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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

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

Grade:12

1 Answers

Riddhish Bhalodia
askIITians Faculty 434 Points
5 years ago
512-791_2016-01-13-000851.jpg
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
r = 2\sqrt{pq}

now putting this in equation (1)
we get
area(ABCD) = p+q+2\sqrt{pq} = (\sqrt{p} + \sqrt{q})^2
Hence Proved!!

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