In Trapezium ABCD, diagonals intersect at O. A(∆ABO) is `p` and A(∆CDO) is `q`. Then prove area of Trapezium is (√p+√q)^2

Keep in the mind of the above figure, also I wont write the triangle symbol

given: area (AOB) = p = ax/2 and area (DOC) = q = by/2
proof: area (ABCD) = area(ABD) + area(ABC) – area(AOB) + area (DOC)
= a(x+y)/2 + a(x+y)/2 – p +q = ay + 2p-p+q …....... as ax = 2p
= p+q+ay ….....................(1)
similarly
area (ABCD) = area(DCA) + area(DCB) – area(DOC) + area (AOB)
= p + q + bx ….....................(2)

from (1) and (2) ay=bx = r (say)
r²= aybx = axby = 4pq


now putting this in equation (1)
we get

Hence Proved!!

Approved