SSC CHSL 15 Oct 2020 Shift 1 Paper

Given, Radius $=5\mathrm{cm}$
We know that, the length of two chords are equal, then the perpendicular distance of the chords from the centre are equal.
∴ $\mathrm{OA}=\mathrm{OB}$
Given, $\mathrm{AB}=6\mathrm{cm}$
∴ $\mathrm{OA}+\mathrm{OB}=6$
⇒ $\mathrm{OA}=\mathrm{OB}=3\mathrm{cm}$
$\ln \mathrm{\Delta }\mathrm{OBC}$
${\mathrm{OB}}^2+{\mathrm{BC}}^2={\mathrm{OC}}^2$
⇒ ${\mathrm{BC}}^2=5^2-3^2=25-9=16$
$\therefore \mathrm{BC}=\sqrt{16}=4\mathrm{cm}$ and length of each chord $=2\times 4=8\mathrm{cm}$.