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Question Number 203282 by mr W last updated on 14/Jan/24

Commented by cortano12 last updated on 14/Jan/24

$AD ∥ BC ?$
Commented by mr W last updated on 15/Jan/24

$s q u a r e i s i n s c r i b e d i n r i g h t a n g l e d$ $t r a p e z o i d$
	Answered by behi834171 last updated on 15/Jan/24

	$S P = x$ $\frac{\frac{x}{2} + \frac{\sqrt{3}}{2} x}{2} . 8 = \frac{\sqrt{3} x^{2}}{2} + \frac{1}{2} x^{2} \Rightarrow x = 4$ $S D = D R = y$ $x^{2} = 4 y^{2} \Rightarrow y = 2$ $R C = Q R = 4 \Rightarrow \frac{C R}{D R} \overset{?}{=} 2$ $b u t \bar{} i t s e e m s t h a t s o m e t h i n g w e n t w r o n g!$
	Commented by mr W last updated on 16/Jan/24

	$i n d e e d s o m e t h i n g w e n t w r o n g s i r .$ $p l e a s e r e c h e c k .$
	Answered by mr W last updated on 16/Jan/24

	Commented by mr W last updated on 16/Jan/24

	$A B F E i s a l s o a s q u a r e .$ $\Rightarrow A E = B F = A B = 8$ $F C = 11 - 8 = 3$ $D E = 8 - 6 = 2$ $\frac{C R}{D R} = \frac{F C}{D E} = \frac{3}{2} ✓$
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