Question

"Question 14 In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

Class 9 - Math - Areas of Parallelograms and Triangles Page 164"

Answer 1

Two Triangles on the same base and between the same parallels are equal in area.

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Given,

AP || BQ || CR

To Prove,

ar(∆AQC) = ar(∆PBR)

Proof:

ar(△AQB) = ar(△PBQ) — (i)

(On the same base BQ and between the same parallels AP and BQ.)
also,

ar(△BQC) = ar(△BQR) — (ii)

 (On the same base BQ and between the same parallels BQ and CR.)

Adding (i) and (ii),

ar(△AQB) + ar(△BQC) = ar(△PBQ) + ar(△BQR)

 ar(△ AQC) = ar(△ PBR)

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Hope this will help you...

Answer 2

ar(△AQB) = ar(△PBQ) — (i)

(On the same base BQ and between the same parallels AP and BQ.)

also,

ar(△BQC) = ar(△BQR) — (ii)

 (On the same base BQ and between the sameparallels BQ and CR.)

Adding (i) and (ii),

ar(△AQB) +ar(△BQC) =ar(△PBQ) +ar(△BQR)

 ar(△ AQC) = ar(△ PBR)