Question:
In a triangle \(ABC\), if
\[
r_2+r_3=2R,
\]
then
\[
r+2r_2+2r_3-r_1=
\]
In a triangle \(ABC\), if
\[
r_2+r_3=2R,
\]
then
\[
r+2r_2+2r_3-r_1=
\]
Show Hint
A very important identity involving inradius and exradii is
\[
r_1=r+r_2+r_3.
\]
It frequently simplifies complicated radius expressions instantly.
Updated On: Jun 17, 2026
- \(4R\)
- \(2R\)
- \(4R\cos A\)
- \(4R\cos B\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
The exradii and circumradius satisfy
\[
r_1=4R\sin\frac B2\sin\frac C2\cos\frac A2,
\]
\[
r_2=4R\sin\frac C2\sin\frac A2\cos\frac B2,
\]
\[
r_3=4R\sin\frac A2\sin\frac B2\cos\frac C2.
\]
Also,
\[
r=4R\sin\frac A2\sin\frac B2\sin\frac C2.
\]
A standard identity is
\[
r_1=r+r_2+r_3.
\]
Step 1: Use the given condition. Given \[ r_2+r_3=2R. \]
Step 2: Apply the standard relation. Since \[ r_1=r+r_2+r_3, \] we have \[ r_1=r+2R. \]
Step 3: Substitute into the required expression. \[ r+2r_2+2r_3-r_1. \] Using \[ r_2+r_3=2R, \] \[ =r+4R-r_1. \] Substituting \[ r_1=r+2R, \] \[ =r+4R-(r+2R). \] \[ =2R. \] Conclusion: \[ \boxed{2R} \]
Step 1: Use the given condition. Given \[ r_2+r_3=2R. \]
Step 2: Apply the standard relation. Since \[ r_1=r+r_2+r_3, \] we have \[ r_1=r+2R. \]
Step 3: Substitute into the required expression. \[ r+2r_2+2r_3-r_1. \] Using \[ r_2+r_3=2R, \] \[ =r+4R-r_1. \] Substituting \[ r_1=r+2R, \] \[ =r+4R-(r+2R). \] \[ =2R. \] Conclusion: \[ \boxed{2R} \]
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