Question

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

• A. \(216\)
• B. \(342\)
• C. \(432\)
• D. \(126\)

Hint:

Remember the important identity \[ \boxed{\Delta = r_1r_2r_3 \left( \frac1{r_1}+\frac1{r_2}+\frac1{r_3} \right)} \] which directly gives the area when the three exradii are known.

Answer 1

The Correct Option is A

Step 1: Use the relation between exradii and area. For a triangle, \[ r_1=\frac{\Delta}{s-a},\qquad r_2=\frac{\Delta}{s-b},\qquad r_3=\frac{\Delta}{s-c}, \] where \(s\) is the semi-perimeter. Hence, \[ (s-a)+(s-b)+(s-c)=s. \] Therefore, \[ \frac{\Delta}{r_1}+\frac{\Delta}{r_2}+\frac{\Delta}{r_3}=s. \]

Step 2: Use the formula involving exradii. Also, \[ \Delta^2 = r_1r_2r_3\,r, \] where \(r\) is the inradius. Since \[ \Delta=rs, \] we get \[ \Delta = r_1r_2r_3 \left( \frac1{r_1}+\frac1{r_2}+\frac1{r_3} \right). \]

Step 3: Calculate the area. Substituting \[ r_1=12,\qquad r_2=18,\qquad r_3=36, \] we obtain \[ \Delta = 12\times18\times36 \left( \frac1{12}+\frac1{18}+\frac1{36} \right). \] Now, \[ \frac1{12}+\frac1{18}+\frac1{36} = \frac{3+2+1}{36} = \frac16. \] Hence, \[ \Delta = 12\times18\times6 = 216. \] Therefore, \[ \boxed{216}. \] Thus, \[ \boxed{(A)} \] is the correct answer.