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An equilateral triangle that has an area of 9root3 is

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An equilateral triangle that has an area of 9root3 is [#permalink]

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27 Sep 2006, 18:02
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. An equilateral triangle that has an area of 9root3 is inscribed in a circle. What is the area of the circle?

A. 6 Pi B. 9 Pi C. 12 Pi D. 9 Pi Root3 E. 18 Pi Root3
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27 Sep 2006, 18:40
draw a perrpendicular bisector from any side to the center of circle and join vertices of any side with the center, you form a 30-60-90 triangle, one of the side of which is 6/2 where 6 is the side of equilateral triangle obtained using triangle area

that will give the hypotenuse of that triange or radius = 6/sqrt(3)

Area = pi * 36/3 = 12 pi
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Re: PS- area of circle [#permalink]

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27 Sep 2006, 18:59
lan583 wrote:
. An equilateral triangle that has an area of 9root3 is inscribed in a circle. What is the area of the circle?

A. 6 Pi B. 9 Pi C. 12 Pi D. 9 Pi Root3 E. 18 Pi Root3

c. 12pi

aet= area equ tria with side a
hence height h= sqrt3*a/2
aet = 9sqrt3 = (1/2)* (a*sqrt3*a/2)
= (1/2)*(a^2sqrt3/2) = a^2sqrt3/4
hence 9 = a^2/4
a = 6

2 ways to proceed from here -
1-
If an equilateral triangle is inscribed in a circle, then the square on the side of the triangle is triple the square on the radius of the circle.

so, a^2 = 3r^2
r^2 = a^2/3 = 36/3 = 12

area circle = 12pi

2-
since the equilateral triangle is inscribed in a circle, the centre of the circle bisects the height from vertex:opposite base in a ratio of 2:1

hence radius r = (2/3)*sqrt3*a/2 = 2sqrt3
area circle = pi(2sqrt3)^2 = 12pi
Re: PS- area of circle   [#permalink] 27 Sep 2006, 18:59
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