What is the dimension of the largest square that will fit in a 6' circle?

Hello.

The dimensions of the square will be 4.2426" by 4.2426" (i.e., 18 square inches).

If the circle has a diameter of 6", then the diagonal of the largest
inscribed square will also be 6".  The inscribed square may be looked
at as two right triangles with 6" hypotenuse. Of course, the
Pythagorean theorem tells us that the squares of the sides of a right
triangle must add up to the square of the hypotenuse.

Here, the hypotenuse is 6", so the square of the hypotenuse is 36.

2a^2=36
a^2 = 18
a = square root of 18 = 4.2426

The dimensions of the square and the circle here are the same as in
this problem on mathforum.org:
The Circle and The Square
http://mathforum.org/lucentpow/solutions/solution.ehtml?puzzle=34

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search strategy:
"18 square inches" circle

I hope this helps.

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Clarification of Answer by juggler-ga on 02 Apr 2004 13:34 PST

Oops, I notice that your question asked for 6 feet.

Thus, everything in my answer should be FEET, not inches.

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Clarification of Answer by juggler-ga on 02 Apr 2004 13:36 PST

The dimensions of the square will be 4.2426' by 4.2426' (i.e., 18 square FEET).

If the circle has a diameter of 6', then the diagonal of the largest
inscribed square will also be 6'.  The inscribed square may be looked
at as two right triangles with 6' hypotenuse. Of course, the
Pythagorean theorem tells us that the squares of the sides of a right
triangle must add up to the square of the hypotenuse.

Here, the hypotenuse is 6', so the square of the hypotenuse is 36.

2a^2=36
a^2 = 18
a = square root of 18 = 4.2426