Question

PQRS is a parallelogram. A is any point on SR. PA is produced to meet QR produced at B. Prove that area (ΔQAR) = area (ΔSAB).

Answer 1

We know that,

If a triangle and parallelogram are drawn on the same base and between the same parallel lines, then the area of the triangle is half of the area of the parallelogram.

Since ΔPAQ and || gm PQRS are on the same base PQ and between the same || lines PQ and RS.

⇒ ar(ΔPAQ) = `1/2` ar(|| gm PQRS)   ...(i)

From figure, ar(|| gm PQRS) = ar(ΔPAS) + ar(ΔPAQ) + ar(ΔQAR)

Using (i), ar(|| gm PQRS) = ar(ΔPAS) + `1/2` ar(|| gm PQRS) + ar(ΔQAR)

⇒ `1/2` ar(|| gm PQRS) = ar(ΔPAS) + ar(ΔQAR)   ...(ii)

Also, ΔPBS and || gm PQRS are on the same base PS and between the same || lines PS and BQ.

⇒ ar(ΔPBS) = `1/2` ar(|| gm PQRS)

⇒ ar(ΔPAS) + ar(ΔSAB) = `1/2` ar(|| gm PQRS)   ...(iii)

From (ii) and (iii), we get

ar(ΔPAS) + ar(ΔQAR) = ar(ΔPAS) + ar(ΔSAB)

⇒ ar(ΔQAR) = ar(ΔSAB)

Hence proved.