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Question Number 166640 by cherokeesay last updated on 23/Feb/22

	Answered by som(math1967) last updated on 24/Feb/22
	$Area with circle =2×{(π/6)×4^2 −(1/2)×((√3)/4)×4^2 }cm^2 =(((16π)/3) −4(√3))cm^2 let radius of circle =r ∴(4−r)^2 =r^2 +2^2 16−8r+r^2 =r^2 +4 ⇒r=(3/2)cm ∴ area of hached region =(((16π)/3) −4(√3)) −π×((3/2))^2 =(((37π)/(12)) −4(√3))cm^2$
	$A r e a w i t h c i r c l e$ $= 2 \times {\frac{π}{6} \times 4^{2} - \frac{1}{2} \times \frac{\sqrt{3}}{4} \times 4^{2}} {c m}^{2}$ $= (\frac{16 π}{3} - 4 \sqrt{3}) {c m}^{2}$ $l e t r a d i u s o f c i r c l e = r$ $∴ {(4 - r)}^{2} = r^{2} + 2^{2}$ $16 - 8 r + r^{2} = r^{2} + 4$ $\Rightarrow r = \frac{3}{2} c m$ $∴ a r e a o f h a c h e d r e g i o n$ $= (\frac{16 π}{3} - 4 \sqrt{3}) - π \times {(\frac{3}{2})}^{2}$ $= (\frac{37 π}{12} - 4 \sqrt{3}) {c m}^{2}$
	Commented by cherokeesay last updated on 24/Feb/22

	$v e r y n i c e$ $t h a n k y o u s i r .$
	Commented by Tawa11 last updated on 26/Feb/22

	$Great sir$
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