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 )


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Question Number 140533 by liberty last updated on 09/May/21

$A triangle is inscribed in a circle .$ $the vertices of the triangle divided$ $the circumference of the circle$ $into three area of length 6, 8, 10$ $units then the area of triangle$ $is equal to . . .$ $(a) \frac{64 \sqrt{3} (\sqrt{3} + 1)}{π^{2}} (c) \frac{36 \sqrt{3} (\sqrt{3} - 1)}{π^{2}}$ $(b) \frac{72 \sqrt{3} (\sqrt{3} + 1)}{π^{2}} (d) \frac{36 \sqrt{3} (\sqrt{3} + 1)}{π^{2}}$
	Answered by mr W last updated on 09/May/21

	$p e r i m e t e r o f c i r c l e = 6 + 8 + 10 = 24$ $r a d i u s o f c i r c l e = \frac{24}{2 π} = \frac{12}{π}$ $a r e a o f t r i a n g l e :$ $\frac{1}{2} r^{2} (\sin \frac{6 \times 2 π}{24} + \sin \frac{8 \times 2 π}{24} + \sin \frac{10 \times 2 π}{24})$ $= \frac{1}{2} \times \frac{144}{π^{2}} \times (\sin \frac{π}{2} + \sin \frac{2 π}{3} + \sin \frac{5 π}{6})$ $= \frac{1}{2} \times \frac{144}{π^{2}} \times (1 + \frac{\sqrt{3}}{2} + \frac{1}{2})$ $= \frac{36 \sqrt{3} (\sqrt{3} + 1)}{π^{2}}$
	Commented by mr W last updated on 09/May/21

	$n o a n s w e r i s c o r r e c t! m a y b e t y p o i n$ $t h e q u e s t i o n, s i n c e (c) a n d (d) a r e t h e$ $s a m e .$
	Commented by liberty last updated on 09/May/21

	$oo yes . right$
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