In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6 cm respectively. Calculate the distance between the chords, if they are on

(i) the same side of the centre

(ii) the opposite sides of the centre.

(i) From figure,

AB and CD are chords of length 8 cm and 6 cm, respectively.

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

∴ AE = BE = $\dfrac{8}{2}$ = 4 cm and,

CF = FD = $\dfrac{6}{2}$ = 3 cm.

Radius = OA = OC = 5 cm.

In right angle triangle OAE,

⇒ OA² = AE² + OE² (By pythagoras theorem)

⇒ OE² = OA² - AE²

⇒ OE² = 5² - 4²

⇒ OE² = 25 - 16 = 9

⇒ OE = $\sqrt{9}$ = 3 cm.

In right angle triangle OCF,

⇒ OC² = CF² + OF² (By pythagoras theorem)

⇒ OF² = OC² - CF²

⇒ OF² = 5² - 3²

⇒ OF² = 25 - 9 = 16

⇒ OF = $\sqrt{16}$ = 4 cm.

Distance between two chords = EF = OF - OE = 4 - 3 = 1 cm.

Hence, distance between two chords = 1 cm.

(ii) From figure,

AB and CD are chords of length 8 cm and 6 cm respectively.

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

∴ AE = BE = $\dfrac{8}{2}$ = 4 cm and,

CF = FD = $\dfrac{6}{2}$ = 3 cm.

Radius = OA = OC = 5 cm.

In right angle triangle OAE,

⇒ OA² = AE² + OE² (By pythagoras theorem)

⇒ OE² = OA² - AE²

⇒ OE² = 5² - 4²

⇒ OE² = 25 - 16 = 9

⇒ OE = $\sqrt{9}$ = 3 cm.

In right angle triangle OCF,

⇒ OC² = CF² + OF² (By pythagoras theorem)

⇒ OF² = OC² - CF²

⇒ OF² = 5² - 3²

⇒ OF² = 25 - 9 = 16

⇒ OF = $\sqrt{16}$ = 4 cm.

Distance between two chords = EF = OF + OE = 4 + 3 = 7 cm.

Hence, distance between two chords = 7 cm.