Plane Geometry Solutions
Topics || Problems
Calculate the area of the circle circumscribing an isosceles right triangle having an area of 242 mm2.
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
\(A_{tri} = \frac{1}{2}a^2\)
\(242 = \frac{1}{2}a^2\)
\(a = 22 mm\)
\(a^2 +a^2 = d^2\)
\(d = 22\sqrt{2}\)
\(A_{circle} = \frac{ \pi d^2}{4}\)
\(A_{circle} = \frac{ \pi (22\sqrt{2})^2}{4}\)
\(A_{circle} = 760.27 mm^2\)