Question

In $\triangle ABC$, if a: b: $c =4: 5: 6$, then the ratio of the circum radius to its inradius is

• A. $16: 7$
• B. $7: 16$
• C. $4: 5$
• D. $5: 4$

Answer 1

Answer: (A)

We have,
$a: b: c=4: 5: 6$
$\Rightarrow a=4 K, b=5 K, c=6 K$
Now, $R=\frac{a b c}{4 \Delta}$ and $r=\frac{\Delta}{s}$
$\Rightarrow \frac{R}{r}=\frac{\left(\frac{a b c}{4 \Delta}\right)}{\left(\frac{\Delta}{s}\right)}=\frac{a b c s}{4 \Delta^{2}}$
Now, $\Delta^{2}=s(s-a)(s-b)(s-c)$ and
and
$s=\frac{a+b+c}{2}=\frac{4 K+5 K+6 K}{2}=\frac{15 K}{2} $
$\therefore \frac{R}{r}=\frac{\left[a b c \frac{(a+b+c)}{2}\right]}{[4 s(s-a)(s-b)(s-c)]}$
$=\frac{a b c}{8(s-a)(s-b)(s-c)} $
$=\frac{4 K \times 5 K \times 6 K}{4\left(\frac{15 K}{2}-4 K\right)\left(\frac{15 K}{2}-5 K\right)\left(\frac{15 K}{2}-6 K\right)} $
$=\frac{120 K^{3}}{4 \times \frac{7 K}{2} \times \frac{5 K}{2} \times \frac{3 K}{2}}=\frac{16}{7}$
Required ratio $=16: 7$