45-45-90 (Inscribed circle in a square)

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A circle is inscribed in a square. The question asks for the area of the circle. The only information given is that the diagonal of the square is 2 root 6.

So to find the side of the square, which equals the diameter of the circle, we would say that the hypotenuse x root 2=2 root 6 and solve for x. I lack the simple foundations to solve this correctly.

Divide both sides by root 2 gives us: x= 2 root 6/root 2

How do we solve this and simplify this further in order to plug it into our area of circle (Pie R squared)?

Thanks in advance!

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by PGPEX_aspirant » Wed Apr 21, 2010 7:09 pm
Okay. In a square with side 'a', the length of the diagnol is a*(SQRT 2)

So in this case, since the length of diagnol is 2(SQRT 6), the length of the side of the square comes out as 2(SQRT 3)

Now, in this case, side of square = diameter of the circle = 2(SQRT 3)

Do radius of circle = 2 *(SQRT 3) /2 = SQRT 3

Area of circle = Pie(SQRT 3)^2 = 3Pie

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by tpr-becky » Wed Apr 21, 2010 9:17 pm
in order to get the root 2 out of the denominator you have to multiply the whole fraction by root2/root 2 - which leaves you with 2(root6)(root2)/2 - the 2's cancel leaving you with a side of (root6)(root2) - you can multiply under the square root so that is (root12) - which can be factored to (root4)(root3) - which is 2(root3). This is the side of the square and thus the diameter of the circle.

radius is 1/2 diameter so radius is (root3)

area of a circle is Pi(r^2) - so it is Pi(root3)^2 - or 3(pi)
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