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Question Number 121328 by Maclaurin Stickker last updated on 06/Nov/20

Commented by Maclaurin Stickker last updated on 06/Nov/20

$T h e s q u a r e h a s s i d e 1 . W h a t i s t h e$ $r e d a r e a ?$
	Answered by TANMAY PANACEA last updated on 06/Nov/20

	$\frac{1}{4} t h c i r c l e c e n t r e D (0, 0) x^{2} + y^{2} = 1^{2} s$ $\frac{1}{4} t h c i r c l e c e n t r e C (1, 0) {(x - 1)}^{2} + y^{2} = 1$ $s e m i c i r c l e c e n t r e (\frac{1}{2}, 1) {(x - \frac{1}{2})}^{2} + {(y - 1)}^{2} = {(\frac{1}{2})}^{2}$ $★ y = 1 + \sqrt{\frac{1}{4} - {(x - \frac{1}{2})}^{2}} ★$ $p o i n t F s o l v e$ ${(x - 1)}^{2} + y^{2} = 1 a n d {(x - \frac{1}{2})}^{2} + {(y - 1)}^{2} = \frac{1}{4}$ $x^{2} - 2 x + y^{2} = 0$ $x^{2} - x + \frac{1}{4} + y^{2} - 2 y + 1 = \frac{1}{4}$ $- x + 2 x - 2 y + 1 = 0 x = 2 y - 1$ $4 y^{2} - 4 y + 1 - 4 y + 2 + y^{2} = 0$ $5 y^{2} - 8 y + 3 = 0$ $5 y^{2} - 5 y - 3 y + 3 = 0$ $5 y (y - 1) - 3 (y - 1) = 0$ $y = 1 a n d \frac{3}{5} s o x = 1 a n d \frac{1}{5}$ $p o i n t F (\frac{1}{5}, \frac{3}{5}) G (\frac{4}{5}, \frac{3}{5})$ $p o i n t E$ $s o l v e x^{2} + y^{2} = 1$ ${(x - 1)}^{2} + y^{2} = 1$ $x^{2} = x^{2} - 2 x + 1 \to x = \frac{1}{2}$ $E (\frac{1}{2}, \frac{\sqrt{3}}{2})$ $s o a r e a o f (r e d p o r t i o n + s m a l l c i r c l e) =$ $\int_{\frac{1}{5}}^{\frac{1}{2}} \sqrt{1 - {(x - 1)}^{2}} d x + \int_{\frac{1}{2}}^{\frac{4}{5}} \sqrt{1 - x^{2}} d x - \int_{\frac{1}{5}}^{\frac{4}{5}} 1 + \sqrt{\frac{1}{4} - {(x - \frac{1}{2})}^{2}} d x$ $l e t t h e v a l u e o f t h i s i n t r e g a l s a y = p$ $a r e a o f s m a l l c i r c l e i s Q$ $s o a n s w e r w i l l b e = P - Q$
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