The area of any cyclic quadrilateral ABCD is given by `A^(2) = (s -a) (s-b) (s-c) (s-d)`, where `2s = a + b ++ c + d, a, b, c and d` are the sides of
The area of any cyclic quadrilateral ABCD is given by `A^(2) = (s -a) (s-b) (s-c) (s-d)`, where `2s = a + b ++ c + d, a, b, c and d` are the sides of the quadrilateral
Now consider a cyclic quadrilateral ABCD of area 1 sq. unit and answer the following question
The minium perimeter of the quadrilateral is
A. 4
B. 2
C. 1
D. none of these
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Correct Answer - A
Using A.M. `ge` G.M for `s -a, s-b, s-c, s-d`, we have
`(s-a + s-b + s-c + s-d)/(4) ge sqrtA`
or `(4s -2s)/(4) ge 1`
or `2s ge 4`
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