The length of two parallel chords of a circle of radius 5 cm are 6 cm and 8 cm in the same side of the centre. The distance between them is

Let AB and CD are the two parallel chords

From the figure,

In $$\triangle\ $$OFD,

OD = radius = 5 cm

FD = 3 cm

$$\text{FD}^2+\text{OF}^2=\text{OD}^2$$

$$=$$>  $$3^2+\text{OF}^2=5^2$$

$$=$$>  $$9+\text{OF}^2=25$$

$$=$$>  $$\text{OF}^2=25-9$$

$$=$$>  $$\text{OF}^2=16$$

$$=$$>  $$\text{OF}=$$  4 cm

In $$\triangle\ $$OBE,

OB = radius = 5 cm

EB = 4 cm

$$\text{EB}^2+\text{OE}^2=\text{OB}^2$$

$$=$$> $$4^2+\text{OE}^2=5^2$$

$$=$$> $$16+\text{OE}^2=25$$

$$=$$> $$\text{OE}^2=25-16$$

$$=$$> $$\text{OE}^2=9$$

$$=$$> $$\text{OE}=$$ 3 cm

$$\therefore\ $$The distance between the parallel chords = OF - OE = 4 - 3 = 1 cm

Hence, the correct answer is Option C