Question

Radius of a circle with centre O is 25 cm. Find the distance of a chord from the centre if length of the chord is 48 cm.

Answer 1

Perpendicular drawn from the centre of the circle to the chord bisects the chord.
So, $\mathrm{PD}=\frac{1}{2}\mathrm{CD}=\frac{48}{2}=24\mathrm{cm}$
In $△$OPD,
We apply the Pythagoras theorem,
$$\begin{gathered} {\mathrm{OP}}^2+{\mathrm{PD}}^2={\mathrm{OD}}^2 \\ \Rightarrow {\mathrm{OP}}^2+{24}^2={25}^2 \\ \Rightarrow {\mathrm{OP}}^2={25}^2-{24}^2=625-576=49 \\ \Rightarrow {\mathrm{OP}}^2=49 \\ \Rightarrow \mathrm{OP}=7\mathrm{cm} \end{gathered}$$
Hence, the distance of the chord CD from the centre O is 7 cm.