## RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2

Other Exercises

- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

Question 1.

BD and CE are bisectors of ∠B and ∠C of an isosceles ∠ABC with AB = AC. Prove that BD = CE.

Solution:

Given : In ∆ABC, AB = AC

BD and CE are the bisectors of ∠B and ∠C respectively

To prove : BD = CE

Proof: In ∆ABC, AB = AC

∴ ∠B = ∠C (Angles opposite to equal sides)

∴ \(\frac { 1 }{ 2 }\) ∠B = \(\frac { 1 }{ 2 }\) ∠C

∠DBC = ∠ECB

Now, in ∆DBC and ∆EBC,

BC = BC (Common)

∠C = ∠B (Equal angles)

∠DBC = ∠ECB (Proved)

∴ ∆DBC ≅ ∆EBC (ASA axiom)

∴ BD = CE

Question 2.

In the figure, it is given that RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that: ∆RBT = ∆SAT.

Solution:

Given : In the figure, RT = TS

∠1 = 2∠2 and ∠4 = 2∠3

To prove : ∆RBT ≅ ∆SAT

Proof : ∵ ∠1 = ∠4 (Vertically opposite angles)

But ∠1 = 2∠2 and 4 = 2∠3

∴ 2∠2 = 2∠3 ⇒ ∠2 = ∠3

∵ RT = ST (Given)

∴∠R = ∠S (Angles opposite to equal sides)

∴ ∠R – ∠2 = ∠S – ∠3

⇒ ∠TRB = ∠AST

Now in ∆RBT and ∆SAT

∠TRB = ∠SAT (prove)

RT = ST (Given)

∠T = ∠T (Common)

∴ ∆RBT ≅ ∆SAT (SAS axiom)

Question 3.

Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at O.

Solution:

Given : Two lines AB and CD intersect each other at O such that AD = BC and AD \(\parallel\)

BC

To prove : AB and CD bisect each other

i. e. AO = OB and CO = OD

Proof: In ∆AOD and ∆BOC,

AD = BC (Given)

∠A = ∠B (Alternate angles)

∠D = ∠C (Alternate angles)

∴ ∆AOD ≅ ∆BOC (ASA axiom)

AO = OB and AO = OC (c.p.c.t.)

Hence AB and CD bisect each other.

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2 are helpful to complete your math homework.

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