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Page No 123:

Question 1:

Mark the correct alternative in each of the following:

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

(a) LK

(b) KM

(c) LM

(d) None of these

Answer:

It is given that

As triangles are congruent, same sides will be equal.

So

Hence (c).

Page No 123:

Question 2:

If ΔABC ΔABC is isosceles with

(a) AB = AC

(b) AB = BC

(c) AC = BC

(d) None of these

Answer:

It is given that and is isosceles

Since triangles are congruent so as same side are equal

Hence (a).

Page No 123:

Question 3:

If ΔABC  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:

If and is not congruent to

Since and compare corresponding sides you will see

(As )

Hence (a) , is not true.

Page No 123:

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:

In and

It is given that

Since two sides and an angle are equal so it obeys

Hence (a).

Page No 123:

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:

In and

It is given that

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

Hence (b).



Page No 124:

Question 6:

In ΔPQR ΔEFD then ED =

(a) PQ

(b) QR

(c) PR

(d) None of these

Answer:

If

We have to find

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

Hence (c).

Page No 124:

Question 7:

If ΔPQR ΔEFD, then ∠E =

(a) ∠P

(b) ∠Q

(c) ∠R

(d) None of these

Answer:

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)

Page No 124:

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:

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).

Page No 124:

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:

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).

Page No 124:

Question 10:

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

(a) SAS

(b) SSA

(c) ASA

(d) SSS

Answer:

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

Page No 124:

Question 11:

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

(a) 50°

(b) 60°

(c) 70°

(d) 80°
 

Answer:

We have to find

Since triangles are congruent 

So

Now in

(By property of triangle)

Hence (b) .

Page No 124:

Question 12:

If ABC and DEF are two triangles such that ΔABC 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:

It is given thatand

So and

Now, in triangle ABC,

Therefore,

Hence the correct option is (c).

Page No 124:

Question 13:

In the given figure, ABBE and FEBE. If BC = DE and AB = EF, then ΔABD is congruent to

(a) ΔEFC

(b) ΔECF

(c) ΔCEF

(d) ΔFEC
 

Answer:

It is given that

And

(Given)

(Given)

So (from above)

Hence 

From (d).

Page No 124:

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:

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).

Page No 124:

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:

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) .

Page No 124:

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:

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) .



Page No 125:

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:

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) 

Page No 125:

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:

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) .

Page No 125:

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:

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) .

Page No 125:

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:

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) .

Page No 125:

Question 21:

Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If AD is median of ∆ABC , then perimeter of ∆ABC is greater than 2 AD.
Statement-2 (Reason): The sum of any two sides of a triangle is greater than the third side.

Answer:

Statement-1 
Given: AD is median of ABC

In ABD,
AB+BD>AD        .....(1) [sum of any two sides of a triangle is greater than the third side]
Similarly, in ACD
AC + CD > AD     .....(2)
Now, on adding (1) and (2), we get 
AB + BD + AC + CD > 2AD
AB + BC + CA > 2AC          (BD+CD=BC)
Thus, it is true. 

Statement-2: True, the sum of any two sides of a triangle is greater than the third side. 

So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Page No 125:

Question 22:

Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): In an equilateral triangle ABC, if AD is the median, then AB + AC > 2 AD.
Statement-2 (Reason): In a right triangle, hypotenuse is the longest side.

Answer:

Statement-1
Given that AD is the median of ABC

In ABD,
AB+BD>AD        .....(1) [sum of any two sides of a triangle is greater than the third side]
Similarly, in ACD
AC + CD > AD     .....(2)
Now, on adding (1) and (2), we get 
AB + BD + AC + CD > 2AD
AB + BC + CA > 2AC          (BD+CD=BC)
Thus, it is true. 

Statement-2:
Let us consider a right-angled triangle ABC, right-angled at B.

In ∆ABC,

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

∠A + 90o + ∠C = 180°

∠A + ∠C = 90°

Hence, the other two angles have to be acute (i.e., less than 90o ).

Thus, ∠B is the largest angle in ∆ABC.

So, ∠B > ∠A and ∠B > ∠C

Therefore, AC > BC and AC > AB [Using theorem 7.7 of triangles, in any triangle, the side opposite to the larger (greater) angle is longer.]

Therefore, AC is the largest side in ∆ABC.

However, AC is the hypotenuse of ∆ABC.

Therefore, the hypotenuse is the longest side in a right-angled triangle.

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

Page No 125:

Question 23:

Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): The sum of any two sides of a triangle is greater than the third side.
Statement-2 (Reason): It is possible to construct a triangle with lengths of its sides as 4 cm, 3 cm, and 7 cm.

Answer:

Statement-1
True, the sum of any two sides of a triangle is greater than the third side.

Statement-2
Let AB = 4cm, BC = 3 cm and CA = 7 cm
Now, AB + BC = 4 + 3 = 7 cm = CA
Thus, it is not possible to construct a triangle with sides of 4 cm, 3 cm, and 7 cm.

Hence, the correct answer is option  (a).

Page No 125:

Question 24:

Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If ∆ABC ≅ ∆RPQ, then BC = QR 
Statement-2 (Reason): Corresponding parts of two congruent triangle are equal.

Answer:

Statement-1 
If ∆ABC ≅ ∆RPQ.

We know that the corresponding parts of two congruent triangles are equal.(Statement-2)

∴ AB = RP,
BC = PQ
and CA = QR

Thus, Statement-1 is false.

So, ​Statement-1 is false, Statement-2 is true.

Hence, the correct answer is option (d).

Page No 125:

Question 25:

Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has the following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the right choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are concurrent.
Statement-2 (Reason): If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent.

Answer:

Statement-1:
Yes, the statement is true because the two triangles can be congruent by either AAS or ASA congruence criteria.
If any two angles and a non-included side of one triangle are equal to the corresponding angles and side of the other triangle, then the two triangles are concurrent by AAS criteria.

Statement-2: 
True, If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent by AAS or ASA criteria.

Hence, the correct answer is an option (a).


 



Page No 126:

Question 26:

Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c), and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) ​Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If two sides and the included angle of one triangle are equal to the corresponding sides and the included angle of the other triangle, then the two triangles are concurrent
Statement-2 (Reason): If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles are congruent.

Answer:

Statement-1:
If two sides and the included angle of one triangle are equal to the corresponding sides and the included angle of the other triangle, then the two triangles are concurrent by SAS criteria.
Thus, it is true

Statement-2: 
The two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle, i.e., the SAS rule.
According to the SAS rule: If any two sides and the angle included between the sides of one triangle are equivalent to the corresponding two sides and the angle between the sides of the second triangle, then the two triangles are said to be congruent by the SAS rule.
Therefore, the angles should be included between the angles of their two given sides.
Thus, the given
Statement is false.

Hence, the correct answer is option (c).
 



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