^{2}. This means that radius of the given circle is sqrt(16) = 4 inches

^{2}, the are of the square is 32 inches

^{2}.

I don't know how to start.

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Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...

The formula for the area of a circle is π r^{2} . This means that radius of the given circle is sqrt(16) = 4 inches

Consequently, the diameter of the circle is 8 inches. This means that the diagonal of the largest possible inscribed square is also 8 inches. From the Pythagorean theorem, we figure the side of the largest square to be (8 inches)/sqrt(2). Since the area of a square is (length of side)^{2} , the are of the square is 32 inches^{2} .

Wow! So many similar solutions using the diagonal of the square.

The equation of the circle is 16π = πr^{2}, so r^{2} = 16 and r=4. Great! so let's assume the circle is centered at the origin. Then x^{2} + y^{2} = r^{2} = 16

Assume we rotate the square so that it's sides are parallel with the x and y axes. Then each side of the square is a chord which intersects a 90º arc, and the top side of the square intersects the arc at 45º and 135º, or at (-2√2, 2√2) and (2√2, 2√2). So the length of a side of the square is 4√2.

So the area of the square with side 4√2 is 16*2 = 32. Pythagorean theorem is not required.

You used the Pythagorean Theorem when you used the formula for a circle, which is derived using the Pythagorean Theorem.

And when you used the side lengths of the 45°-45°-90° Special Triangle; also derived using the Pythagorean Theorem.

[Re: "Pythagorean theorem is not required."]

Hi Julian;

Area of circle=(pi)(radius)^{2}

Area=(pi)(16)

radius=√16=4

diameter is twice the radius.

diameter=(2)(4)=8

This is the diagonal of the square.

Pythagorean Theorem is...

a^{2}+b^{2}=c^{2}

However, because this is a square, a=b...

2a^{2}=c^{2}

c is the diameter of the circle.

a^{2 }is also the area of the square.

The formula for the area of the largest square within a circle is...

Area of square=a^{2}=(1/2)(diameter of circle)^{2}

(1/2)(8)^{2}

(1/2)(64)

32

Curt J. | Math/Science/General Ed Tutor in WaikeleMath/Science/General Ed Tutor in Waikele

Julian, to visualize this problem, draw a square within the circle where each of the four points are on the edge of the circle (like this http://www.yogaflavoredlife.com/wp-content/uploads/2010/09/square-circle.gif).

Area of a circle: A = πr^{2}

Thus,

16π = πr^{2}

16 = r^{2}

r = √16

r = 4

Double the radius, and you have the diameter (d=8), which also happens to be the distance from one corner of the square to the other (like this: http://mathcentral.uregina.ca/QQ/database/QQ.09.04/bob1.1.gif).

Note you've just made a triangle out of your square, and you know the length of the hypotenuse, 8/√n!

Using the Pythagorean Theorem (A^{2} + B^{2} = C^{2}), you can solve for the length of the sides of the square (and since it's a square, and thus has equal sides, A^{2} + B^{2} = A^{2} + A^{2} = 2A^{2}

Thus,

2A^{2} = C^{2}

2A^{2} = 8^{2}

2A^{2} = 64

EDIT: Whoops, didn't notice the pi in the original problem. Good call, Richard!

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