Question

Find the length of a chord that is at a distance of 4 cm from the centre of a circle of radius 6 cm.

Answer 1

Distance of the chord from the centre = OC = 4 cm (Given)

Radius of the circle = OA = 6 cm (Given)

In $\Delta$OCA:

Using Pythagoras theorem,

OA${}^2$ = AC${}^2$ + OC${}^2$

36 = AC${}^2$ + 16

AC2 = 36 – 16 = 20

AC = $\sqrt{20}$ = 4.47

AC = 4.47 cm

As, perpendicular from the centre to chord bisects the chord.

Therefore, AC = BC = 4.47 cm

$\Rightarrow$ AB = AC + BC = 4.47 + 4.47 = 8.94 cm