

The length of two sides of a triangle is

$7 \,cm$ and

$9 \,cm$ . the length of the third side may lie between

0%
$1 \,cm$ and $10 \,cm$
0%
$2 \,cm$ and $8 \,cm$
0%
$3 \,cm$ and $16 \,cm$ `
0%
$1 \,cm$ and $16 \,cm$

Explanation

Length of two sides is greater than third side in a trinagle.
Hence,

$7 + 9 = 16 \,cm$ and

$9 - 7 = 2$
The length of the third side is lie between

$2$ and

$16$ .

$\Delta ABC$ and

$\Delta DBC$ are on the same base

$BC, AB = DC$ and

$AC = DB$ , then which of the following gives a congruence relationship?

0%
$\Delta ABC\cong \Delta DBC$
0%
$\Delta ABC \cong \Delta CBD$
0%
$\Delta ABC\cong \Delta DCB$
0%
$\Delta ABC\cong \Delta BCD$

Explanation

$\triangle ABC$ and

$\triangle DCB,$

$AC = DB$ [Given]

$AB = DC$ [Given]

$BC = CB$ [Common]

So, by

$SSS$ rule of congruence,

$\Delta ABC = \Delta DCB$

Hence, option

$C$ is correct.

If length of two sides of a triangle are

$6 \,cm$ and

$10 \,cm$ , then the length of the third side can be

0%
$3 \,cm$
0%
$4 \,cm$
0%
$2 \,cm$
0%
$6 \,cm$

Explanation

Let

$y \, cm$ be the third side of triangle.

Sum of two sides of triangle is always greater than

$3^{rd}$ side

Hence there are three possibilities,

$\Rightarrow 6+y > 10\cdots (1)$

$\Rightarrow 10+y > 6\cdots(2)$

$\Rightarrow 10+6 > y\cdots(3)$

From

$1,2,3$

$y>4, y>-4, y<16$

As, length cannot be negative

So,

$4 < y >16$

Therefore, from option third side of a triangle can be

$6 \, cm$ .

Hence, option

$D$ is correct.

Which of the following can be the length of the third side of triangle whose two sides measure

$18 cm$ and

$14 cm$ ?

0%
$4 \,cm$
0%
$3 \,cm$
0%
$5 \,cm$
0%
$32 \,cm$

Explanation

Let

$y \,cm$ be the third side of the triangle.

Hence,

$\Rightarrow 18+14>y$

$\Rightarrow 32>y$

$18-14 <y$

$y>4 \ and\ 32>y$

$4<y<32$

Therefore, the third side of a triangle can be

$5 cm$ .

All congruent triangles are equal in area.

0%
True
0%
False

All triangles are congruent.

0%
True
0%
False

Write true or False and justify your answer in each of the following:
If the angle between two tangents drawn from a point P to a circle of radius 'a' and centre O is

$60^\circ,$ then

$OP = a\sqrt{3}.$

0%
True
0%
False

Explanation

PT and OT are tangent and radius respectively at contact point T.

$\therefore \angle OTP = 90^\circ$

$\Rightarrow \Delta OTP$ is right angle

$\Delta$ at T

$\Delta OPT \cong \Delta OPR$ [By SSS criterion of congruence]

$\Rightarrow \angle OPT = \angle OPR = \dfrac{1}{2} \times 60^\circ = 30^\circ$

$\therefore$ In right angle

$\Delta OPT,$

$sin 30^\circ = \dfrac{OT}{OP} \Rightarrow$

$\Rightarrow OP = 2a$

Choose the correct alternative answer for the following question.

$\ln \Delta \mathrm{PQR},$ If

$\angle \mathrm{R}>\angle \mathrm{Q}$ then

$\ldots \ldots$

0%
$\mathrm{QR}>\mathrm{PR}$
0%
$P Q>P R$
0%
$\mathrm{PQ}<\mathrm{PR}$
0%
$\mathrm{QR}<\mathrm{PR}$

Explanation

$\ln \Delta \mathrm{PQR},$

$\angle \mathrm{R}>\angle \mathrm{Q}$

$\therefore \mathrm{PQ}>\mathrm{PR} \quad$ (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence, the correct answer is

$\mathrm{PQ}>\mathrm{PR}$ .

$\ln \Delta \mathrm{TPQ}, \angle \mathrm{T}=65^{\circ}, \angle \mathrm{P}=95^{\circ}$ which of the following is a true statement?

0%
$\mathrm{PQ}<\mathrm{TP}$
0%
$P Q< T Q$
0%
$\mathrm{TQ}<\mathrm{TP}<\mathrm{PQ}$
0%
$\mathrm{PQ}<\mathrm{TP}<\mathrm{TQ}$

Explanation

$\ln \Delta T P Q, \angle T=65^{\circ}, \angle P=95^{\circ}$
A triangle can have a maximum one obtuse angle.

