SSC CHSL 19 Mar 2020 Shift 3 Paper

Given, Let the radius of the circle be $\mathit{r}\mathrm{cm}$.
We know that in a circle, perpendicular from centre of the circle to the chord bisects the chord.
⇒ $\mathrm{AC}=\mathrm{BC}=\frac{16}{2}=8\mathrm{cm}$
∴ $\ln \mathrm{\Delta }\mathrm{OAC}$
${\mathrm{AC}}^2+{\mathrm{OC}}^2={\mathrm{OA}}^2$ $[\because$ by Pythagoras theorem]
⇒ $8^2+6^2={\mathit{r}}^2$
⇒ ${\mathit{r}}^2=64+36=100$
⇒ $\mathit{r}=\sqrt{100}=10\mathrm{cm}$