Question

O is centre of the circle. Find the length of radius, if the chord of length 24 cm is at a distance of 9 cm from the centre of the circle.

Answer 1

Join OA.

Let the perpendicular drawn from point O to the chord AB be P.
We know that the perpendicular drawn from the centre of the circle to the chord bisects the chord.
So, $\mathrm{AP}=\frac{\mathrm{AB}}{2}=\frac{24}{2}=12\mathrm{cm}$
In $△$OPA,
We apply the Pythagoras theorem,
$$\begin{gathered} {\mathrm{OP}}^2+{\mathrm{AP}}^2={\mathrm{OA}}^2 \\ \Rightarrow 9^2+{12}^2={\mathrm{OA}}^2 \\ \Rightarrow {\mathrm{OA}}^2=81+144=225 \\ \Rightarrow \mathrm{OA}=\sqrt{225}=15\mathrm{cm} \end{gathered}$$
Hence, the radius of the circle is 15 cm.