A chord of length 40 cm is drawn at a distance of 15 cm from the centre of a circle, then the radius of the circle will be :

1. 12 cm

2. 15 cm

3. 17 cm

4. 25 cm

Let the chord be AB and the perpendicular from the center O meet the chord at M.

Since the perpendicular from the center bisects the chord AB.

AM = $\dfrac{AB}{2} = \dfrac{40}{2}$ = 20 cm

In the right-angled triangle △OMA:

Using the Pythagorean theorem:

OA² = AM² + OM²

OA² = 20² + 15²

OA² = 225 + 400

OA² = 625

OA = $\sqrt{625}$ = 25 cm.

Hence, option 4 is the correct option.