Question

In a circle of radius 17 cm, two parallel chords of lengths 30 cm and 16 cm are drawn.

Prove that:
(i) distance between the chords, if both the chords are on the opposite sides of the centre, is 23 cm, and
(ii) distance between the chords, if both the chords are on the same side of the centre is 7 cm.

Answer 1

(i)

Radius of the circle C = 17 cm

Length of chord AB = 30 cm

Length of chord CD = 16 cm

Draw OM $\perp$ AM and OP $\perp$ CD and join OA and OC.

$\because$ The perpendicular from O, bisects the chord,

$\therefore AM=\frac{30}{2}=15cmandCP=\frac{16}{2}=8cm$

In right $\Delta OAM$,

$O{A}^2=O{M}^2+A{M}^2$

$\Rightarrow (17{)}^2=O{M}^2+{15}^2\Rightarrow 289=O{M}^2+225$

$\Rightarrow O{M}^2=289-225=64=(8{)}^2$

$\therefore$ OM = 8 cm . . . (i)

(ii)

In right $\Delta OCP$,

$O{C}^2=O{P}^2+C{P}^2$

$\Rightarrow (17{)}^2=O{P}^2+(8{)}^2\Rightarrow 289=O{P}^2+64$

$\therefore O{P}^2=289-64=225=(15{)}^2$

$\therefore$ OP = 15 cm. . . .(ii)

Now in figure (ii) PM = OP - OM = 15 - 8 = 7 cm

and figure (i) PM = OP + OM = 15 + 8 = 23 cm