In this chapter, we provide RD Sharma Solutions for Chapter 12 Heron’s Formula Ex 12.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 12 Heron’s Formula Ex 12.1 Maths pdf, free RD Sharma Solutions for Chapter 12 Heron’s Formula Ex 12.1 Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 12 |

Chapter Name | Heron’s Formula |

Exercise | Ex 12.1 |

**RD Sharma Solutions for Class 9 Chapter 12 Heron’s Formula Ex 12.1 Download PDF**

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)

∴ 12 ∠B = 12 ∠C

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2 – 1

∠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.

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2 – 2

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 ∥

BC

RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2 – 3

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.

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