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Question Number 211953 by efronzo1 last updated on 25/Sep/24

Commented by BHOOPENDRA last updated on 25/Sep/24

$h = \frac{\sqrt{2} (\sqrt{a b} - a + b - c)}{\sqrt{(\sqrt{a b}) + b}}$
Commented by BHOOPENDRA last updated on 26/Sep/24

	Answered by BHOOPENDRA last updated on 26/Sep/24

	$△ E F G \sim △ B C G$ $a r e a r a t i o [E F G : B C G] = a : b$ $a n d a l t i t u d e P G : G Q = \sqrt{a} : \sqrt{b}$ $P G = \sqrt{a} n, G Q = \sqrt{b} n$ $P Q = (\sqrt{a} + \sqrt{b}) n$ $a r e a o f B C G = \frac{B C . G Q}{2} \Rightarrow \frac{(\sqrt{a} + \sqrt{b}) \sqrt{b} n^{2}}{2} = b$ $n = \sqrt{\frac{2 \sqrt{b}}{(\sqrt{a} + \sqrt{b})}} \Rightarrow P Q = \sqrt{2 (\sqrt{a b} + b)}$ $A r e a o f A B F = A B C D - a - b - c - \sqrt{a b}$ $A B F = \frac{h \sqrt{2 (\sqrt{a b} + b)}}{2}$ $h = \frac{\sqrt{2} (\sqrt{a b} - a + b - c)}{\sqrt{(\sqrt{a b} + b)}}$
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