A chord of length 16 cm is at a distance 6 cm from the center of the circle. Find the length of chord of the same circle which is at a distance of 8 cm from the center.

From figure,

AB is the chord which is at a distance 6 cm from the center so OC = 6 cm.

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

∴ CB = AC = $\dfrac{16}{2}$ = 8 cm.

In right angle triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ OA² = 6² + 8²

⇒ OA² = 36 + 64

⇒ OA² = 100

⇒ OA = $\sqrt{100}$ = 10 cm.

Radius = 10 cm,

∴ OD = 10 cm.

From figure,

In right angle triangle ODF,

⇒ OD² = OF² + DF² (By pythagoras theorem)

⇒ DF² = OD² - OF²

⇒ DF² = 10² - 8²

⇒ DF² = 100 - 64 = 36

⇒ DF = $\sqrt{36}$ = 6 cm.

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

∴ DE = DF + FE = 6 cm + 6 cm = 12 cm.

Hence, the length of chord which is at a distance of 8 cm from the center of the circle = 12 cm.