RD Sharma 2020 solution class 9 chapter 12 Congruent Triangles MCQS

MCQS

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Question 1:

Mark the correct alternative in each of the following:

If ABC $\cong$ ΔLKM, then side of ΔLKM equal to side AC of ΔABC is

(a) LK

(b) KM

(c) LM

(d) None of these

Answer 1:

It is given that

As triangles are congruent, same sides will be equal.

So

Hence (c).

Question 2:

If ΔABC $\cong$ ΔABC is isosceles with

(a) AB = AC

(b) AB = BC

(c) AC = BC

(d) None of these

Answer 2:

It is given that and is isosceles

Since triangles are congruent so as same side are equal

Hence (a).

Question 3:

If ΔABC $\cong$ ΔPQR and ΔABC is not congruent to ΔRPQ, then which of the following is not true:

(a) BC = PQ

(b) AC = PR

(c) AB = PQ

(d) QR = BC

Answer 3:

If and is not congruent to

Since and compare corresponding sides you will see

(As )

Hence (a) , is not true.

Question 4:

In triangles ABC and PQR three equality relations between some parts are as follows:
AB = QP, ∠B = ∠P and BC = PR

State which of the congruence conditions applies:

(a) SAS

(b) ASA

(c) SSS

(d) RHS
 

Answer 4:

In and

It is given that

Since two sides and an angle are equal so it obeys

Hence (a).

Question 5:

In triangles ABC and PQR, if ∠A = ∠R, ∠B = ∠P and AB = RP, then which one of the following congruence conditions applies:

(a) SAS

(b) ASA

(c) SSS

(d) RHS

Answer 5:

In and

It is given that

Since given two sides and an angle are equal so it obeys

Hence (b).

Question 6:

In ΔPQR $\cong$ ΔEFD then ED =

(a) PQ

(b) QR

(c) PR

(d) None of these

Answer 6:

If

We have to find

Since, as in congruent triangles equal sides are decided on the basis of “how they are named”. 

Hence (c).

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Question 7:

If ΔPQR $\cong$ ΔEFD, then ∠E =

(a) ∠P

(b) ∠Q

(c) ∠R

(d) None of these

Answer 7:

If

Then we have to find

From the given congruence, as equal angles or equal sides are decided by the location of the letters in naming the triangles. 

Hence (a)

Question 8:

In a ΔABC, if AB = AC and BC is produced to D such that ∠ACD = 100°, then ∠A =

(a) 20°

(b) 40°

(c) 60°

(d) 80°

Answer 8:

In the triangle ABC it is given that 

We have to find

Now (linear pair)

Since

So, (by isosceles triangle)

This implies that

Now, 

(Property of triangle)

Hence (a).

Question 9:

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is

(a) 100°

(b) 120°

(c) 110°

(d) 130°

Answer 9:

Let be isosceles triangle 

Then

Now it is given that vertex angle is 2 times the sum of base angles 

(As)

Now 

(Property of triangle)

(Since, and )

Hence (b).

Question 10:

Which of the following is not a criterion for congruence of triangles?

(a) SAS

(b) SSA

(c) ASA

(d) SSS

Answer 10:

(b) ,as it does not follow the congruence criteria.

Question 11:

In the given figure, the measure of ∠B'A'C' is

(a) 50°

(b) 60°

(c) 70°

(d) 80°
 

Answer 11:

We have to find

Since triangles are congruent 

So

Now in

(By property of triangle)

Hence (b) .

Question 12:

If ABC and DEF are two triangles such that ΔABC $\cong$ ΔFDE and AB = 5cm, ∠B = 40°

(a) DF = 5cm, ∠F = 60°

(b) DE = 5cm, ∠E = 60°

(c) DF = 5cm, ∠E = 60°

(d) DE = 5cm, ∠D = 40°

Answer 12:

It is given thatand

So and

Now, in triangle ABC,

Therefore,

Hence the correct option is (c).

Question 13:

In the given figure, AB ⊥ BE and FE ⊥ BE. If BC = DE and AB = EF, then ΔABD is congruent to

(a) ΔEFC

(b) ΔECF

(c) ΔCEF

(d) ΔFEC
 

Answer 13:

It is given that

And

(Given)

(Given)

So (from above)

Hence 

From (d).

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Question 14:

In the given figure, if AE || DC and AB = AC, the value of ∠ABD is

(a) 70°

(b) 110°

(c) 120°

(d) 130°
 

Answer 14:

We have to find the value of in the following figure.

It is given that

(Vertically apposite angle)

Now (linear pair) …… (1)

Similarly (linear pair) …… (2)

From equation (1) we have

Now (same exterior angle)

(Interior angle)

Now 

So

Since

Hence (b).

