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In Fig.9.28, AP || BQ || CR. Prove that ar(△AQC) = ar(△PBR). Q.14

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I want to know the best answer of the question from Areas of Parallelograms and Triangles chapter of class 9th ncert math. The question from exercise 9.3of math. Give me the easy way for solving this question of  14 In Fig.9.28, AP || BQ || CR. Prove that ar(△AQC) = ar(△PBR).

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  1. Given,

    AP || BQ || CR

    To Prove,

    ar(AQC) = ar(PBR)

    Proof:

    ar(△AQB) = ar(△PBQ) — (i) (Since they are on the same base BQ and between the same parallels AP and BQ.)

    also,

    ar(△BQC) = ar(△BQR) — (ii) (Since they are 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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