Q84 / 125 · BITSAT 2011 · Mathematics
In an equilateral triangle, the in radius, circumradius and one of the ex-radii are in the ratio
- A\(2: 3: 5\)
- Correct answerB\(1: 2: 3\)
- C\(3: 7: 9\)
- D\(3: 7: 9\)
Solution
We have \(\Delta=\frac{\sqrt{3}}{4} \mathrm{a}^2, \mathrm{~s}=\frac{3 \mathrm{a}}{2}\)
\(\therefore \quad r=\frac{\Delta}{\mathrm{s}}=\frac{\mathrm{a}}{2 \sqrt{3}}, \mathrm{R}=\frac{\mathrm{abc}}{4 \Delta}=\frac{\mathrm{a}^3}{\sqrt{3} \mathrm{a}^2}=\frac{\mathrm{a}}{\sqrt{3}}\) and \(r_1=\frac{\Delta}{s-a}=\frac{\sqrt{3} / 4 a^2}{a / 2}=\frac{\sqrt{3}}{2} a\)
Hence, \(\mathrm{r}: \mathrm{R}: \mathrm{r}_1=\frac{\mathrm{a}}{2 \sqrt{3}}: \frac{\mathrm{a}}{\sqrt{3}}: \frac{\sqrt{3}}{2} \mathrm{a}=1: 2: 3\)
Hence, the correct answer is option: 'B'.