Question

ABCD is a trapezium in which AB//DC.If P is the point of intersection of diagonals AC and BD such that area of triangle APB =32 cm^2 and area of triangle DPC =50 cm^2, then find the area of trapezium ABCD

Answer 1

The area of Trapezium = 1/2(AB+CD).H 

Let AB = 4x & CD = 5x (since you already found that sides are in 4:5) 

Let the heights be h1 & h2 respectively. 

Then H = h1 + h2 

Area = 1/2 * (4x+5x)(h1+h2) 
= 1/2 * 9x * h1 + 1/2 * 9x * h2 

1/2 * 4x*h1 = 32 
1/2 * 9x*h1 = 32 * 9/4 = 72 

1/2 * 5x * h2 = 50 
1/2 * 9x * h2 = 50 * 9/5 = 90 

Area = 72 + 90 = 162