SSC CHSL 12 Oct 2020 Shift 3 Paper

Two concentric circles with center $\mathrm{O}$, AC is the chord of smaller circle which ouches the larger circle.
Radius of smaller circle $\left(\mathrm{OB}\right)=5\mathrm{cm}$
Radius of larger circle $\left(\mathrm{OA}\right)=13\mathrm{cm}$
We know that,
$\mathrm{OB}\perp \mathrm{AC}$
and $\mathrm{AB}=\frac{1}{2}\mathrm{AC}$
∴ In right $\mathrm{\Delta }\mathrm{OBA}$,
${\mathrm{OA}}^2={\mathrm{OB}}^2+{\mathrm{AB}}^2$
(by Pythagoras theorem)
⇒ ${\mathrm{AB}}^2={\mathrm{OA}}^2-{\mathrm{OB}}^2$
⇒ ${\mathrm{AB}}^2={\left(13\right)}^2-5^2$
⇒ ${\mathrm{AB}}^2=169-25$
⇒ ${\mathrm{AB}}^2=144$
⇒ $\mathrm{AB}=12\mathrm{cm}$
Now, $\mathrm{AC}=2\times \mathrm{AB}=2\times 12=24\mathrm{cm}$