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

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(C) Let the sides be $a = 13, b = 14, c = 15$. Semi-perimeter $s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21$. Area of the triangle $\Delta = \sqrt{s

Vedclass · Accepted Answer

$61$

Vedclass · Answer

(C) Let the sides be $a = 13, b = 14, c = 15$. Semi-perimeter $s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21$. Area of the triangle $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 	imes 8 	imes 7 	imes 6} = \sqrt{7056} = 84$. Circumradius $R = \frac{abc}{4\Delta} = \frac{13 	imes 14 	imes 15}{4 	imes 84} = \frac{2730}{336} = \frac{65}{8}$. Inradius $r = \frac{\Delta}{s} = \frac{84}{21} = 4$. Now,$8R - r = 8 	imes \left(\frac{65}{8}ight) - 4 = 65 - 4 = 61$.

The lengths of the sides of a triangle are $13, 14, \text{ and } 15$. If $R$ and $r$ respectively denote the circumradius and inradius of this triangle,then $8R - r = $

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