In a triangle ABC, Let <span class="math-tex">\(C = \frac {\pi} 2\)</span>, If r is the inradius and R is circumradius of the triangle ABC, then 2(r + R) equals
Step-by-step Solution:
\begin{aligned} &\text{We know that:} \\ &R = \frac{abc}{4\Delta} \quad \text{(Equation 1)} \\ &\text{Also,} \quad r = \frac{\Delta}{s} = (s - c) \tan \frac{C}{2} \quad \text{(Equation 2)} \\ &\text{Given } \angle C = \frac{\pi}{2}, \\ &\text{Hence,} \quad \Delta = \frac{b \times a}{2} \quad \text{(Equation 3)} \\ &\text{Substituting Equation (3) in Equation (1), we get:} \\ &R = \frac{abc}{4 \times \frac{b \times a}{2} \times c} \\ &\Rightarrow R = \frac{c}{2} \quad \text{(Equation 4)} \\ &\text{Now, the semi-perimeter is:} \\ &s = \frac{a + b + c}{2} \\ &\text{Substituting in Equation (2), we get:} \\ &r = \frac{a + b - c}{2} \quad \text{(Equation 5)} \\ &\text{Now, calculating } 2(R + r): \\ &2(R + r) = 2 \left(\frac{c}{2} + \frac{a + b - c}{2} \right) \\ &= 2 \times \frac{c + a + b - c}{2} \\ &= a + b \\ &\therefore 2(R + r) = a + b. \end{aligned}