Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.

Let $r$ be the radius and let $d_1$ and $d_2$ be the perpendicular distances from the centre to the chords of length 10 cm and 24 cm respectively.

Half-chord for 10 cm chord = 5 cm;

For 24 cm chord = 12 cm.

$$r^2 = d_1^2 + 25$$

$$\Rightarrow d_1^2 = r^2 - 25 \quad \text{...(i)}$$

$$r^2 = d_2^2 + 144$$

$$\Rightarrow d_2^2 = r^2 - 144 \quad \text{...(ii)}$$

Since, the longer chord (24 cm) is closer to the centre, $d_2 < d_1$.

Both chords are on the same side, so distance between them = $d_1 - d_2 = 7$.

$\text{So}, d_{1} = d_{2} + 7 \text{...} (iii)$

Substituting eq. (iii) in eq. (i),

$$(d_2 + 7)^2 = r^2 - 25$$

$d_{2}^{2} + 14d_{2}^{2} + 49 = r^{2} - 25$ $(r^{2} - 144) + 14d_{2}^{2} + 49 = r^{2} - 25$ [From eq. (ii)]
$14d_{2}^{2} - 95 = -25$ $14d_{2}^{2} = 70$ $\rightarrow d_{2}^{2} = 5$
Then $r^{2} = 5^{2} + 144 = 25 + 144 = 169$, So, $r = 13$ cm.
Radius of the circle $= 13$ cm.