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Chapter 7 Triangles Ex-7.2 Interview Questions Answers

Question 1 :

In an isosceles triangle ABC, with AB = AC, the bisectorsof B and C intersect each other at O. Join A to O.Show that:

(i) OB = OC (ii) AO bisects A

Answer 1 :

(i) It is given that in triangle ABC, AB = AC

ACB = ABC (Angles opposite to equal sides of atriangle are equal)

 ACB = ABC

OCB = OBC

OB = OC (Sides opposite to equal angles of a triangle arealso equal)

(ii) In ΔOAB and ΔOAC,

AO =AO (Common)

AB = AC (Given)

OB = OC (Proved above)

Therefore, ΔOAB ΔOAC (BySSS congruence rule)

BAO = CAO (CPCT)

AO bisects A. 

Question 2 :

In ΔABC, AD is the perpendicular bisector of BC (see thegiven figure). Show that ΔABC is an isosceles triangle in which AB = AC.

Answer 2 :

In ΔADC and ΔADB,

AD = AD (Common)

ADC =ADB (Each 90º)

CD = BD (AD is the perpendicular bisector of BC)

ΔADC ΔADB (By SAS congruence rule)

AB = AC (By CPCT)

Therefore, ABC is an isosceles triangle inwhich AB = AC.

Question 3 :

ABC is an isosceles triangle in which altitudes BE and CFare drawn to equal sides AC and AB respectively (see the given figure). Showthat these altitudes are equal.

Answer 3 :

In ΔAEB and ΔAFC,

AEB and AFC (Each 90º)

A = A (Common angle)

AB = AC (Given)

ΔAEB ΔAFC (By AAS congruence rule)

BE = CF (By CPCT)

Question 4 :

ABC is a triangle in which altitudes BE and CF to sides ACand AB are equal (see the given figure). Show that

(i) ABE  ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

Answer 4 :

(i) In ΔABE and ΔACF,

AEB = AFC (Each90º)

A = A (Common angle)

BE = CF (Given)

ΔABE ΔACF (By AAS congruence rule)

(ii) It has already been proved that

ΔABE ΔACF

AB = AC (By CPCT)

Question 5 :

ABC and DBC are two isosceles triangles on the same baseBC (see the given figure). Show that ABD = ACD.

Answer 5 :


Let us join AD.

In ΔABD and ΔACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common side)

ΔABD ΔACD (By SSS congruence rule)

ABD = ACD (By CPCT)

Question 6 :

ΔABC is an isosceles triangle in which AB = AC. Side BA isproduced to D such that AD = AB (see the given figure). Show that BCD is a right angle.

Answer 6 :

In ΔABC,

AB = AC (Given)

ACB = ABC (Angles opposite to equal sides of atriangle are also equal)

In ΔACD,

AC = AD

ADC = ACD (Angles opposite to equal sides of atriangle are also equal)

In ΔBCD,

ABC + BCD + ADC = 180º(Angle sum property of a triangle)

ACB + ACB +ACD + ACD = 180º

2(ACB + ACD) = 180º

2(BCD) = 180º

BCD = 90º

Question 7 :

ABC is a right angled triangle in which A = 90º and AB = AC. Find B and C.

Answer 7 :

It is given that

AB = AC

C = B (Angles opposite to equal sides are also equal)

In ΔABC,

A + B + C = 180º (Angle sum property of a triangle)

90º + B + C = 180º

90º + B + B = 180º

2 B = 90º

B = 45º

B = C = 45º

Question 8 :

Show that the angles of an equilateraltriangle are 60º each.

Answer 8 :

Let us consider that ABC is an equilateral triangle.

Therefore, AB = BC = AC

AB = AC

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

Also,

AC = BC

B = A (Angles opposite to equal sides of a triangle are equal)

Therefore, we obtain

A = B = C

In ΔABC,

A + B + C = 180°

A + A + A = 180°

3A = 180°

A = 60°

A = B = C = 60°

Hence, in an equilateral triangle, all interior anglesare of measure 60º


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Chapter 7 Triangles Ex-7.2 Contributors

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