Question

In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2. DP produced meets AB produces at Q. Given the area of triangle CPQ = 20 cm².

Calculate:

1. area of triangle CDP,
2. area of parallelogram ABCD.
   

Answer 1

i. In ΔBPQ and ΔCPD

∠BPQ = ∠CPD   ...(Vertically opposite angles)

∠BQP = ∠PDC  ...(Alternate angles)

ΔBPQ ~ ΔCPD   ...(AA similarity)

∴ `(BP)/(PC) = (PQ)/(PD) = (BQ)/(CD) = 1/2`   ...`(∵ (BP)/(PC) = 1/2)`

Also, `(ar(ΔBPQ))/(ar(ΔCPD)) = ((BP)/(PC))^2`

`=> 10/(ar(ΔCPD)) = 1/4`   ...`[ar(ΔBPQ) = 1/2 xx ar(ΔCPQ), ar(CPQ) = 20]`

`=>` ar(∆CPD) = 40 cm²

ii. In ∆BAP and ∆AQD

As BP || AD, corresponding angles are equal

∠QBP = ∠QAD

∠BQP = ∠AQD  ...(Common)

∆BQP ~ ∆AQD   ...(AA similarity)

∴ `(AQ)/(BQ) = (QD)/(QP) = (AD)/(BP) = 3`   ...`(∵ (BP)/(PC) = (PQ)/(PD) = 1/2 => (PQ)/(QD) = 1/3)`

Also, `(ar(ΔAQD))/(ar(ΔBQP))=((AQ)/(BQ))^2`

`=> (ar(ΔAQD))/10 = 9`

`=>` ar(∆AQD) = 90 cm²

∴ ar(ADPB) = ar(∆AQD) – ar(∆BQP)

= 90 cm² – 10 cm²

= 80 cm²

ar(ABCD) = ar(∆CDP) + ar(ADPB)

= 40 cm² + 80 cm²

= 120 cm²