Question:
In a triangle \(ABC\), let \(\angle C=\dfrac{\pi}{2}\). If \(r\) and \(R\) are respectively the inradius and circumradius of \(ABC\), then \(R+r=\)
In a triangle \(ABC\), let \(\angle C=\dfrac{\pi}{2}\). If \(r\) and \(R\) are respectively the inradius and circumradius of \(ABC\), then \(R+r=\)
Show Hint
For a right-angled triangle:
\[
R=\frac{\text{hypotenuse}}{2}
\]
and
\[
r=\frac{a+b-c}{2}
\]
where \(a\) and \(b\) are the legs and \(c\) is the hypotenuse.
Updated On: Jun 26, 2026
- \(\dfrac{a-b}{2}\)
- \(\dfrac{a+b}{2}\)
- \(a+b\)
- \(a-b\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Use the circumradius formula for a right triangle.
Since \[ \angle C=\frac{\pi}{2}, \] the side \(c\) is the hypotenuse.
For a right-angled triangle, \[ R=\frac{c}{2} \]
Step 2: Use the inradius formula.
For a right triangle with sides \(a,b,c\), \[ r=\frac{a+b-c}{2} \]
Step 3: Add \(R\) and \(r\).
Therefore, \[ R+r = \frac{c}{2} + \frac{a+b-c}{2} \] \[ = \frac{a+b}{2} \]
Step 4: Final conclusion.
Hence, \[ \boxed{\frac{a+b}{2}} \] Therefore, the correct option is \[ \boxed{\left(2\right)\ \frac{a+b}{2}} \]
Since \[ \angle C=\frac{\pi}{2}, \] the side \(c\) is the hypotenuse.
For a right-angled triangle, \[ R=\frac{c}{2} \]
Step 2: Use the inradius formula.
For a right triangle with sides \(a,b,c\), \[ r=\frac{a+b-c}{2} \]
Step 3: Add \(R\) and \(r\).
Therefore, \[ R+r = \frac{c}{2} + \frac{a+b-c}{2} \] \[ = \frac{a+b}{2} \]
Step 4: Final conclusion.
Hence, \[ \boxed{\frac{a+b}{2}} \] Therefore, the correct option is \[ \boxed{\left(2\right)\ \frac{a+b}{2}} \]
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