Geometry - Triangle circumscribed in a circle

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Geometry - Triangle circumscribed in a circle [#permalink]

23 Jan 2010, 11:50

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What is the measure of the radius of the circle that circumscribes a triangle whose sides measure 9, 40 and 41?
6
4
24.5
20.5
12.5
The correct choice is (D) and the correct answer is 20.5.

Please somebody explain how....

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Re: Geometry - Triangle circumscribed in a circle [#permalink]

23 Jan 2010, 12:18

shuj00 wrote:

Triangle with sides 9, 40 and 41 is a right triangle

40^2+9^2 = 41^2

A right triangle incribed inside a circle - longest side of the triangle will be the diameter of the circle.

so dia of circle is 41 and so the redius is 20.5

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Re: Geometry - Triangle circumscribed in a circle [#permalink]

23 Jan 2010, 12:24

First, you can determine that you have a right triangle:

40^2 + 9^2 = 41^2

Because you have a right triangle, the 'circumcenter' of is on the hypotenuse. What this means is that the center of the circle lies on the hypotenuse of the right triangle. Because the 'circumcenter' is determined by the perpendicular bisectors of the midpoint of each side of the right triangle, we know that the circumcenter is at the midpoint of the hypotenuse.

Because we know that the hypotenuse of the right triangle goes through the center of the circle, the length of the hypotenuse is also the diameter of the circle. So, the radius is half the length of this side of the triangle.

Thus, 41 / 2 = 20.5.

There is a much better explanation on Wikipedia - search for 'Circumscribed circle' (I can't post links yet).
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Re: Geometry - Triangle circumscribed in a circle [#permalink] 23 Jan 2010, 12:24

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Geometry - Triangle circumscribed in a circle

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Geometry - Triangle circumscribed in a circle

Geometry - Triangle circumscribed in a circle

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