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# In Fig.9.29, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums. Q.16

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How i solve the question of class 9th ncert math of Areas of Parallelograms and Triangles chapter of exercise 9.3of question no 16. I think it is very important question of class 9th give me the tricky way for solving this question In Fig.9.29, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

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### 1 Answer

1. Given,

ar(â–³DRC) = ar(â–³DPC)

ar(â–³BDP) = ar(â–³ARC)

To Prove,

ABCD and DCPR are trapeziums.

Proof:

ar(â–³BDP) = ar(â–³ARC)

â‡’ ar(â–³BDP) â€“ ar(â–³DPC) = ar(â–³DRC)

â‡’ ar(â–³BDC) = ar(â–³ADC)

ar(â–³BDC) = ar(â–³ADC).

âˆ´, ar(â–³BDC) and ar(â–³ADC) are lying in-between the same parallel lines.

âˆ´, AB âˆ¥ CD

ABCD is a trapezium.

Similarly,

ar(â–³DRC) = ar(â–³DPC).

âˆ´, ar(â–³DRC) andar(â–³DPC) are lying in-between the same parallel lines.

âˆ´, DC âˆ¥ PR

âˆ´, DCPR is a trapezium.

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