SSC CHSL 17 Mar 2020 Shift 3 Paper

AB is the chord of length $24\mathrm{cm}$. OC is the distance of chord from centre $=5\mathrm{cm}$ We know that, perpendicular from centre to the chord, bisect the chord. ∴ $\mathrm{AC}=\mathrm{CB}=\frac{24}{2}=12\mathrm{cm}$
Now, in right $\mathrm{\Delta }\mathrm{OAC}$,
${\left(\mathrm{OA}\right)}^2={\mathrm{OC}}^2+{\mathrm{AC}}^2$ (by Pythagoras theorem)
⇒ ${\left(\mathrm{OA}\right)}^2={\left(5\right)}^2+{\left(12\right)}^2$
⇒ $\mathrm{OA}=\sqrt{25+144}=\sqrt{169}=13\mathrm{cm}$
∴ Area of circle $=\mathrm{\pi }{\mathit{r}}^2=\mathrm{\pi }{\left(\mathrm{OA}\right)}^2$
$=\mathrm{\pi }\times {\left(13\right)}^2=\frac{22}{7}\times 169$
$=531.14{\mathrm{cm}}^2$