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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

Semicircleareawithradiusa=12πa2Whitepart=5sectorswithradiusa/2&2equilateraltriangleswithsidesa/2=5×16π(a2)2+2×34(a2)2={5×16π+2×34}(a2)2=(5π6+32)(a24)A=Semicirclewhitearea=12πa2(5π6+32)(a24)=a2{12π(5π6+32)(14)}=a2{12π(5π24+38)}=a2{π25π2438}=a2{12π5π3324}=7π3324a2

Commented by I want to learn more last updated on 17/Aug/20

Thanks sir,  i appreciate.

Thankssir,iappreciate.

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