$\angle T<\angle P$

$\mathrm{PQ}<\mathrm{TQ} \quad$ (The side opposite to the greater angle is greater than the side opposite to the smaller angle)
Hence, the correct answer is

$\mathrm{PQ}<\mathrm{TQ}$ .

$AB=QR,BC=PR$ and

$CA=PQ$ then

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

Explanation

$\triangle ABC$ and

$\triangle QRP$

AB=QR

BC=PRAB=QR,BC=PR

and CA=PQCA=PQ

By SSS congruency,

$\triangle ABC\cong \triangle QRP$

$\triangle CBA\cong \triangle PRQ$

In triangles

$ABC$ and

$DEF$ ,

$AB=FD$ and

$\angle A=\angle D$ . Two triangles are congruent by SAS criterion if

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

Explanation

$\triangle ABC$ and

$\triangle DEF$

$AB=DF$

$\angle A=\angle D$

for congruence by SAS rule

$AC=DE$

1863856_1876416_ans_6f62c0688d2d4528851274d53898994f.PNG

$\Delta PQR$ , if

$\angle R > \angle Q$ , then

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

$AB = QR; BC = PR$ and

$CA = PQ$ , then

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

In figure, if

$AB = AC$ and

$\angle B = 70^{0}$ , the value of

$\angle A$ will be:

0%
$70^{0}$
0%
$40^{0}$
0%
$55^{0}$
0%
$90^{0}$

In figure, if

$AB = AC$ , then the value of

$\angle C$ is equal to:

0%
$60^{0}$
0%
$36^{0}$
0%
$72^{0}$
0%
$108^{0}$

In two triangles

$ABC$ and

$DEF$ , if

$AC = DF, BC = EF$ and

$\angle ABC = \angle DEF = 90^{0}$ then two triangles are said to be congruent by:

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

Let ABC is a triangle. Then, which of the following is possible?

0%
AC + BC > AB
0%
AB + BC > AC
0%
AB + AC > BC
0%
None of these.

Is it possible to have a triangle with the following sides?
3 cm, 6 cm, 8 cm

0%
Yes
0%
No
0%
May be
0%
Information insufficient.

The lengths of two sides of a triangle are 6 cm and 8 cm. Which of the following can be length the third side of the triangle?

0%
1 cm
0%
5 cm
0%
15 cm
0%
None of the above

AM is a median of a

$\triangle ABC$ . Which of the following is true?

0%
AB + BM + MC > AC
0%
AB + BC + AM > AC
0%
AM + BC + CA > 2 AB
0%
None of these.

Let two sides of a triangle are 2 cm and 5 cm. Which of the following can be the length of its third side?

0%
6 cm
0%
7 cm
0%
2 cm
0%
8 cm

Given

$\triangle ABC$ in the figure. Then which is NOT true for a, b, c?
(Assume a > c > b)

0%
a + b > c
0%
c + b > a
0%
a + c > b
0%
None of these

Given

$\triangle ABC$ in the figure. Then which is true for a, b, c?

0%
a + b > c
0%
b + c > a
0%
a + c > b
0%
None of these

The equation of circumcircle of an

equilateral triangle is $x^2 + y^2 + 2gx + 2fy + c = 0$ and one vertex

of the triangle is $(1, 1)$ . The equation of incircle of the triangle is

0%
$4(x^2+y^2)=g^2+f^2$
0%
$4(x^2+y^2) +8gx+8fy=(1-g)(1+3g)+(1-f)(1+3f)$
0%
$4(x^2+y^2) +8gx+8fy=g^2+f^2$
0%
None of these

The distance from 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 and III

The three sides of a triangle are

$10,100$ and

$x$ . Which one of the following is correct?

0%
$10< x< 100$
0%
$90< x< 110$
0%
$90\le x\le 110$
0%
$90\le x<110$

In fig 3, if AB

$=$ AC and AP

$=$ AQ, then by which congruence criterion

$\Delta PBC \cong \Delta QCB$

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

$\triangle PQR$ , if the measure of

$\angle Q$ is

$50^o$ and the measure of

$\angle P$ is

$p^o$ , and if

$\overline{PQ}$ is longer than

$\overline{PR}$ , then

0%
0 < p < 40
0%
0 < p < 80
0%
40 < p < 80
0%
80 < p < 90
0%
80 < p < 130

Explanation

$PQ>PR$ (given)

$\Rightarrow \angle R>50^o$ -----------(1)

Also,

$\angle P+\angle Q+\angle R=180^o$

$\Rightarrow P+50^o+\angle R=180^o$

$\Rightarrow \angle R=130-p >50$ (using 1)

$\Rightarrow 130-p>50$

$\Rightarrow p<80$

$\therefore 0<p<80$

2114949_534975_ans_01d9fdea5232464ab8b0cec71609b6a6.png

In a

$\triangle PQR,$ if

$\angle R>\angle P,$ then which of the following cannot be true ?

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

CBSE Questions for Class 9 Maths Triangles Quiz 7 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Triangles Quiz 7 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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