In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1:2.
DP produced meets AB produced at Q. Given ar (CPQ) = 20 cm^2. Calculate :
(i) ar (CDP) (ii) ar (ll gm ABCD).
Answers
Step-by-step explanation:
ABCD is a parallelogram. P is a point on BC such that BP:PC=1:2. DP produced meets AB produced at Q and area of ∆CPQ=20cm^2, calculate area of ∆CDP?
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Ar( tri CPQ) = 20 cm² BP : PC = 1:2 . BP= x , PC = 2x so, AD = 3x
TO FIND : ar( triangle CDP)
CONSTRUCTION: Draw QR // BC, meeting extended DC at R. So, AQRD is a parallelogram. RQ = AD = 3x
Triangle PBQ ~ Tri PCD ( by AA similarity)
So, ratio of the areas of similar triangles = the ratio of the squares of the corresponding sides
=> area( tri PBQ) /area( tri PCD) = x² / 4x²
tri CPD ~ tri RQD ( by AA similarity)
=> tri CPD/ tri RQD = 2:3
Ar tri CPD = 4x² => ar tri RQD = 9x²
=> ar trap RQPC = 5x²
=> ar parallelogram RQBC = 5x² + x² = 6x²
=> ar tri BQC = 6x²/2 = 3x²
3x² - x² = 20
=> 2x² = 20
=> x² = 10
=> 4x² 40 cm²
=> ar ( tri CPD) = 40 cm²
