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Question Number 108483 by I want to learn more last updated on 17/Aug/20

Commented by Rasheed.Sindhi last updated on 17/Aug/20

Commented by Rasheed.Sindhi last updated on 17/Aug/20
$Semicircle area with radius a =(1/2)πa^2 White part= 5 sectors with radius a/2 & 2 equilateral triangles with sides a/2 =5×(1/6)π((a/2))^2 +2×((√3)/4)((a/2))^2 ={5×(1/6)π+2×((√3)/4)}((a/2))^2 =(((5π)/6)+((√3)/2))((a^2 /4)) A=Semicircle−white area =(1/2)πa^2 −(((5π)/6)+((√3)/2))((a^2 /4)) =a^2 {(1/2)π−(((5π)/6)+((√3)/2))((1/4))} =a^2 {(1/2)π−(((5π)/(24))+((√3)/8))} =a^2 {(π/2)−((5π)/(24))−((√3)/8)} =a^2 {((12π−5π−3(√3))/(24))} =((7π−3(√3))/(24))a^2$
$S e m i c i r c l e a r e a w i t h r a d i u s a$ $= \frac{1}{2} π a^{2}$ $W h i t e p a r t =$ $5 s e c t o r s w i t h r a d i u s a / 2$ $&$ $2 e q u i l a t e r a l t r i a n g l e s w i t h$ $s i d e s a / 2$ $= 5 \times \frac{1}{6} π {(\frac{a}{2})}^{2} + 2 \times \frac{\sqrt{3}}{4} {(\frac{a}{2})}^{2}$ $= {5 \times \frac{1}{6} π + 2 \times \frac{\sqrt{3}}{4}} {(\frac{a}{2})}^{2}$ $= (\frac{5 π}{6} + \frac{\sqrt{3}}{2}) (\frac{a^{2}}{4})$ $A = S e m i c i r c l e - w h i t e a r e a$ $= \frac{1}{2} π a^{2} - (\frac{5 π}{6} + \frac{\sqrt{3}}{2}) (\frac{a^{2}}{4})$ $= a^{2} {\frac{1}{2} π - (\frac{5 π}{6} + \frac{\sqrt{3}}{2}) (\frac{1}{4})}$ $= a^{2} {\frac{1}{2} π - (\frac{5 π}{24} + \frac{\sqrt{3}}{8})}$ $= a^{2} {\frac{π}{2} - \frac{5 π}{24} - \frac{\sqrt{3}}{8}}$ $= a^{2} {\frac{12 π - 5 π - 3 \sqrt{3}}{24}}$ $= \frac{7 π - 3 \sqrt{3}}{24} a^{2}$
Commented by I want to learn more last updated on 17/Aug/20

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