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A circle inscribed in an equilateral triangle with side

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A circle inscribed in an equilateral triangle with side [#permalink] New post 31 Dec 2005, 22:18
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A circle inscribed in an equilateral triangle with side length 20. What is the area of inscribed circle?


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Re: PS: Circle in Eqvilateral Triangle [#permalink] New post 31 Dec 2005, 22:27
vivek123 wrote:
A circle inscribed in an equilateral triangle with side length 20. What is the area of inscribed circle?
No OA available. Just try!

draw a triangle by adding the mid point of each side of the equilateral triangle. this new trangle, also equilateral trangle, is inscribed in the circle. the side of the new equi trangle is 10.
so r = a/sqrt 3 = 10/sqrt(3)

Area of the circle = pi (10/sqrt(3))^2 = (100/3) (pi)
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 [#permalink] New post 01 Jan 2006, 16:45
another way to solve this could be...

draw perpendicular bisectors of each angle. They should meet at the centroid of the triangle which is the center of the circle. the radius of the circle is 1/3 the length of the bisector.

or

length of bisector x = [(20)^2-(10)^2]^1/2 = 300^1/2

area = pi*300*1/3*1/3 = 100pi/3
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Re: PS: Circle in Eqvilateral Triangle [#permalink] New post 01 Jan 2006, 16:48
Professor wrote:
vivek123 wrote:
A circle inscribed in an equilateral triangle with side length 20. What is the area of inscribed circle?
No OA available. Just try!

draw a triangle by adding the mid point of each side of the equilateral triangle. this new trangle, also equilateral trangle, is inscribed in the circle. the side of the new equi trangle is 10.
so r = a/sqrt 3 = 10/sqrt(3)

Area of the circle = pi (10/sqrt(3))^2 = (100/3) (pi)



Pls explain "so r = a/sqrt 3" how u got this with working
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 [#permalink] New post 01 Jan 2006, 17:09
for a equilateral triangle with side 20, the height of the equilateral triangle is 10sqrt3.

the radius of the inscribed circle within this equilateral traingle is 1/3 of the height of the traingle

so radius is (10sqrt3)/3 = 10/(sqrt3)

area of the inscribed circle = pi* [10/(sqrt3)]^2
= pi*100/3
  [#permalink] 01 Jan 2006, 17:09
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