Question

in the given figure, ABCD is a parallelogramin which p and q are lying on the sides AB and BC respectively then ar(Δ AQD) is equal to

Answer 1

Given, ABCD is a parallelogram in which points P and Q trisects the BC.

⇒ BP = PQ = QC

To prove:



Construction: Through P and Q, draw PR and QS parallel to AB and CD.

Proof:

Now, ΔAPD and ΔAQD lie on the same base AD and between the same parallel AD and BC.

∴ ar (ΔAPD) = ar (ΔAQD)

⇒ ar (ΔAPD) - ar (ΔAOD) = ar (ΔAQD) - ar (ΔAOD)  [on subtracting ar (ΔAOD) from both sides]

 ⇒ ar (ΔAPO) =  ar (ΔOQD)  ..... (1)

 ⇒ ar (ΔAPO) + ar (ΔOPQ) =  ar (ΔOQD) + ar (ΔOPQ)  [on adding ar (ΔOPQ) on both sides]

 ⇒ ar (ΔAPQ) = ar (ΔDPQ)  ..... (2)

Again, ΔAPQ and parallelogram PQSR are on the same base PQ and between same parallels PQ and AD

 .... (3)

Now,



Using (2), (3) and (4), we get



[Hence proved]