NCERT Solutions of Class 9 maths exercise 7.3 of chapter 7- Triangles - Future Study Point

NCERT Solutions of Class 9 maths exercise 7.3 – Triangle

NCERT Solutions of Class 9 maths exercise 7.3 of chapter 7- Triangles are created by futurestudypoint.com for helping the class 9 students of CBSE. These NCERT solutions are the solutions of unsolved questions of exercise 7.3 of chapter 7- Triangles of class 9 Maths NCERT textbook. The questions of Exercise 7.3 -Triangle are based on the congruency of the triangle when the median of a triangle, and the perpendicular bisector of the triangle are given. All the questions are solved by a step-by-step method so that each student could clear the concept of each question.

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Exercise 7.1- Triangles

Exercise 7.2- Triangle

Exercise 7.4- Triangle

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

Q1. ΔABC and ΔDBC are two isosceles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that

(i) ΔABD ≅ ACD

(ii)ΔABP ≅ ACP

(iii)AP bisects ∠A and as well as ∠D

(iv) AP is the perpendicular bisector of BC

Ans.

GIVEN: ΔABC and ΔDBC are two isosceles with a common base BC

In which AB = AC and AD = DC

(i) TO PROVE: ΔABD ≅ ACD

PROOF:AB = AC (given)

BD = CD (given)

ΔABD ≅ ACD (SSS rule)

Hence proved

(ii) TO PROVE:ΔABP ≅ ACP

PROOF: AB = AC (given)

AP = AP(common)

Since we have already proved ΔABD ≅ ACD

∴∠BAP = ∠PAC (By CPCT)

ΔABP ≅ ACP(SAS rule)

Hence proved

(iii TO PROVE: AP bisects ∠A and as well as ∠D

PROOF:In ΔDPB and ΔDPC

Since we have proved ΔABD ≅ ACD and ΔABP ≅ ACP

∴ BD = DC (By CPCT) and BP = PC (CPCT)

DP = DP (common)

ΔDPB ≅ΔDPC

∠DPB =∠DPC (By CPCT)…(i)

And it is already proved ΔABD ≅ ACD

From (i) and (ii) it is clear that AP bisects ∠A and as well as ∠D

Hence proved

(iv) TO PROVE: AP is the perpendicular bisector of BC

PROOF: Since we have proved ΔABP ≅ ACP

Therefore ∠APB = ∠APC (By CPCT)

BP = PC(By CPCT)…..(i)

∠APB + ∠APC = 180°

∠APB +∠APB = 180°

2 ∠APB = 180°

∠APB = 90°

∴ AP ⊥ BC……(ii)

From (i) and (ii), it is cleared that AP is a perpendicular bisector of BC

Hence proved

Q2.AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC  (ii) AD bisects ∠A.

Ans.

GIVEN: An isosceles triangle ΔABC  in which AB = AC

AD is the altitude of the isosceles triangle

TO PROVE: (i) AD bisects BC

In ΔABD and ΔACD

AB = AC (given)

ΔABD ≅ ΔACD(RHS rule)

BD = DC (By CPCT)

(ii) TO PROVE:  AD bisects ∠A.

Since we have proved ΔABD ≅ ΔACD

Q3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR see the given figure. Show that.

(i) ΔABM ≅Δ PQN

(ii)ΔABC ≅ ΔPQR

Ans.

GIVEN: In ΔABC and ΔPQR

AB = PQ

BC = QR

AM = PN

TO PROVE: (i) ΔABM ≅Δ PQN

PROOF: AB = PQ (given)

AM = PN (given)

BC = QR (given)

Since AM is the median of the triangle ΔABC and PN is the median of the triangle ΔPQR

∴BC/2 = QR/2

BM = QN

ΔABM ≅Δ PQN (SSS rule)

Hence proved

TO PROVE (ii)ΔABC ≅ ΔPQR

PROOF: Since we have already proved ΔABM ≅Δ PQN

∴∠ABM = ∠PQN (by CPCT)

AB = PQ (given)

BC = QR(given)

ΔABM ≅Δ PQN (SAS rule)

Hence proved

Q4. BE and CF are two equal altitudes of a triangle ABC. Using the RHS congruence rule, prove that the triangle ABC is isosceles.

Ans.

GIVEN: In ΔABC, in which two altitudes BE and CF are equal

BE = CF

∠CFB = ∠CEB = 90°

TO PROVE: triangle ABC is isosceles

In ΔCFB and ΔBEC

The hypotenuse of both right triangles is common

BC = BC (common)

∠CFB = ∠CEB = 90°(BE and CF are altitudes of both triangles)

BE = CF (equal)

ΔΔCFB≅ ΔBEC (RHS rule)

∠B = ∠C (by CPCT)

AB = AC (opposite sides of equal triangles)

Q5. ABC is an isoscles triangle with AB = AC . Draw AP⊥ BC to show that ∠B = ∠C.

Ans.

GIVEN: In triangle ABP and triangle ACP

AB = AC

∠APB = ∠APC = 90° (AP ⊥ BC)

TO PROVE:∠B = ∠C

PROOF: In ΔABP and ΔACP

AB = AC (given)

AP = AP (common)

∠APB = ∠APC = 90°(given)

ΔABP ≅ ΔACP (RHS rule)

∠B = ∠C  (by CPCT)

Hence proved

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