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Question Number 206111 by cortano21 last updated on 07/Apr/24

Commented by cortano21 last updated on 07/Apr/24


	Answered by nikif99 last updated on 07/Apr/24

	$L e t a : s i d e o f s q u a r e .$ $P R = \frac{a}{4}$ $I n o r t h o g o n a l a x i s s y s t e m y A x :$ $E q u a t i o n o f A S : y = 2 x$ $E q . o f D B : y = - x + a$ $(x, y) = (Q_{x}, Q_{y}) = (\frac{a}{3}, \frac{2 a}{3})$ $Q T = Q_{y} - T_{y} = \frac{2 a}{3} - \frac{a}{2} = \frac{a}{6}$ $A_{P Q R} = \frac{1}{2} (P R) (Q T) = \frac{1}{2} \cdot \frac{a}{4} \cdot \frac{a}{6} = 20 \Rightarrow$ $a^{2} = 960$
	Commented by nikif99 last updated on 07/Apr/24

	Answered by mr W last updated on 07/Apr/24

	Commented by mr W last updated on 07/Apr/24

	$F P = D E = \frac{D C}{2}$ $F R = \frac{D E}{2} = R P$ $\frac{R Q}{Q E} = \frac{R P}{D E} = \frac{1}{2}$ $\Rightarrow [P Q R] = \frac{1}{3} [E R P] = \frac{1}{3} \times \frac{[A B C D]}{16}$ $\Rightarrow [A B C D] = 48 \times [P Q R] = 48 \times 20 = 960$
	Answered by A5T last updated on 07/Apr/24

	Commented by A5T last updated on 07/Apr/24

	${D D}^{'} = x \Rightarrow R P = \frac{x}{2} \Rightarrow [{D D}^{'} Q] = 2^{2} \times 20 = 80$ $\Rightarrow [Q A B] = 2^{2} \times 80 = 320$ $\Rightarrow [A R P B] = 300 = 2 x^{2} - (\frac{x^{2}}{4} + \frac{x^{2}}{2}) = \frac{5 x^{2}}{4} \Rightarrow x^{2} = 240$ $[A B C D] = 4 x^{2} = 960$
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