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Question Number 186111 by Rupesh123 last updated on 01/Feb/23

Commented by Rupesh123 last updated on 01/Feb/23
A right angle is rotated and translated as shown in the diagram
Commented by mr W last updated on 01/Feb/23

${i t}^{'} s n o t a f i x e d v a l u e, i t d e p e n d s o n t h e$ $s h a p e o f t h e t r i a n g l e .$
	Answered by HeferH last updated on 01/Feb/23

	Commented by HeferH last updated on 01/Feb/23

	$I c o n s i d e r t h e e n t i r e d i a g r a m a s$ $‘ ‘ e n t i r e {s h a p e}^{″} :)$
	Commented by HeferH last updated on 01/Feb/23

	$A r e a s h a d e d : 2 S$ $A r e a o f t h e e n t i r e s h a p e : 6 S$ $\frac{2 S}{6 S} = \frac{1}{3}$
	Commented by mr W last updated on 01/Feb/23

	$\frac{2 S}{4 S} = \frac{1}{2}$
	Answered by mr W last updated on 01/Feb/23

	Commented by mr W last updated on 01/Feb/23

	$θ = t h e s m a l l e r a n g l e$ $t o t a l a r e a o f t r i a n g l e$ $A = \frac{c^{2} \sin θ \cos θ}{2}$ $c a s e 0 < θ ⩽ 30 °$ $\frac{x}{c (1 - \sin θ)} = \frac{\sin (\frac{π}{2} - θ)}{\sin (\frac{π}{2} - 2 θ)} = \frac{\cos θ}{\cos 2 θ}$ $A_{S} = \frac{c^{2} {(1 - \sin θ)}^{2} \sin θ \cos θ}{2 \cos 2 θ}$ $A_{1} = A - A_{S}$ $\frac{A_{1}}{A} = 1 - \frac{A_{S}}{A} = 1 - \frac{{(1 - \sin θ)}^{2}}{\cos 2 θ} ⩽ \frac{1}{2}$ $c a s e 30 ° ⩽ θ ⩽ 45 °$ $A_{S} = \frac{{(c \cos θ)}^{2} \tan θ}{4} = \frac{c^{2} \sin θ \cos θ}{4} = \frac{A}{2}$ $A_{1} = \frac{A}{2}$ $\Rightarrow \frac{A_{1}}{A} = \frac{1}{2}$
	Commented by mr W last updated on 01/Feb/23

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