Question:
In a \(\triangle ABC\), \(a:b:c=4:5:6\). The ratio of the radius of the circumcircle to that of the incircle is:
In a \(\triangle ABC\), \(a:b:c=4:5:6\). The ratio of the radius of the circumcircle to that of the incircle is:
Show Hint
For any triangle,
\[
\Delta=rs
\]
and
\[
\Delta=\frac{abc}{4R}
\]
These two formulas are very useful for finding the relation between the inradius and circumradius.
Updated On: Jun 26, 2026
- \(7:16\)
- \(17:16\)
- \(16:17\)
- \(16:7\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Take the sides proportional to the given ratio.
Let \[ a=4k,\quad b=5k,\quad c=6k \] The semi-perimeter is \[ s=\frac{4k+5k+6k}{2} =\frac{15k}{2} \]
Step 2: Find the area using Heron's formula.
Using \[ \Delta=\sqrt{s(s-a)(s-b)(s-c)} \] we get \[ \Delta= \sqrt{\frac{15k}{2}\cdot\frac{7k}{2}\cdot\frac{5k}{2}\cdot\frac{3k}{2}} \] \[ =\frac{k^2}{4}\sqrt{15\cdot7\cdot5\cdot3} \] \[ =\frac{15\sqrt7}{4}k^2 \]
Step 3: Find the inradius \(r\).
We know \[ \Delta=rs \] Hence \[ r=\frac{\Delta}{s} =\frac{\frac{15\sqrt7}{4}k^2}{\frac{15k}{2}} \] \[ r=\frac{\sqrt7}{2}k \]
Step 4: Find the circumradius \(R\).
Using \[ \Delta=\frac{abc}{4R} \] \[ R=\frac{abc}{4\Delta} \] Substituting, \[ R=\frac{(4k)(5k)(6k)} {4\left(\frac{15\sqrt7}{4}k^2\right)} \] \[ R=\frac{120k^3}{15\sqrt7\,k^2} \] \[ R=\frac{8k}{\sqrt7} \]
Step 5: Find the required ratio.
\[ \frac{R}{r} = \frac{\frac{8k}{\sqrt7}} {\frac{\sqrt7}{2}k} \] \[ =\frac{16}{7} \] Therefore, \[ R:r=16:7 \]
Step 6: Final conclusion.
Hence, \[ \boxed{16:7} \]
Let \[ a=4k,\quad b=5k,\quad c=6k \] The semi-perimeter is \[ s=\frac{4k+5k+6k}{2} =\frac{15k}{2} \]
Step 2: Find the area using Heron's formula.
Using \[ \Delta=\sqrt{s(s-a)(s-b)(s-c)} \] we get \[ \Delta= \sqrt{\frac{15k}{2}\cdot\frac{7k}{2}\cdot\frac{5k}{2}\cdot\frac{3k}{2}} \] \[ =\frac{k^2}{4}\sqrt{15\cdot7\cdot5\cdot3} \] \[ =\frac{15\sqrt7}{4}k^2 \]
Step 3: Find the inradius \(r\).
We know \[ \Delta=rs \] Hence \[ r=\frac{\Delta}{s} =\frac{\frac{15\sqrt7}{4}k^2}{\frac{15k}{2}} \] \[ r=\frac{\sqrt7}{2}k \]
Step 4: Find the circumradius \(R\).
Using \[ \Delta=\frac{abc}{4R} \] \[ R=\frac{abc}{4\Delta} \] Substituting, \[ R=\frac{(4k)(5k)(6k)} {4\left(\frac{15\sqrt7}{4}k^2\right)} \] \[ R=\frac{120k^3}{15\sqrt7\,k^2} \] \[ R=\frac{8k}{\sqrt7} \]
Step 5: Find the required ratio.
\[ \frac{R}{r} = \frac{\frac{8k}{\sqrt7}} {\frac{\sqrt7}{2}k} \] \[ =\frac{16}{7} \] Therefore, \[ R:r=16:7 \]
Step 6: Final conclusion.
Hence, \[ \boxed{16:7} \]
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