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Question Number 206292 by cortano21 last updated on 11/Apr/24

	Answered by A5T last updated on 11/Apr/24

	$L e t A E = x; B E = y; B F = v; F C = w$ $S + 39 = \frac{(2 v + w) (x + y)}{2} = \frac{2 v (x + y)}{2} + 15 \Rightarrow S = v x$ $w x = 54 - S = 30 - y w \Rightarrow y w = S - 24$ $\frac{[B F D]}{[D F C]} = \frac{v}{w} \Rightarrow [B F D] = \frac{15 v}{w}$ $\frac{[E D B]}{[A D E [} = \frac{y}{x} \Rightarrow E D B = \frac{27 y}{x}$ $\Rightarrow \frac{15 v}{w} + \frac{27 y}{x} = S + 12 \Rightarrow \frac{15 S + 27 (S - 24)}{54 - S} = S + 12$ $\Rightarrow S = 36$
	Answered by mr W last updated on 11/Apr/24

	Commented by mr W last updated on 11/Apr/24

	$(a + b) c = 2 A \Rightarrow a c = 2 A - b c$ $b (c + d) = 2 C \Rightarrow b d = 2 C - b c$ $a d = 2 B$ $a b c d = (2 A - b c) (2 C - b c)$ $2 B (b c) = 4 A C - 2 (A + C) (b c) + {(b c)}^{2}$ $4 A C - 2 (A + B + C) (b c) + {(b c)}^{2} = 0$ $\Rightarrow b c = (A + B + C) - \sqrt{{(A + B + C)}^{2} - 4 A C}$ $S = (a + b) (c + d) - (A + B + C)$ $S = \frac{4 A C}{b c} - (A + B + C)$ $S = \frac{4 A C}{(A + B + C) - \sqrt{{(A + B + C)}^{2} - 4 A C}} - (A + B + C)$ $S = (A + B + C) + \sqrt{{(A + B + C)}^{2} - 4 A C} - (A + B + C)$ $\Rightarrow S = \sqrt{{(A + B + C)}^{2} - 4 A C}$ $= \sqrt{{(27 + 12 + 25)}^{2} - 4 \times 27 \times 15} = 36$
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