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Question Number 211167 by cherokeesay last updated on 30/Aug/24

Commented by Frix last updated on 30/Aug/24

$12$
	Answered by som(math1967) last updated on 30/Aug/24

	Commented by cherokeesay last updated on 30/Aug/24

	$n i c e!$ $t h a n k y o u s i r!$
	Commented by som(math1967) last updated on 30/Aug/24

	$l e t A B = x, A D = y$ $\Rightarrow A E = \frac{18}{y}$ $∴ B E = x - \frac{18}{y}$ $C F = \frac{10}{x} \Rightarrow B F = y - \frac{10}{x}$ $∴ (x - \frac{18}{y}) (y - \frac{10}{x}) = 8$ $\Rightarrow x y - 10 - 18 + \frac{180}{x y} = 8$ $\Rightarrow {(x y)}^{2} - 36 x y + 180 = 0$ $\Rightarrow (x y - 30) (x y - 6) = 0$ $\Rightarrow x y = 30 [x y \neq 6 ∵ x y > 9]$ $O r a n g e a r e a = 30 - (9 + 5 + 4)$ $= 12 s q u n i t$
	Answered by mr W last updated on 30/Aug/24

	$A_{o r a n g e} = \sqrt{{(A + B + C)}^{2} - 4 A C}$ $= \sqrt{{(9 + 4 + 5)}^{2} - 4 \times 9 \times 5} = 12$
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