what is the area of the largest quadrileteral that can be inscribed in a circle with an area of 64pie?
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The largest quadrilateral that can be inscribed in a circle is the square whose diagonal is the diameter of the circle.
Since the circle's area is: `A = 64pi,`
This means that `64 = r^2,`
so the radius of the circle is `8.`
Now, that makes the length of the diagonal of the square equivalent to: `16.`
The diagonal is the hypotenuse length of a `45-45-90 triangle.`
When given the hypotenuse of this special triangle the length of the sides are: `16 /sqrt(2)`
Simplified this gives the square a side length of `8sqrt(2).`
Since this isa square and all side lengths are equal to find area of square: `A = s^2.`
So `A = (8sqrt(2))^2`
So the area of the square (the largest) is `128.`
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