Question

In a triangle \(ABC\), if \(r_1=12\), \(r_2=18\) and \(r_3=36\), then \(b=\)

• A. \(12\)
• B. \(6\)
• C. \(24\)
• D. \(18\)

Hint:

For exradius problems, use \(r_1=\frac{\Delta}{s-a}\), \(r_2=\frac{\Delta}{s-b}\), \(r_3=\frac{\Delta}{s-c}\), along with Heron's formula.

Answer 1

The Correct Option is C

Step 1: Use exradius formulas.
\[ r_1=\frac{\Delta}{s-a},\quad r_2=\frac{\Delta}{s-b},\quad r_3=\frac{\Delta}{s-c} \] Given, \[ r_1=12,\quad r_2=18,\quad r_3=36 \] So, \[ s-a=\frac{\Delta}{12},\quad s-b=\frac{\Delta}{18},\quad s-c=\frac{\Delta}{36} \]

Step 2: Add all three equations.
We know that \[ (s-a)+(s-b)+(s-c)=s \] Therefore, \[ \frac{\Delta}{12}+\frac{\Delta}{18}+\frac{\Delta}{36}=s \] \[ \Delta\left(\frac{3+2+1}{36}\right)=s \] \[ \frac{\Delta}{6}=s \] Hence, \[ \Delta=6s \]

Step 3: Use Heron's formula.
By Heron's formula, \[ \Delta^2=s(s-a)(s-b)(s-c) \] Substituting, \[ \Delta^2=s\cdot \frac{\Delta}{12}\cdot \frac{\Delta}{18}\cdot \frac{\Delta}{36} \] \[ \Delta^2=\frac{s\Delta^3}{7776} \] Dividing by \(\Delta^2\), \[ 1=\frac{s\Delta}{7776} \] Using \(\Delta=6s\), \[ 1=\frac{6s^2}{7776} \] \[ s^2=1296 \] \[ s=36 \]

Step 4: Find \(b\).
Since \[ s-b=\frac{\Delta}{18} \] and \[ \Delta=6s=6(36)=216, \] we get \[ s-b=\frac{216}{18}=12 \] So, \[ 36-b=12 \] \[ b=24 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{24} \]