Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.                                                                                                                                                                                                                                                         [Page No. 105]

Let O be the centre, AB be the chord, and M be the foot of perpendicular from O to AB.

Then OM = 6 cm and OA = 7 cm.

In right triangle OMA:

$$AM^2 = OA^2 - OM^2 = 7^2 - 6^2 = 49 - 36 = 13$$

$$AM = \sqrt{13} \text{ cm}$$

Since, the perpendicular from the centre of a circle to a chord bisects the chord.

$$\text{Therefore, } AB = 2AM = 2\sqrt{13} \text{ cm}$$

Hence, the length of the chord is $2\sqrt{13}$ cm.