Question 15:

In the given figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is

(a) 52°

(b) 76°

(c) 156°

(d) 104°
 

Answer 15:

We are given that;

, is isosceles 

And

We are asked to find angle x

From the figure we have

Therefore,

Since, so

Now 

Hence (d) .

Question 16:

In the given figure, if AC is bisector of ∠BAD such that AB = 3 cm and AC =  5 cm, then CD =

(a) 2 cm

(b) 3 cm

(c) 4 cm

(d) 5 cm
 

Answer 16:

It is given that

, is bisector of

We are to find the side CD

Analyze the figure and conclude that

(As in the two triangles are congruent)

In

So 

Hence (c) .

Question 17:

D, E, F are the mid-point of the sides BC, CA and AB respectively of ΔABC. Then ΔDEF is congruent to triangle

(a) ABC

(b) AEF

(c) BFD, CDE

(d) AFE, BFD, CDE

Answer 17:

It is given that, and are the mid points of the sides, andrespectively of

(By mid point theorem)

(As it is mid point)

Now in and

(Common)

(Mid point)

(Mid point)

Hence (d) 

Question 18:

ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, ∠BAD =

(a) 55°

(b) 70°

(c) 35°

(d) 110°
 

Answer 18:

It is given that, AB=AC and Ad is the median of BC

We know that in isosceles triangle the median from he vertex to the unequal side divides it into two equal part at right angle.

Therefore, 

(Property of triangle)

Hence (a) .

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Question 19:

In the given figure, X is a point in the interior of square ABCD. AXYZ is also a square. If DY = 3 cm and AZ = 2 cm, then BY =

(a) 5 cm

(b) 6 cm

(c) 7 cm

(d) 8 cm

Answer 19:

In the following figure we are given

Where ABCD is a square and AXYZ is also a square

We are asked to find BY

From the above figure we have XY=YZ=AZ=AX

Now in the given figure

So,

Now inn triangle ΔAXB

So 

Hence (c) .

Question 20:

In the given figure, ABC is a triangle in which ∠B = 2∠C. D is a point on side BC such that AD bisects ∠BAC and AB = CD. BE is the bisector of ∠B. The measure of ∠BAC is

(a) 72°

(b) 73°

(c) 74°

(d) 95°
 

Answer 20:

It is given that

, 

AB = CD 

We have to find

Now AB = CD

AB = BD

Now the triangle is isosceles 

Let

So

Now 

Since 

Hence (a) .

Question 1:

If AB = QR, BC = PR and CA = PQ, then triangle _________ ≅ triangle _________.

Answer 1:

It is given that, AB = QR, BC = PR and CA = PQ.

So, A ↔ Q, B ↔ R and C ↔ P

∴ ∆ABC ≅ ∆QRP

If AB = QR, BC = PR and CA = PQ, then triangle __ABC__ ≅ triangle __QRP__.

Question 2:

In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom, if __________.

Answer 2:

SAS congruence axiom states that two triangles are congruent if two sides and the included angle of one are equal to the corresponding sides and the included angle of the other triangle.

In ∆ABC, ∠A is included between the sides AB and AC.

In ∆DEF, ∠D is included between the sides DF and DE.



∴ ∆ABC ≅ ∆DEF by SAS axiom if AC = DE.

In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom, if __AC = DE__.

Question 3:

In ∆ABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are __________ but not __________.

Answer 3:

If two sides of a triangle are equal, then it is an isosceles triangle.

In ∆ABC, AB = AC

∴ ∆ABC is an isosceles triangle.

In ∆ABC,

AB = AC    (Given)

∴ ∠C = ∠B       (In a triangle, equal sides have equal angles opposite to them)

It is given that, ∠C = ∠P and ∠B = ∠Q.

∴ ∠P = ∠Q

In ∆PQR,

∠P = ∠Q        (Proved)

∴ QR = PR     (In a triangle, equal sides have equal angles opposite to them)

So, ∆PQR is an isosceles triangle.

However, it cannot be proved that the corresponding sides of ∆ABC are congruent to the corresponding sides of ∆PQR. Hence, the triangles are not congruent.

In ∆ABC and ∆PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are __isosceles__ but not __congruent__.

Question 4:

In ∆PQR, ∠P = ∠R, QR = 4 cm and PR = 5 cm. Then PQ = ________.

Answer 4:

In ∆PQR,

∠P = ∠R       (Given)

∴ QR = PQ    (Sides opposite to the equal angles of a triangle are equal)

⇒ PQ = 4 cm       (QR = 4 cm)

In ∆PQR, ∠P = ∠R, QR = 4 cm and PR = 5 cm. Then PQ = ___4 cm___.

Question 5:

In ∆ABC, AB = AC and ∠B = 50°. Then, ∠C = __________.

