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Question Number 140533 by liberty last updated on 09/May/21
A triangle is inscribed in a circle. the vertices of the triangle divided the circumference of the circle into three area of length 6,8,10 units then the area of triangle is equal to... (a) ((64(√3)((√3)+1))/π^2 ) (c) ((36(√3)((√3)−1))/π^2 ) (b) ((72(√3)((√3)+1))/π^2 ) (d) ((36(√3)((√3)+1))/π^2 )
$$\mathrm{A}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}. \\ $$$$\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{divided} \\ $$$$\mathrm{the}\:\mathrm{circumference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle} \\ $$$$\mathrm{into}\:\mathrm{three}\:\mathrm{area}\:\mathrm{of}\:\mathrm{length}\:\mathrm{6},\mathrm{8},\mathrm{10} \\ $$$$\mathrm{units}\:\mathrm{then}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{triangle} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to}… \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{64}\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\pi^{\mathrm{2}} }\:\:\:\:\left(\mathrm{c}\right)\:\frac{\mathrm{36}\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\pi^{\mathrm{2}} } \\ $$$$\left(\mathrm{b}\right)\:\frac{\mathrm{72}\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\pi^{\mathrm{2}} }\:\left(\mathrm{d}\right)\:\frac{\mathrm{36}\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\pi^{\mathrm{2}} } \\ $$
Answered by mr W last updated on 09/May/21
perimeter of circle=6+8+10=24 radius of circle=((24)/(2π))=((12)/π) area of triangle: (1/2)r^2 (sin ((6×2π)/(24))+sin ((8×2π)/(24))+sin ((10×2π)/(24))) =(1/2)×((144)/π^2 )×(sin (π/2)+sin ((2π)/3)+sin ((5π)/6)) =(1/2)×((144)/π^2 )×(1+((√3)/2)+(1/2)) =((36(√3)((√3)+1))/π^2 )
$${perimeter}\:{of}\:{circle}=\mathrm{6}+\mathrm{8}+\mathrm{10}=\mathrm{24} \\ $$$${radius}\:{of}\:{circle}=\frac{\mathrm{24}}{\mathrm{2}\pi}=\frac{\mathrm{12}}{\pi} \\ $$$${area}\:{of}\:{triangle}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \left(\mathrm{sin}\:\frac{\mathrm{6}×\mathrm{2}\pi}{\mathrm{24}}+\mathrm{sin}\:\frac{\mathrm{8}×\mathrm{2}\pi}{\mathrm{24}}+\mathrm{sin}\:\frac{\mathrm{10}×\mathrm{2}\pi}{\mathrm{24}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{144}}{\pi^{\mathrm{2}} }×\left(\mathrm{sin}\:\frac{\pi}{\mathrm{2}}+\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{6}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{144}}{\pi^{\mathrm{2}} }×\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{36}\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\pi^{\mathrm{2}} } \\ $$
Commented by mr W last updated on 09/May/21
no answer is correct! maybe typo in the question, since (c) and (d) are the same.
$${no}\:{answer}\:{is}\:{correct}!\:{maybe}\:{typo}\:{in} \\ $$$${the}\:{question},\:{since}\:\left({c}\right)\:{and}\:\left({d}\right)\:{are}\:{the} \\ $$$${same}. \\ $$
Commented by liberty last updated on 09/May/21
oo yes. right
$$\mathrm{oo}\:\mathrm{yes}.\:\mathrm{right} \\ $$

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