Question:
In a triangle \(ABC\), if
\[
r_1=36,\quad r_2=18 \quad \text{and} \quad r_3=12,
\]
then
\[
s=
\]
In a triangle \(ABC\), if
\[
r_1=36,\quad r_2=18 \quad \text{and} \quad r_3=12,
\]
then
\[
s=
\]
Show Hint
For exradii \(r_1,r_2,r_3\) and inradius \(r\),
remember:
\[
\frac1r=\frac1{r_1}+\frac1{r_2}+\frac1{r_3}.
\]
This identity is very useful in triangle geometry problems.
Updated On: Jun 22, 2026
- \(6\)
- \(8\)
- \(16\)
- \(36\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use the exradius formulas.
For a triangle with semiperimeter \(s\) and area \(\Delta\), the exradii are given by \[ r_1=\frac{\Delta}{s-a}, \] \[ r_2=\frac{\Delta}{s-b}, \] \[ r_3=\frac{\Delta}{s-c} \] Also, the inradius satisfies \[ r=\frac{\Delta}{s} \]
Step 2: Use the standard identity involving exradii.
A known relation is \[ \frac1r=\frac1{r_1}+\frac1{r_2}+\frac1{r_3} \] Substituting the given values: \[ \frac1r=\frac1{36}+\frac1{18}+\frac1{12} \]
Step 3: Take LCM and simplify.
LCM of \(36,18,12\) is \(36\).
So, \[ \frac1r= \frac1{36}+\frac2{36}+\frac3{36} \] \[ =\frac6{36} \] \[ =\frac16 \] Hence, \[ r=6 \]
Step 4: Use the relation between semiperimeter and exradii.
For a triangle, \[ s^2=r_1r_2r_3\left(\frac1r\right) \] Substituting values, \[ s^2=36\cdot18\cdot12\cdot \frac16 \] \[ =36\cdot18\cdot2 \] \[ =1296 \]
Step 5: Find \(s\).
\[ s=\sqrt{1296} \] \[ s=36 \]
Step 6: Final conclusion.
Therefore, \[ \boxed{36} \]
For a triangle with semiperimeter \(s\) and area \(\Delta\), the exradii are given by \[ r_1=\frac{\Delta}{s-a}, \] \[ r_2=\frac{\Delta}{s-b}, \] \[ r_3=\frac{\Delta}{s-c} \] Also, the inradius satisfies \[ r=\frac{\Delta}{s} \]
Step 2: Use the standard identity involving exradii.
A known relation is \[ \frac1r=\frac1{r_1}+\frac1{r_2}+\frac1{r_3} \] Substituting the given values: \[ \frac1r=\frac1{36}+\frac1{18}+\frac1{12} \]
Step 3: Take LCM and simplify.
LCM of \(36,18,12\) is \(36\).
So, \[ \frac1r= \frac1{36}+\frac2{36}+\frac3{36} \] \[ =\frac6{36} \] \[ =\frac16 \] Hence, \[ r=6 \]
Step 4: Use the relation between semiperimeter and exradii.
For a triangle, \[ s^2=r_1r_2r_3\left(\frac1r\right) \] Substituting values, \[ s^2=36\cdot18\cdot12\cdot \frac16 \] \[ =36\cdot18\cdot2 \] \[ =1296 \]
Step 5: Find \(s\).
\[ s=\sqrt{1296} \] \[ s=36 \]
Step 6: Final conclusion.
Therefore, \[ \boxed{36} \]
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