Question:
In \(\triangle ABC\), if
\[
r_1=\frac{21}{2},\qquad r_2=12,\qquad r_3=14,
\]
then \(\Delta=\)
In \(\triangle ABC\), if
\[
r_1=\frac{21}{2},\qquad r_2=12,\qquad r_3=14,
\]
then \(\Delta=\)
Show Hint
For problems involving exradii, remember the important identities
\(\Delta^2=r\,r_1r_2r_3\) and
\(\frac1r=\frac1{r_1}+\frac1{r_2}+\frac1{r_3}\). Together they allow direct computation of the area.
Updated On: Jul 29, 2026
- \(84\)
- \(72\)
- \(64\)
- \(96\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
If \(r_1,r_2,r_3\) are the exradii of a triangle and \(\Delta\) is its area, then
\[
\Delta^2=r\,r_1r_2r_3,
\]
where \(r\) is the inradius.
Also,
\[
\frac{1}{r}
=
\frac{1}{r_1}
+
\frac{1}{r_2}
+
\frac{1}{r_3}.
\]
Step 1: Find the inradius \(r\). Given, \[ r_1=\frac{21}{2}, \qquad r_2=12, \qquad r_3=14. \] Using \[ \frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}, \] \[ \frac{1}{r} = \frac{2}{21} + \frac{1}{12} + \frac{1}{14}. \] Taking LCM \(84\), \[ \frac{1}{r} = \frac{8+7+6}{84} = \frac{21}{84} = \frac14. \] Hence, \[ r=4. \]
Step 2: Use the area formula. \[ \Delta^2 = r\,r_1r_2r_3. \] Substituting the values, \[ \Delta^2 = 4\times\frac{21}{2}\times12\times14. \] \[ = 2\times21\times12\times14. \] \[ = 7056. \] \[ \Delta=\sqrt{7056}=84. \] Therefore, \[ \boxed{\Delta=84} \] \[ \boxed{\text{Answer = (A)}} \]
Step 1: Find the inradius \(r\). Given, \[ r_1=\frac{21}{2}, \qquad r_2=12, \qquad r_3=14. \] Using \[ \frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}, \] \[ \frac{1}{r} = \frac{2}{21} + \frac{1}{12} + \frac{1}{14}. \] Taking LCM \(84\), \[ \frac{1}{r} = \frac{8+7+6}{84} = \frac{21}{84} = \frac14. \] Hence, \[ r=4. \]
Step 2: Use the area formula. \[ \Delta^2 = r\,r_1r_2r_3. \] Substituting the values, \[ \Delta^2 = 4\times\frac{21}{2}\times12\times14. \] \[ = 2\times21\times12\times14. \] \[ = 7056. \] \[ \Delta=\sqrt{7056}=84. \] Therefore, \[ \boxed{\Delta=84} \] \[ \boxed{\text{Answer = (A)}} \]
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