In the given figure, the radius of the inner circle is 40 cm and the lengths of PQ and RS are 90 cm and 64 cm respectively. What is the radius of the outer circle?

We start by drawing OM perpendicular to PQ.

We know that the perpendicular drawn from the centre of a circle to the chord bisects the chord.

$\begin{array}{rcl}\therefore \mathrm{RM}& =& \frac{1}{2}\mathrm{RS}\\ & =& \frac{1}{2}\times 64\\ & =& 32 \mathrm{cm}\end{array}$

$\begin{array}{rcl}\mathrm{PM}& =& \frac{1}{2}\mathrm{PQ}\\ & =& \frac{1}{2}\times 90\\ & =& 45 \mathrm{cm}\end{array}$

On applying Pythagoras theorem to ΔORM, we obtain

OM² + RM² = OR²

⇒ OM² + (32 cm)² = (40 cm)²

⇒ OM² = (1600 − 1024) cm² = 576 cm²

⇒ OM = 24 cm

Similarly, in ΔOPM

OP² = OM² + PM²

⇒ OP² = (24 cm)² + (45 cm)² = (576 + 2025) cm² = 2601 cm²

∴OP = 51 cm

Thus, the radius of the outer circle is 51 cm.