Question

In a triangle ABC, let \(\angle C = \pi/2\). If r and R are respectively inradius and circumradius of ABC, then R + r =

• A. \( \frac{a-b}{2} \)
• B. \( \frac{a+b}{2} \)
• C. \( a+b \)
• D. \( a-b \)

Hint:

For a right-angled triangle (right angle at C, hypotenuse c, legs a, b):
Circumradius \(R = c/2\).
Inradius \(r = (a+b-c)/2\).
Therefore, \(R+r = \frac{c}{2} + \frac{a+b-c}{2} = \frac{a+b}{2}\).

Answer 1

The Correct Option is B

Given a right-angled triangle ABC, with \(\angle C = \pi/2 = 90^\circ\). Side c (hypotenuse) = AB. Sides a, b are AC, BC (or vice-versa). Circumradius (R): For a right-angled triangle, the circumcenter is the midpoint of the hypotenuse, and the circumradius is half the length of the hypotenuse. So, \(R = \frac{c}{2}\). Inradius (r): For a right-angled triangle, the inradius is given by the formula \(r = \frac{a+b-c}{2}\), where a and b are the legs and c is the hypotenuse. (This can be derived from Area \(\Delta = rs\) and \(\Delta = (1/2)ab\). \(s = (a+b+c)/2\). So \((1/2)ab = r(a+b+c)/2 \Rightarrow r = ab/(a+b+c)\). Also, \( (a+b-c)/2 = (a+b-\sqrt{a^2+b^2})/2 \). It is known that \(r = (a+b-c)/2\)). Now, we need to find \(R+r\): \(R+r = \frac{c}{2} + \frac{a+b-c}{2}\) \(R+r = \frac{c + a + b - c}{2}\) \(R+r = \frac{a+b}{2}\). This matches option (b). \[ \boxed{\frac{a+b}{2}} \]