A triangular board has sides 9 cm, 10 cm, and 17 cm. A circle passes through all three vertices. Find the circumradius.
10.625 cm
Using Heron's formula, the semi-perimeter is \(s = \frac{9 + 10 + 17}{2} = 18\).
The area is \(\text{Area} = \sqrt{18 \times 9 \times 8 \times 1} = \sqrt{1296} = 36\) square cm.
The circumradius formula is \(R = \frac{a \times b \times c}{4 \times \text{Area}}\).
Substituting the values gives \(R = \frac{9 \times 10 \times 17}{4 \times 36} = \frac{1530}{144} = 10.625\) cm.
Hence, the circumradius of the triangle is 10.625 cm.
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