A chord of length 24 cm is at a distance of 5 cm from the center of the circle. Find the length of the chord of the same circle which is at a distance of 12 cm from the centre.

Let AB be the chord of the circle with center O at the distance of 5 cm from the center of the circle.

We know that,

Perpendicular from center to chord, bisects the chord.

∴ AC = $\dfrac{AB}{2} = \dfrac{24}{2}$ = 12 cm.

In right angled triangle OAC,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OA² = OC² + AC²

⇒ OA² = 5² + 12²

⇒ OA² = 25 + 144

⇒ OA² = 169

⇒ OA = $\sqrt{169}$ = 13 cm.

From figure,

OD = OA = 13 cm (Radius of same circle)

Let chord DE be at the distance of 12 cm from the center.

In right angled triangle OFD,

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ OD² = OF² + FD²

⇒ 13² = 12² + FD²

⇒ FD² = 13² - 12²

⇒ FD² = 169 - 144

⇒ FD² = 25

⇒ FD = $\sqrt{25}$ = 5 cm.

⇒ DE = 2 × FD = 2 × 5 = 10 cm.

Hence, length of chord at a distance of 12 cm from the centre equals to 10 cm.