Question

In $△ABC,R,r,r_1,r_2,r_3$ denote the circumradius, inradius, the exradii opposite to the vertices $A,B,C$ respectively. Given that $r_1:r_2:r_3=1:2:3$.
The value of $R:r$ is

• A. $5:2$
• B. $5:4$
• C. $5:3$
• D. $3:2$

Answer 1

The correct option is A $5:2$

$\frac{△}{s-a}:\frac{△}{s-b}:\frac{△}{s-c}=1:2:3$
Let $\frac{△}{s-a}=\frac{2△}{s-b}=\frac{3△}{s-c}=\frac{△}{6k}\Rightarrow \frac{1}{s-a}=\frac{1}{6k},\frac{1}{s-b}=\frac{1}{3k},\frac{△}{s-c}=\frac{1}{2k}\Rightarrow s-a=6k,s-b=3k,s-c=2k\Rightarrow s=11k\therefore a=5k,b=8k,c=9k$
Hence, ratio of sides is $5:8:9$
Now,
Area of the triangle, $△=k^2\sqrt{11\times 6\times 3\times 2}=6\sqrt{11}{k}^2$
So, $r=\frac{△}{s}=\frac{6}{\sqrt{11}}k$
And, $R=\frac{abc}{4△}=\frac{5\times 8\times 9}{4\times 6\times \sqrt{11}}k=\frac{15}{\sqrt{11}}k\therefore R:r=5:2$