Answer 5:

In ∆ABC,

AB = AC           (Given)

∴ ∠C = ∠B        (Angles opposite to the equal sides of a triangle are equal)

⇒ ∠C = 50°       (∠B = 50°)

In ∆ABC, AB = AC and ∠B = 50°. Then, ∠C = __50°__.

Question 6:

In ∆ABC, AB = BC and ∠B = 80°. Then ∠A = __________.

Answer 6:

In ∆ABC,

AB = BC           (Given)

∴ ∠C = ∠A        ..... (1)         (Angles opposite to the equal sides of a triangle are equal)

Now,

∠A + ∠B + ∠C = 180°       (Angle sum property of triangle)

∴ ∠A + 80° + ∠A = 180°

⇒ 2∠A = 180° − 80° = 100°

⇒ ∠A = $\frac{100°}{2}$ = 50°

In ∆ABC, AB = BC and ∠B = 80°. Then ∠A = __50°__.

Question 7:

If in ∆PQR, ∠P = 70° and ∠R = 30°, then the longest side of ∆PQR is ___________.

Answer 7:

In ∆PQR,

$\angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180°$          (Angle sum property of a triangle)

$\therefore 70°+\angle \mathrm{Q}+30°=180°$

$\Rightarrow \angle \mathrm{Q}+100°=180°$

$\Rightarrow \angle \mathrm{Q}=180°-100°=80°$



So, ∠Q is the greatest angle in the ∆PQR.

We know that, in a triangle the greater angle has the longer side opposite to it.

∴ PR is the longest side of ∆PQR.

If in ∆PQR, ∠P = 70° and ∠R = 30°, then the longest side of ∆PQR is ___PR___.

Question 8:

In ∆PQR, if ∠R > ∠Q, then __________.

Answer 8:

In ∆PQR,

∠R > ∠Q        (Given)

∴ PQ > PR     (In a triangle, the greater angle has the longer side opposite to it)

In ∆PQR, if ∠R > ∠Q, then ___PQ > PR___.

Question 9:

D is a point on side BC of a ∆ABC such that AD bisects ∠BAC. Then ________.

Answer 9:

In ∆ABC, AD is the bisector of ∠A.

∴ ∠BAD = ∠CAD             .....(1)

We know that exterior angle of a triangle is greater than each of interior opposite angle.

In ∆ABD,

∠ADC >  ∠BAD

⇒ ∠ADC > ∠CAD        [Using (1)]

In ∆ADC,

∠ADC > ∠CAD

∴ AC > CD           (In a triangle, the greater angle has the longer side opposite to it)

Similarly, AB > BD

D is a point on side BC of a ∆ABC such that AD bisects ∠BAC. Then ___AC > CD and AB > BD___.

Question 10:

Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle lies between _________ and _________.

Answer 10:

In a triangle, the sum of two sides is greater than the third side.

∴ 5 cm + 1.5 cm > Third side

Or Third side < 6.5 cm

In a triangle, the difference of two sides is less than the third side.

∴ 5 cm − 1.5 cm < Third side

Or Third side > 3.5 cm

So,

3.5 cm < Third side < 6.5 cm

Thus, the length of the third side of the triangle lies between 3.5 cm and 6.5 cm.

Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle lies between __3.5 cm__ and _6.5 cm _.
 

Question 11:

If AD is a median of ∆ABC, then the perimeter of ∆ABC cannot be less than or equal to ___________.

Answer 11:

AD is the median of the ∆ABC.

In ∆ABD,

AB + BD > AD      .....(1)          (In a triangle, the sum of any two sides is greater than the third side)

In ∆ACD,

CD + CA > AD      .....(2)          (In a triangle, the sum of any two sides is greater than the third side)

Adding (1) and (2), we get

AB + BD + CD + CA > AD + AD

⇒ AB + BC + CA > 2AD                     (BC = BD + CD)

⇒ Perimeter of ∆ABC > 2AD

Or the perimeter of ∆ABC cannot be less than or equal to 2AD

If AD is a median of ∆ABC, then the perimeter of ∆ABC cannot be less than or equal to ___2AD___.

Question 12:

If ∆PQR ≅ ∆EDF, then PR = ___________.

Answer 12:

If ∆PQR ≅ ∆EDF, then

P ↔ E, Q ↔ D and R ↔ F 

So, PR = EF,  PQ = ED and QR = DF       (If two triangles are congruent, then their corresponding sides are congruent)

If ∆PQR ≅ ∆EDF, then PR = ____EF____.

Question 13:

It is given that ∆ABC ≅ ∆RPQ, then BC = __________.

Answer 13:

If ∆ABC ≅ ∆RPQ, then

A ↔ R, B ↔ P and C ↔ Q

So, AB = RP, BC = PQ and CA = QR       (If two triangles are congruent, then their corresponding sides are congruent)

It is given that ∆ABC ≅ ∆RPQ, then BC = ___PQ___.