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Question Number 39811 by ajfour last updated on 11/Jul/18

Commented by ajfour last updated on 11/Jul/18

$T h e o u t e r f r a m e i s a u n i t s q u a r e .$ $F i n d a r e a s o f r e d, g r e e n, a n d$ $b l u e r e g i o n s i n t e r m s o f α .$
	Answered by MrW3 last updated on 11/Jul/18

	$\tan β = \frac{1 - 1 \times \tan α}{1} = 1 - \tan α$ $\Rightarrow β = \tan^{- 1} (1 - \tan α)$ $\Rightarrow \sin β = \frac{1 - \tan α}{\sqrt{1 + {(1 - \tan α)}^{2}}}$ $\Rightarrow \cos β = \frac{1}{\sqrt{1 + {(1 - \tan α)}^{2}}}$ $s i d e l e n g t h o f b l u e s q u a r e :$ $1 \times \tan α \times \cos β = \frac{\tan α}{\sqrt{1 + {(1 - \tan α)}^{2}}}$ $\Rightarrow A_{B l u e} = \frac{\tan^{2} α}{1 + {(1 - \tan α)}^{2}}$ $R e d t r i a n g l e :$ $s i d e a = 1 \times \sin α$ $s i d e b w i t h$ $\frac{b}{\sin α} = \frac{1}{\sin (α + β)}$ $\Rightarrow b = \frac{1 \times \sin α}{\sin (α + β)} = \frac{\sin α \sqrt{1 + {(1 - \tan α)}^{2}}}{\sin α + \cos α (1 - \tan α)} = \tan α \sqrt{1 + {(1 - \tan α)}^{2}}$ $\Rightarrow A_{R e d} = \frac{1}{2} \times \sin α \times \tan α \sqrt{1 + {(1 - \tan α)}^{2}} \times \cos (α + β)$ $\frac{1}{2} \times \sin α \times \tan α \times [\cos α - \sin α (1 - \tan α)]$ $\Rightarrow A_{R e d} = \frac{(2 - \sin 2 α) \tan^{2} α}{4}$ $G r e e n t r i a n g l e :$ $s i d e c w i t h$ $c = 1 \times \sin β - b = \frac{1 - \tan α}{\sqrt{1 + {(1 - \tan α)}^{2}}} - \tan α \sqrt{1 + {(1 - \tan α)}^{2}}$ $= \frac{1 - \tan α - \tan α [1 + {(1 - \tan α)}^{2}]}{\sqrt{1 + {(1 - \tan α)}^{2}}}$ $= \frac{1 - 2 \tan α - \tan α {(1 - \tan α)}^{2}}{\sqrt{1 + {(1 - \tan α)}^{2}}}$ $s i d e d w i t h$ $d = c \tan γ = c \tan (α + β) = c \frac{\tan α + (1 - \tan α)}{1 - \tan α (1 - \tan α)}$ $= \frac{1 - 2 \tan α - \tan α {(1 - \tan α)}^{2}}{(1 - \tan α + \tan^{2} α) \sqrt{1 + {(1 - \tan α)}^{2}}}$ $A_{G r e e n} = \frac{c d}{2} = \frac{1}{2} \times \frac{1 - \tan α - \tan α [1 + {(1 - \tan α)}^{2}]}{\sqrt{1 + {(1 - \tan α)}^{2}}} \times \frac{1 - 2 \tan α - \tan α {(1 - \tan α)}^{2}}{(1 - \tan α + \tan^{2} α) \sqrt{1 + {(1 - \tan α)}^{2}}}$ $\Rightarrow A_{G r e e n} = \frac{{[1 - 2 \tan α - \tan α {(1 - \tan α)}^{2}]}^{2}}{2 (1 - \tan α + \tan^{2} α) [1 + {(1 - \tan α)}^{2}]}$
	Commented by MrW3 last updated on 11/Jul/18

	Commented by ajfour last updated on 12/Jul/18

	$G r e a t! (p a t i e n c e a n d i d e a b o t h), S i r$ $T h a n k y o u . I w i l l a l s o t r y .$
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