Question

In a circle of radius 17 cm, two parallel chords of lengths 30 cm and 16 cm are drawn. Select the statements that are true.

• A. Distance between the chords if both the chords are on the opposite sides of the centre is 23 cm.
• B. Distance between the chords if both the chords are on the same side of the centre is 7 cm.
• C. Distance between the chords if both the chords are on the opposite sides of the centre is 28 cm.
• D. Distance between the chords if both the chords are on the same side of the centre is 17 cm.

Answer 1

Answer: (A), (B)

The correct options are
A

Distance between the chords if both the chords are on the opposite sides of the centre is 23 cm.

B

Distance between the chords if both the chords are on the same side of the centre is 7 cm.

(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