A park is designed in the shape of a right-angled triangle, where the two shorter sides measure 13 m and 84 m. A circular track with a uniform width of 2 m is constructed such that its outer boundary coincides with the circumcircle of the triangular park. Determine the perimeter of the inner boundary of this circular track.
81\(\pi\) m
Find the hypotenuse: \(h = \sqrt{13^2 + 84^2} = \sqrt{169 + 7056} = \sqrt{7225} = 85\) m.
For a right-angled triangle, the circumradius equals half the hypotenuse: \(R = \frac{85}{2} = 42.5\) m.
The track has a uniform width of 2 m, so the inner radius = \(42.5 - 2 = 40.5\) m.
Perimeter of the inner boundary = \(2\pi r = 2\pi \times 40.5 = 81\pi\) m.
Hence, the perimeter of the inner boundary is \(81\pi\) m.
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