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Square inscribed in a circle (Powerprep Practice Test)

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Junior | Next Rank: 30 Posts
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Location: Kingston, Jamaica
by jcan-gmatter » Fri Oct 05, 2007 11:38 am
Hello,

How do I approach the following problem? I was using Brahmagupta's formula but I'm getting 16Pi as the answer, which is incorrect. That is

Square Root (S-a) (S-b) (S-c) (S-d)

where S = a + b + c + d /2

See attached image.
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by mschling52 » Fri Oct 05, 2007 11:54 am
I would use the formula (a^2)+(b^2) = (c^2) (pythagorean theorem), where a and b are the legs of a right triangle and c is the hypotense.

If you draw a right triangle by connecting 2 corners of the square, you will note that the hypotenuse of the right triangle you create is also the diameter of the circle. Each side of the square is length 4 (since the area is 16). Therefore, the length of the hypotenuse of the right triangle (and the diameter of the circle) can be solved by

(4^2)+(4^2) = c^2
32 = c^2
sqrt(32) = c
4sqrt(2) = c (you could also get this quicker by knowing the ratios for a 45-45-90 triangle)

Since we know the diameter of the circle is 4sqrt(2), we also know the radius is 2sqrt(2). Then we can calculate the area of the circle as

pi*(2sqrt(2))^2 = 8*pi
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Junior | Next Rank: 30 Posts
Posts: 17
Joined: Thu Jul 05, 2007 9:08 am
Location: Kingston, Jamaica

Didn't think of the triangle angle

by jcan-gmatter » Fri Oct 05, 2007 12:13 pm
Hello,

Thank you for the response. Looked at the figure for quite a bit but didn't figure that I could use the triangle route. Will definitely keep this in mind.

/Thanks.
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