Question

In the given figure, $O$ is the centre of the circle $AB$ and $CD$ are diameters perpendicular to each other. Find the ratio of the perimeter of the semi-circle $ADB$ and triangle $ADB$

• A. $(\pi +2)(\sqrt{2}+1):2$
• B. $(\pi +2)(\sqrt{2}-1):2$
• C. $\lambda (\sqrt{2}-1):2$
• D. $\pi (\sqrt{2}+1):2$

Answer 1

The correct option is D $\pi (\sqrt{2}+1):2$

Let radius of circle $=r$

in right triangle $ADO$

$A{D}^2=O{A}^2+O{D}^2$

$A{D}^2=r^2+r^2$

$A{D}^2=2r^2⟹AD=\sqrt{2}r$

Perimeter of Semicircle arc $ADB=\pi r$

Perimeter of $\Delta ADB=AD+BD+AB=\sqrt{2}r+2r+\sqrt{2}r=2r(\sqrt{2}+1)$

Ratio=$\frac{\pi r}{2r(\sqrt{2}+1)}=\frac{\pi }{2(\sqrt{2}+1)}$

$⟹\frac{\pi }{2}(\sqrt{2}-1)$