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Question Number 15642 by mrW1 last updated on 12/Jun/17

	Answered by ajfour last updated on 12/Jun/17

	$R e q u i r e d A r e a = S$ $l e t O b e o r i g i n .$ $A (a, 0); B (0, b); C (- c, 0);$ $D (0, - d); E (a + p, - d - q)$ $\frac{q + d}{p} = \frac{b}{a}, \frac{p + a}{q} = \frac{c}{d}$ $\Rightarrow a q + a d = b p . . . (i)$ $d p + a d = c q . . . (i i)$ $o r a c q + a d c = b p c a n d$ $a d p + a^{2} d = a c q$ $a d d i n g : a d c + a d p + a^{2} d = b p c$ $p (b c - a d) = a d (c - a)$ $p = \frac{a d (c + a)}{b c - a d} . . . (i i i)$ $(i) - (i i) g i v e s :$ $a q - d p = b p - c q$ $o r q (a + c) = p (b + d)$ $q = (\frac{b + d}{a + c}) p$ $s u b s t i t u t i n g f o r$ $p = \frac{a d (c + a)}{b c - a d}; q = \frac{a d (b + d)}{b c - a d}$ $N o w, S + \frac{p (d + q)}{2} + \frac{q (a + p)}{2}$ $= (a + p) (d + q)$ $\Rightarrow 2 S + d p + p q + a q + p q$ $= 2 a d + 2 a q + 2 d p + 2 p q$ $o r S = \frac{1}{2} (a q + d p + 2 a d)$ $S = \frac{1}{2} [\frac{a^{2} d (b + d)}{b c - a d} + \frac{{a d}^{2} (c + a)}{b c - a d} + 2 a d]$ $= \frac{a d}{2 (b c - a d)} (a b + a d + c d + a d + 2 b c - 2 a d)$ $S = \frac{a d}{2 (b c - a d)} (a b + 2 b c + c d)$ $\frac{c}{a} = \frac{A r e a (Δ B O C)}{A r e a (Δ A O B)} = 2$ $\frac{d}{b} = \frac{A r e a (Δ C O D)}{A r e a (Δ B O C)} = \frac{3}{2}$ $\Rightarrow \frac{a d}{b c} = \frac{(a / c)}{(b / d)} = \frac{(1 / 2)}{(2 / 3)} = \frac{3}{4}$ $a d = \frac{3}{4} (b c)$ $f r o m a r e a s \frac{a b}{2} = 1 \Rightarrow a b = 2$ $\frac{b c}{2} = 2 \Rightarrow b c = 4$ $\frac{c d}{2} = 3 \Rightarrow c d = 6$ $a n d a s a d = \frac{3}{4} (b c) \Rightarrow$ $a d = \frac{3}{4} \times 4 = 3$ $s u b s t i t u t i n g f o r t h e s e i n$ $e x p r e s s i o n o b t a i n e d f o r S,$ $S = \frac{a d}{2 (b c - a d)} (a b + 2 b c + c d)$ $= \frac{3}{2 (4 - 3)} \times (2 + 8 + 6)$ $S = 24 s q . u n i t s .$
	Commented by mrW1 last updated on 12/Jun/17

	$24 is correct!$
	Commented by ajfour last updated on 23/May/18

	$t h a n k s S i r . W h a t a n i c e q u e s t i o n,$ $S i r, w h e r e d o y o u g e t t h e m f r o m ?$
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