Question

In the figure ABCD is a parallelogram. If PC:PA=1:3 and ar(∆ BPC)=16cm^(2) then find ar( ∆ ADP).​

Answer 1

Step-by-step explanation:

ABCD is a parallelogram

AC is a diagonal

Diagonal of a Parallelogram divides the parallelogram into two equal area triangles

=> Area of triangle AB = Area of AADC

PC / P * A = 1/3 (P is point on AC) => Area of A BPC =(1/4) Area of A ABC

=>16=(1/4) Area of triangle ABC

=> Area of Delta*A * B * C = 64 cm ^ 2

Area of triangle ADC= Area of Delta ABC => Area of triangle ADC=64 cm^ 2

Area of triangle ADP=(3/4) Area of AADC

=> Area of triangle ADP=(3/4)64

=> Area of triangle ADP=48 cm^ 2

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