RBSE Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 7 Triangles Exercise 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:
(i) OB = OC
(ii) AO bisects ∠A
Answer:

(i) In ∆ABC, we have: AB = AC
∴ ∠B = ∠C
(∵ Angles opposite to equal sides are equal.)
or \(\frac{1}{2}\)∠B = \(\frac{1}{2}\)∠C
i.e. ∠OBC = ∠OCB ……….. (1)
(∵ OB and OC bisect ∠s B and C respectively.
∴ ∠OBC = \(\frac{1}{2}\)∠B and ∠OCB = \(\frac{1}{2}\)∠C)
⇒ OB = OC …….. (2) (∵ Sides opp. to equal ∠s are equal.)

(ii) Now, in ∆s ABO and ACO, we have:

∴ By SAS criterion of congruence, we have:
∆ABO ≅ ∆ACO
∠BAO = ∠CAO (CPCT)
⇒ AO bisects ∠BAC.

Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ABC is an isosceles triangle in which AB = AC.
Answer:

In ∆s ABD and ACD, we have:
DB = DC (Given)
∠ADB = ∠ADC (∵ AD ⊥ BC, so each = 90°)
AD = AD (Common)
∴ By SAS criterion of congruence, we have:
∆ABD ≅ ∆ACD
So, AB = AC
(∵ Corresponding parts of congruent triangles are equal.)
Hence, ∆ABC is an isosceles.

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.

Answer:
Let BE ⊥ AC and CF ⊥ AB.
In ∆s ABE and ACF, we have :
∠AEB = ∠AFC (∵ Each = 90°)
∠A = ∠A (Common)
and, AB = AC (Given)
∴ By AAS criterion of congruence, we have:
∆ABE ≅ ∆ACF
So, BE = CF
(∵ Corresponding parts of congruent triangles are equal.)

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that:

(i) ∆ABE ≅ ∆ACF
(ii) AB = AC, i.e. ABC is an isosceles triangle.
Answer:
(i) In ∆s ABE and ACF, we have:
∠AEB = ∠AFC (∵ Each = 90°)
∠BAE = ∠CAF (Common)
and, BE = CF (Given)
∴ By AAS criterion of congruence, we have:
∆ABE ≅ ∆ACF
So, AB = AC
(∵ Corresponding parts of congruent triangles are equal.)
Hence, ∆ABC is isosceles.

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.

Answer:
In ∆ABC, we have: AB = AC
∠ABC = ∠ACB ……… (1)
(∵ Angles opposite to equal sides are equal.)
In ADBC, we have : BD = CD
∠DBC = ∠DCB …………… (2)
(∵ Angles opposite to equal sides are equal.)
Adding (1) and (2), we have :
∠ABC + ∠DBC = ∠ACB + ∠DCB
⇒ ∠ABD = ∠ACD.

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.
Answer:

In ∆ABC, we have : AB = AC
∴ ∠ACB = ∠ABC …………… (1)
(∵ Angles opp. to equal sides are equal.)
Now, AB = AD (Given)
∴ AD = AC (∵ AD = AC)
∠ACD = ∠ADC ………. (2)
(∵ Angles opp. to equal sides are equal.)
Adding (1) and (2), we get:
∠ACB + ∠ACD = ∠ABC + ∠ADC
or ∠BCD = ∠ABC + ∠BDC (∵ ∠ADC = ∠BDQ)
So, ∠BCD + ∠BCD = ∠ABC + ∠BDC + ∠BCD
(Adding ∠BCD on both sides)
2 ∠BCD = 180° (Angle sum property)
or ∠BCD = 90°
Hence, ∠BCD is a right angle.
Hence proved.

Question 7.
ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C
Answer:
Given, AB = AC and ∠A = 90°
∴ ∠C = ∠B (Angles opposite to equal sides of a triangle)
In ∆ABC, ∠A + ∠B + ∠C = 180° (Angle sum property)
⇒ 90° + ∠B + ∠B = 180°
⇒ 2 ∠B = 180° – 90°
⇒ 2 ∠B = 90°
⇒ ∠B = 45°
∴ ∠B = ∠C = 45°

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Answer:
Let us take ∆ABC is an equilateral triangle.
Then, AB = BC = AC
Now, AB = AC ⇒ ∠C = ∠B (Angle opposite to equal sides of a triangle)
Again, AC = BC ⇒ ∠B = ∠A (Angle opposite to equal sides of a triangle)
So we have ∠A = ∠B = ∠C
Now in AABC ∠A + ∠B + ∠C = 180° (Angle sum property)
⇒ ∠A + ∠A + ∠A = 180°
⇒ 3 ∠A = 180°
⇒ ∠A = 60°
∴ ∠A = ∠B = ∠C = 60°
Hence, all interior angles in an equilateral triangle are of 60°.
Hence proved.