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Radius of Circle

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by kartikshah » Fri Jul 27, 2012 6:11 pm
What is the radius of the circle that circumscribes the triangle, measure of whose sides is 9, 40 and 41?

A. 20

B. 20.5

C. 4

D. 8

E. 45
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by Birottam Dutta » Fri Jul 27, 2012 6:22 pm
The triangle specified here is a right angled triangle (9^2 + 40^2 = 41^2).

Always remember that in a right triangle, the hypotenuse is the diameter of the circumcircle and the centre is exactly the mid point of the hypotenuse.

So, diameter here is 41 and so, radius = 41/2 = 20.5

Answer B!
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by tisrar02 » Fri Jul 27, 2012 6:35 pm
Birottam Dutta wrote:The triangle specified here is a right angled triangle (9^2 + 40^2 = 41^2).

Always remember that in a right triangle, the hypotenuse is the diameter of the circumcircle and the centre is exactly the mid point of the hypotenuse.

So, diameter here is 41 and so, radius = 41/2 = 20.5

Answer B!

Quick question but how did you know right off the bat, that this triangle HAD to be a right triangle?
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by tutorphd » Fri Jul 27, 2012 7:36 pm
The general problem of this type requires quite a messy trigonometry. GMAT problems never require trigonometry. Instead, they exploit special simplified cases that are solvable with simpler methods.

If the problem seems too complicated in general, always check for a special case: whether the triangle is right (Pythagorean triplets, 30-60-90, 45-45-90), isosceles, or equilateral.
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