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Question Number 138175 by ajfour last updated on 10/Apr/21

Commented by ajfour last updated on 10/Apr/21

$I f t h e y e l l o w a n d b l u e a r e a s a r e$ $e q u a l, f i n d e d g e l e n g t h o f t h e$ $b l u e s q u a r e . (r a d i u s o f c i r c l e = 1)$
	Answered by mr W last updated on 10/Apr/21

	Commented by mr W last updated on 11/Apr/21

	$p l e a s e s i r! i f o u n d n o b e t t e r w a y t o$ $s o l v e .$
	Commented by mr W last updated on 11/Apr/21

	$y e l l o w s e g m e n t a r e a = \frac{1}{2} (2 θ - \sin 2 θ)$ $= θ - \sin θ \cos θ$ $b l u e s q u a r e a r e a = a^{2}$ $θ - \sin θ \cos θ = a^{2}$ $\Rightarrow a = \sqrt{θ - \sin θ \cos θ}$ $\sqrt{1^{2} - {(\frac{a}{2})}^{2}} + 1 \times \cos θ = a$ $\sqrt{4 - a^{2}} + 2 \cos θ = 2 a$ $\Rightarrow \sqrt{4 - θ + \sin θ \cos θ} + 2 \cos θ = 2 \sqrt{θ - \sin θ \cos θ}$ $\Rightarrow θ \approx 1.3496 (= 77.326 °)$ $\Rightarrow a \approx 1.0656$
	Commented by ajfour last updated on 11/Apr/21

	$T h a n k y o u s i r, t h e p r e v i o u s$ $c u b e - c y l i n d e r q u e s t i o n i w i l l$ $a t t e m p t a g a i n s i r, t o m a t c h w i t h y o u r s o l u t i o n .$
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