Question 270754
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Construct circle radius OB.  Since any tangent is perpendicular to the radius at the point of tangency, OB is perpendicular to BC.  Therefore OBC is a right triangle.  Then the measure of OB can be calculated, with a nod and a tip of the hat to Mr. Pythagoras:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \overline{OB}\ =\ r_O\ =\ \sqrt{65^2\ -\ 60^2]


But the radius of a circle is one half of the measure of the side of a circumscribed square, so the side of the circumscribed square is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2r_O]


And then the area of the circumscribed square is the measure of the side squared or:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(2r_O\right)^2]


You can do your own arithmetic.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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