Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. Class 9

Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. Class 9

Question 1.

Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle. Class 9

Solution:

AB = 24 cm, CD = 10 cm, MN 7 cm

Let OM = x and r = radius of circle.

Since perpendiculars drawn from centre bisects the chords

∴ AM = $\frac{A B}{2}=\frac{24}{2}$ = 12 cm and

CN = $\frac{C D}{2}=\frac{10}{2}$ = 5 cm

In right-angled ∆OMA,

By Baudhayana-Pythagoras theorem,

OA² = OM² + AM²

∴ r² = x² + 12² ...(i)

Similarly, in ∆ONC,

OC² = ON² + CN²

∴ r² = (x + 7)² + 5² ...(ii)

From equations (i) and (ii)

x² + 122 = (x + 7)² + 5²

∴ x² + 144 = x² + 14x + 49 + 25

14x = 70

∴ x = 5 cm

Substituting x = 5 in equation (i), we get

∴ r² = 5² + 12²

∴ r² = 25 + 144

∴ r² = 169

∴ r = 13

∴ Radius of the circle is 13 cm.