Question

In an isosceles triangle with base angle $\alpha ={45}^{\circ }$ and lateral side $b=4,$ then $Rr=$

• A. $\frac{8}{1-\sqrt{2}}$
• B. $\frac{8}{1+\sqrt{2}}$
• C. $\frac{4}{1-\sqrt{2}}$
• D. $\frac{4}{1+\sqrt{2}}$

Answer 1

Answer: (B)

$\frac{8}{1+\sqrt{2}}$

$Rr=\frac{abc}{4\Delta }\times \frac{\Delta }{s}=\frac{abc}{2(a+b+c)}$

$=\frac{b\cdot b\cdot 2b\cos \alpha }{2(b+b+2b\cos \alpha )}=\frac{b^2\cos \alpha }{2(1+\cos \alpha )}$

$=\frac{4^2\cos {45}^{\circ }}{2(1+\cos {45}^{\circ })}=\frac{8}{1+\sqrt{2}}$