Q.5 ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect

at O. The area of the triangle ABO is p and that of the triangle CDO is q.

Prove that the area of the trapezium is (√p +√q)²

Let h and g are the heights of the triangles AOB and COD respectively.Then, 1/2 x AB x h = p and 1/2 x CD x g = q so, AB = 2p/h and 2q/g also, p/q = AB²/CD² so, AB/CD = vp/vq

Area of trapezium = 1/2(sum of parallel sides) x distance between them

                        =1/2 (AB + CD) x (h + g)

                        = 1/2 (vp/vqCD + CD) x (2p/AB +2q/CD)

Now replacing AB by 2p x vq/vp x CD and then solving the equation we get the area as (vp +vq)² .