A circle has a chord of length 48 cm that lies at a distance of 7 cm from its centre. What is the radius of the circle?

The given information can be represented as

Here, O represents the centre of the circle and PQ = 48 cm represents the chord.

OR = 7 cm is the perpendicular from centre O to chord PQ. OQ is a radius of the circle.

The perpendicular drawn from the centre of a circle to a chord bisects the chord.

∴ PR = RQ

⇒ $\mathrm{RQ}=\frac{1}{2}\mathrm{PQ}=\frac{1}{2}\times 48=24 \mathrm{cm}$

Applying Pythagoras Theorem in right-angled triangle ORQ,

$$\begin{gathered} {\mathrm{OQ}}^2={\mathrm{OR}}^2+{\mathrm{RQ}}^2 \\ \Rightarrow {\mathrm{OQ}}^2=7^2+{24}^2=49+576=625 \\ \Rightarrow \mathrm{OQ}=25 \mathrm{cm} \end{gathered}$$

Thus, the radius of the circle is 25 cm.