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ABCD is a trapezium in which AB and CD are the parallel sides. The diagonals meet at point 0. If area of triangle ABO is p and area of triangle COD is q, prove that 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 . where v stands for squareroot
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 .
where v stands for squareroot
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