A chord of length 16 cm is at a distance of 15 cm from centre of the circle. Find the length of chord of same circle which is at 8 cm away from circle.

From figure,

AB is the chord of length 16 cm which is at a distance 15 cm from the center so OC = 15 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 = 15² + 8²

⇒ OA² = 225 + 64

⇒ OA² = 289

⇒ OA = $\sqrt{289}$ ​ ⇒ OA = 17 cm

Radius = 17 cm,

∴ OD = 17 cm.

From figure,

In right angle triangle ODF,

⇒ OD² = OF² + DF² (By pythagoras theorem)

⇒ DF² = OD² - OF²

⇒ DF² = 17² - 8²

⇒ DF² = 289 - 64 = 225

⇒ DF = $\sqrt{225}$ = 15 cm.

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

DF = FE = 15 cm

From figure,

DE = DF + FE = 15 cm + 15 cm = 30 cm.

Hence, the length of chord which is at a distance of 8 cm from the center of the circle = 30 cm.