Question:
In a triangle \(ABC\), if \(r_1=\dfrac{5}{\sqrt2},\ r_2=2\sqrt2,\ r=r\sqrt2\), then \(\dfrac{a+c}{b}=\)
In a triangle \(ABC\), if \(r_1=\dfrac{5}{\sqrt2},\ r_2=2\sqrt2,\ r=r\sqrt2\), then \(\dfrac{a+c}{b}=\)
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Relations involving \(r,\ r_1,\ r_2,\ r_3\) are usually simplified using \(r=\dfrac{\Delta}{s}\).
Updated On: Jun 17, 2026
- \(15\)
- \(12\)
- \(3\)
- \(5\)
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The Correct Option is C
Solution and Explanation
Concept:
Use the relation between exradii and sides of the triangle.
Step 1: Use the standard formulas.
For a triangle, \[ r_1=\frac{\Delta}{s-a},\qquad r_2=\frac{\Delta}{s-b},\qquad r=\frac{\Delta}{s} \] Given, \[ r_1=\frac{5}{\sqrt2},\qquad r_2=2\sqrt2,\qquad r=\sqrt2 \]
Step 2: Find the corresponding ratios.
\[ \frac{r_1}{r} = \frac{s}{s-a} = \frac{5/\sqrt2}{\sqrt2} = \frac52 \] Thus, \[ 2s=5(s-a) \] \[ 5a=3s \] Similarly, \[ \frac{r_2}{r} = \frac{s}{s-b} = \frac{2\sqrt2}{\sqrt2} = 2 \] Hence, \[ s=2(s-b) \] \[ 2b=s \]
Step 3: Find \(\dfrac{a+c}{b}\).
Since \[ 2s=a+b+c \] Using obtained relations, \[ a+c=3b \] Therefore, \[ \frac{a+c}{b}=3 \] Hence, \[ \boxed{3} \]
Step 1: Use the standard formulas.
For a triangle, \[ r_1=\frac{\Delta}{s-a},\qquad r_2=\frac{\Delta}{s-b},\qquad r=\frac{\Delta}{s} \] Given, \[ r_1=\frac{5}{\sqrt2},\qquad r_2=2\sqrt2,\qquad r=\sqrt2 \]
Step 2: Find the corresponding ratios.
\[ \frac{r_1}{r} = \frac{s}{s-a} = \frac{5/\sqrt2}{\sqrt2} = \frac52 \] Thus, \[ 2s=5(s-a) \] \[ 5a=3s \] Similarly, \[ \frac{r_2}{r} = \frac{s}{s-b} = \frac{2\sqrt2}{\sqrt2} = 2 \] Hence, \[ s=2(s-b) \] \[ 2b=s \]
Step 3: Find \(\dfrac{a+c}{b}\).
Since \[ 2s=a+b+c \] Using obtained relations, \[ a+c=3b \] Therefore, \[ \frac{a+c}{b}=3 \] Hence, \[ \boxed{3} \]
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