In a triangle ABC,if a=13, b=14, c=15,then fi | vedclass

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(B) Given sides of the triangle are $a=13, b=14, c=15$. First,calculate the semi-perimeter $s$: $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{

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$\frac{21}{2}$

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(B) Given sides of the triangle are $a=13, b=14, c=15$. First,calculate the semi-perimeter $s$: $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21$. Next,calculate the area of the triangle $\Delta$ using Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 	imes 8 	imes 7 	imes 6} = \sqrt{7056} = 84$. The exradius $r_1$ is given by the formula $r_1 = \frac{\Delta}{s-a}$: $r_1 = \frac{84}{21-13} = \frac{84}{8} = \frac{21}{2}$.

In a $\triangle ABC$,if $a=13, b=14, c=15$,then find the value of $r_1$.

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