• If side lengths of a quadrilateral are a, b, c and d then its area is not constant. By changing internal angles, we can change area of the quadrilateral. For example, take the case of a square with side length 5 whose area is 25 square units. Now keeping the side lengths constant, if we tilt it to turn it into a rhombus, area goes o decreasing which can ultimately minimized to zero when the sides overlap with each other. So, for a given set of four side lengths, Maximum Area is obtained for a cyclic quadrilateral which is given by √ [(s - a)(s - b)(s - c)(s - d)] (BRAHMGUPTA’s formula) , where s is semiperimeter i.e. (a + b + c + d)/2.

Remember that for every four side lengths which form a quadrilateral, a cyclic quadrilateral can be formed.

If D1 and D2 are the two diagonals of a quadrilateral which intersects at angle P, then it’s area is given by ½D1D2sinP. And this area will be largest when diagonals intersect at right angle.

Ptolemy’s inequality

For a quadrilateral with side lengths a, b, c, d and diagonals of length D1 and D2,
we have D1D2 < = ac + bd

Equality arises in case of a cyclic quadrilateral.

If two diagonals divide a quadrilateral in four triangles, as shown,

Whose areas are A, B, C and D in order, then

A × C = B × D

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