The area of a cyclic quadrilateral having lengths $13, 14, 3, 13$ is (approximately).

110

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90

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80

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120

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Solution

The correct option is C $80$
given, $a = 13, b = 14, c = 3, d = 13$
$S = \frac{a + b + c + d}{2} = \frac{13 + 14 + 3 + 13}{2} = 21.5$
$A = \sqrt{(S - a) (S - b) (S - c) (S - d)}$

$= \sqrt{(21.5 - 13) (21.5 - 14) (21.5 - 3) (21.5 - 13)}$

$= \sqrt{(8.5) (7.5) (18.5) (5.5)}$

$\approx 80$

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