Let O be the centre of the circle and AB and CD are two parallel chords of equal side of radius 1.

Solution:

Let radius = r

Distance from center to chord formula:
For a chord of length $l$, distance $d = \sqrt{r^2 - \left(\frac{l}{2}\right)^2}$

For chord AB = 10:
$OP = \sqrt{r^2 - 5^2} = \sqrt{r^2 - 25}$

For chord CD = 24:
$OQ = \sqrt{r^2 - 12^2} = \sqrt{r^2 - 144}$

Given $PQ = 7$ and chords are on same side:
$OP - OQ = 7$

So,
$\sqrt{r^2 - 25} - \sqrt{r^2 - 144} = 7$

Solve:

$\sqrt{r^2 - 25} = 7 + \sqrt{r^2 - 144}$

Square both sides:

$r^2 - 25 = 49 + r^2 - 144 + 14\sqrt{r^2 - 144}$

$-25 = -95 + 14\sqrt{r^2 - 144}$

$70 = 14\sqrt{r^2 - 144}$

$\sqrt{r^2 - 144} = 5$

$r^2 - 144 = 25$
$r^2 = 169$
$r = 13$

Diameter = $2r = 26$

✅ Final Answer: 26