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    Topics || Problems

    Calculate the area of the circle circumscribing an isosceles right triangle having an area of 242 mm2.
    circumscribed circle

    Since the triangle is a right triangle, one side must be the diameter of the circle. Thus,

    \(A_{tri} = \frac{1}{2}ab\)

    But a = b

    \(A_{tri} = \frac{1}{2}a^2\)

    \(242 = \frac{1}{2}a^2\)

    \(a = 22 mm\)

    \(a^2 +a^2 = d^2\)

    \(d = 22\sqrt{2}\)

    \(A_{circle} = \frac{ \pi d^2}{4}\)

    \(A_{circle} = \frac{ \pi (22\sqrt{2})^2}{4}\)

    \(A_{circle} = 760.27 mm^2\)