Question

Two parallogram are given to prove their area of PQRS is equal to area of PXYZ.

Answer 1

Sol: Let QR and BC intersect at O. Join CQ, QB, BR and RC. ar(//gm PQRS) = ar(//gm PABC) ⇒ ar(//gm PQOC) + ar(//gm CORS) = ar(//gm PQOC ) + ar(//gm QABO) ⇒ ar(//gm CORS) = ar(//gm QABO) ⇒1/2 ar(//gm CORS) = 1/2 ar(//gm QABO) ⇒ ar (ΔCOR) = ar (ΔQOB) ⇒ ar (ΔCOR) + ar (ΔCQO)  = ar (ΔQOB) + ar (ΔCQO) ⇒ ar (ΔCQR) = ar (ΔCQB)

ΔCQR and ΔCQB are on the same base CQ and have equal areas.

Therefore, ΔCQR and ΔCQB lie between the same parallels QC and BR. Hence, QC // BR

Answer 2

Let QR and BC intersect at O.

Join CQ, QB, BR and RC.

ar(//gm PQRS) = ar(//gm PABC)

⇒ ar(//gm PQOC) + ar(//gm CORS) = ar(//gm PQOC ) + ar(//gm QABO)

⇒ ar(//gm CORS) = ar(//gm QABO)

⇒1/2 ar(//gm CORS) = 1/2 ar(//gm QABO)

⇒ ar (ΔCOR) = ar (ΔQOB)

⇒ ar (ΔCOR) + ar (ΔCQO)  = ar (ΔQOB) + ar (ΔCQO)

⇒ ar (ΔCQR) = ar (ΔCQB)

ΔCQR and ΔCQB are on the same base CQ and have equal areas.

Therefore, ΔCQR and ΔCQB lie between the same parallels QC and BR.

Hence, QC // BR           for triangle pqrs and pabc samjh lena mere pas solve kiya hua tha to anm change kar lena