The lengths of the sides of a triangle are 13 | vedclass

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(D) Let $a = 13$, $b = 14$, and $c = 15$. First, calculate the semi-perimeter $s$: $s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = 21$. Next, calc

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$69$

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(D) Let $a = 13$, $b = 14$, and $c = 15$. First, calculate the semi-perimeter $s$: $s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = 21$. Next, calculate the area of the triangle $\Delta$ using Heron's formula: $\Delta = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{21(21 - 13)(21 - 14)(21 - 15)} = \sqrt{21 	imes 8 	imes 7 	imes 6} = \sqrt{7056} = 84$. Now, calculate the circumradius $R = \frac{abc}{4\Delta}$: $R = \frac{13 	imes 14 	imes 15}{4 	imes 84} = \frac{2730}{336} = \frac{65}{8}$. Calculate the inradius $r = \frac{\Delta}{s}$: $r = \frac{84}{21} = 4$. Finally, compute $8R + r$: $8R + r = 8 	imes \left(\frac{65}{8}ight) + 4 = 65 + 4 = 69$.

The lengths of the sides of a triangle are $13$, $14$ and $15$. If $R$ and $r$ respectively denote the circumradius and inradius of that triangle, then $8R + r =$

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