In a $\triangle A B C, a: b: c=4: 5: 6$. The ratio of radius of the circumcircle to that of the incircle is
In a $\triangle A B C, a: b: c=4: 5: 6$. The ratio of radius of the circumcircle to that of the incircle is
- $7: 16$
- $17: 16$
- $16: 17$
- $16: 7$
Solution
Given,
$\frac{a}{4}=\frac{b}{5}=\frac{c}{6}=k$ (say)
$\Rightarrow \quad a=4 k, b=5 k, c=6 k$
$s=\frac{a+b+c}{2}=\frac{15 k}{2}$
$\dot{\Delta}=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{\frac{15 k}{2} \cdot \frac{7 k}{2} \cdot \frac{5 k}{2} \cdot \frac{3 k}{2}}=\frac{15 \sqrt{7} k^2}{4}$
Now, $r=\frac{\Delta}{s}$
$=\frac{15 \sqrt{7} k^2 / 4}{15 k / 2}=\frac{\sqrt{7} k}{2}$
and $R=\frac{a b c}{4 \Delta}=\frac{8}{\sqrt{7}} k$
$\therefore \quad \frac{R}{r}=\frac{8 k}{\sqrt{7}} \div \frac{\sqrt{7} k}{2}=\frac{16}{7}$
Asked in: AP EAMCET 2022 (07 Jul Shift 1)
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