ABCD is a trapezium with AB and CD are 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 (vp+vq) 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 .