NCERT Solutions Class 9 Maths exercise 7.2 of the chapter 7-Triangles - Future Study Point

NCERT Solutions Class 9 Maths exercise 7.2 of the chapter 7-Triangles

Exercise 7.2 class 9 maths NCERT

NCERT Solutions Class 9 Maths exercise 7.2 of the chapter 7-Triangles

Exercise 7.2 class 9 maths NCERT

NCERT Solutions Class 9 Maths exercise 7.2 of the chapter 7-Triangles is based on the solutions of the questions related to congruency of triangles.All questions of the exercise 7.2 -Triangles are solved by an expert of CBSE Maths.All students of class 9 are required to go through each solutions for clearing their concept on Triangles.You can study here science and maths NCERT solutions and notes of each chapters from class 9-12, sample papers, solutions of previous year’s question papers,solutions of important questions of science and maths ,articles on science and maths,government entrance exams and other competitive entrance exams and online jobs.

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Exercise 7.1- Triangles

Exercise 7.3 – Triangle

Exercise 7.4-Triangle

NCERT Solutions of Class 9 Science : Chapter 1 to Chapter 15

Q1.In an isosceles triangle ABC, with AB = AC, the bisector of angle ∠B and ∠C, intersect each other at O. Join A to O. Show that:

(i) OB = OC  (ii) AO bisects ∠A

Ans.

Q1 exercise 7.2 class 9 maths

 

GIVEN: In ΔABC

AB = AC

BO is bisector of ∠B and CO is bisector of ∠C

TO PROVE:(i) OB = OC  (ii) AO bisects ∠A

PROOF:In ΔABC

AB = AC (given)

∠ABC = ∠ACB (angles opposite to equal sides)

It is given to us that BO is bisector of ∠B and CO is bisector of ∠C

1/2(∠ABC) = 1/2(∠ACB)

∠OBC = ∠OCB

OB = OC (sides opposite to equal angles)

Hence proved

(ii) OB = OC (proved above)

AO = AO (common)

AB = AC (given)

ΔABO ≅ΔACO (SSS rule)

∠OAB = ∠OCA (by CPCT)

Therefore AO is the bisector of ∠A.

Hence proved

Q2. In ΔABC  , AD is perpendicular bisector of BC (see the given figure). Show that ΔABC  is an isoscles triangle in which AB = AC.

Q2 exercise 7.2 class 9 maths

 

Ans.

GIVEN: In ΔABC  , AD is perpendicular bisector of BC

∴ BD = DC

∠ADB = ∠ADC = 90°

TO PROVE: AB = AC

PROOF:  In ΔABD and  ΔACD

∠ADB = ∠ADC = 90°(given)

AD = AD (common)

BD = DC(given)

ΔABD ≅ ΔACD (SAS rule)

AB = AC(by CPCT)

Hence ΔABC is an isosceles triangle in which AB = AC

Q3. ABC is an isoscles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure).Show that these altitudes are equal.

Exercise 7.2 Q3

 

 

Ans.

GIVEN: ΔABC in which AC = AB

CF ⊥ AB and CE ⊥ AC

∴∠AFC = ∠AEB = 90°

TO PROVE: CF = BE

PROOF: AC = AB (given)

∠AFC = ∠AEB = 90°(given)

∠A = ∠A (common)

ΔAEB ≅ AFC (AAS rule)

BE = CF(by CPCT)

Hence proved

Q4.ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that.

(i) ΔABE ≅ ACF

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

Exercise 7.2 Q3

Ans.

GIVEN: ΔABC in which BE = CF

CF ⊥ AB and CE ⊥ AC

∴∠AFC = ∠AEB = 90°

TO PROVE: AC = AB

(i) PROOF:BE = CF (given)

∠AFC = ∠AEB = 90°(given)

∠A = ∠A (common)

ΔAEB ≅ AFC (AAS rule)

(ii) AC = AB (by CPCT)

Hence proved

Q5. ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ABD = ∠ACD.

Q5. EX.7.2 class 9 maths

Ans.

GIVEN: ΔABC in which AB = AC

ΔDBC in which BD = DC

TO PROVE: ∠ABD = ∠ACD

PROOF: In Δ ABC

AB = AC (given)

∠ABC = ∠ACB ….(i) (angles opposite to equal sides)

BD = DC

∠DBC = ∠DCB….(ii) (angles opposite to equal sides)

Adding equation (i) and (ii)

∠ABC + ∠DBC = ∠ACB + ∠DCB

∠ABD = ∠ACD

Hence proved

Q6. ΔABC is an isosceles triangle in which AB = AC, side BA is produced to D such that AD = AB(see the given figure). Show that ∠BCD is a right angle.

Q6 ex.7.2 class 9 maths

Ans.

GIVEN:AB = AC

AD = AB

TO PROVE: ∠BCD = 90°

PROOF : AB = AC (given)

∠ABC = ∠ACB (angles opposite to equal sides)

Applying the angle sum property in ΔABC

∠ABC + ∠BAC + ∠ACB = 180°

∠BAC = 180°- (∠ABC + ∠ACB)

∠BAC and ∠DAC are lienear pair

So, ∠BAC + ∠DAC = 180°

180°- (∠ABC + ∠ACB) + ∠DAC = 180°

∠DAC = ∠ABC + ∠ACB =∠ACB + ∠ACB =2∠ACB

∠ACB = 1/2 ∠DAC…..(i)

AD = AC (given)

∠ADC = ∠ACD (angles opposite to equal sides)

Applying the angle sum property in ΔABC

∠ADC + ∠ACD + ∠DAC = 180°

∠ACD + ∠ACD + ∠DAC = 180°

2∠ACD + ∠DAC = 180°

∠DAC = 180° – 2∠ACD

∠ACD = 90° – 1/2 ∠DAC……..(ii)

Adding equation (i) and (ii)

∠ACB + ∠ACD  = 1/2 ∠DAC+  90° – 1/2 ∠DAC

∠BCD = 90°

Hence ΔBCD is right triangle

Q7. ABC is a right triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Ans.

Q7 ex. 7.2 class 9 maths

 

We are given triagle ABC in which

∠A = 90°

AB = AC

∠B = ∠C (angles opposite to equal sides)

∠B + ∠C + ∠A  = 180°

∠B + ∠B +90° = 180°

2∠B  + 90° = 180°

2∠B = 90°

∠B = 45°

∠C = 45°

Hence both of the angles ∠B and ∠C are of 45°

Q8. Show that the angles of an equilateral triangle are 60° each.

Ans.

Q8, ex. 7.2 class 9 maths

 

We are given that ΔABC is an equilateral triangle

AB = AC (sides of equilateral triangle)

∠C = ∠B …..(i)(angles opposite to equal sides)

BC = AC (sides of equilateral triangle)

∠B = ∠A …..(ii)(angles opposite to equal sides)

From equation (i) and (ii)

∠A = ∠B = ∠C

Applying angle sum property of the triangle in ΔABC

∠A +∠B +∠C  = 180°

∠A + ∠A + ∠A = 180°

3∠A  = 180°

∠A  = 180°/3

∠A  = 60°

∴∠A  = ∠B =∠C=60°

Therefore angles of an equilateral triangle are 60° each.

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