The centre O of a circle inside a triangle ABC is at a distance of 13 cm from each of the vertices of the triangle. The diameter of the circle is 10 cm and the circle touches only two sides of the triangle, AB and AC.
If x is the perimeter, in cm, of the triangle, then which one of the following is correct?
\(65\text{ cm} < x < 67\text{ cm}\)
Since O is 13 cm from every vertex, O is the circumcentre with circumradius R = 13 cm. The circle of radius 5 cm touches AB and AC only, so the perpendicular distances from O to AB and AC both equal 5 cm, i.e. R cos C = R cos B = 5, giving cos B = cos C = 5/13 and hence B = C (triangle is isosceles with AB = AC). So sin B = 12/13, AB = AC = 2R sin B = 24 cm. Also A = 180 deg - 2B, so sin A = 2 sin B cos B = 120/169 and cos A = 1 - 2cos^2B = 119/169. Side BC = 2R sin A = 2(13)(120/169) = 3120/169 \(\approx 18.46\text{ cm}\). So perimeter \(x = AB+AC+BC = 24+24+18.46 \approx 66.46\text{ cm}\), which lies between 65 cm and 67 cm.
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