The length of the sides of a triangle are $13,14 \& 15$. If $\mathrm{R}$ and $\mathrm{r}$ respectively…

The length of the sides of a triangle are $13,14 \& 15$. If $\mathrm{R}$ and $\mathrm{r}$ respectively denote circumradius and inradius of this triangle, then $8 \mathrm{R}-\mathrm{r}=$
  1. 41
  2. 51
  3. 61
  4. 71

Solution

Given sides of triangle are 13, 14, 15 $ \text { Circumradius }(R)=\frac{a b c}{4 \Delta} $ Inradius $(\mathrm{r})=\frac{\Delta}{5}$ $ \begin{aligned} & \Delta=\sqrt{\left(\frac{13+14+15}{2}\right)\left(\frac{13+14+15}{2}-13\right)\left(\frac{13+14+15}{2}-14\right)\left(\frac{13+14+15}{2}-15\right)} \\ & \Delta=\sqrt{21 \times 8 \times 7 \times 6}=\sqrt{7056}=84 \end{aligned} $ From (i) Circumradius $(\mathrm{R})=\frac{13 \times 14 \times 15}{4 \times 84}=\frac{195}{24}$ Inradius (r) $=\frac{84}{42} \times 2=4$ Now, $8 \mathrm{R}-\mathrm{r}=8 \times \frac{195}{24}-4=65-4=61$

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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