ABCD is a a parallelogram P is a point on BC such that BP:PC=1:2 DP produced meets AB produced - Maths - Area Theorems

Question

ABCD is a a parallelogram P is a point on BC such that BP:PC=1:2 DP
produced meets AB produced at Q given ar(?CPQ)=20centimeter square
Calculate:(i)ar(?CDP) (ii)ar(parallelogram ABCD) Expert please
answer it urgent

Eram Naaz · Accepted Answer

Dear student,
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"https://s3mn%20mnimgs%20com/img/shared/content_ck_images/images/1(40)%20png">

Draw CL ⊥ DQArea of △ CPQ = 12 × base × altitudeArea of △ CPQ = 12 × PQ × CL  
1Area of △ DCP = 12 × base × altitudeArea of △ DCP = 12 × DP × CL  
2Dividing 1 by 2, we get,Area of △ CPQArea of △ DCP  = PQDP  
3Since DC ∥ AQ and BC is a transversal⇒ ∠ DCP = ∠ QBP   Alternate interior anglesIn △ DCP and △ QBP,∠ DCP = ∠QBP  Proved above∠ DPC = ∠ QPB  Vertically opposite angles⇒ △ DCP ~ △ QBP  AA  similarity⇒ PCBP = DPPQ = CDBQ   Corresponding sides of similar △'s are proportional⇒ PCBP = DPPQ⇒ 21 = DPPQ⇒ PQDP = 12  
4Now, from 3, Area of △ CPQArea of △ DCP  = PQDP = 12⇒ 20Area △ DCP = 12⇒ Area △ DCP = 40 cm2

Since ∥gm ABCD & △ DCQ have same base DC & between same parallels DC & AQ, then,area of  ∥gm ABCD = 2 × ar △ DCQ                             = 2 × ar △ DCP + ar △ CPQ                             = 2 × 40 + 20                             = 2 × 60                             = 120 cm2
Regards

ABCD is a a parallelogram P is a point on BC such that BP:PC=1:2. DP produced meets AB produced at Q. given ar(?CPQ)=20centimeter square. Calculate:(i)ar(?CDP) (ii)ar(parallelogram ABCD). Expert please answer it urgent.