Question

In the adjoining figure, ‘O’ is the centre of the circle. If AC = 10 cm and chord AB = 6 cm, find the distance of this chord from the centre of the circle.

Answer 1

We are given: 

O is the centre of the circle.

AC = 10 cm   ...(A chord passing through the circle, hence this is a diameter)

AB = 6 cm   ...(A chord)

OM ⊥ AB, where M is the midpoint of AB.

We are to find the distance from the centre O to chord AB, i.e., length OM.

Step 1: Use key circle properties.

Since AC = 10 cm and it passes through the centre O, it is a diameter.

⇒ Radius `r = (AC)/2`

= `10/2`

= 5 cm

Chord AB = 6 cm

⇒ AM = MB

= `6/2`

= 3 cm 

Step 2: Use Pythagoras theorem in triangle OMA.

In right triangle OMA:

Hypotenuse = OA = 5 cm

Base = AM = 3 cm

Height = OM = ?

Apply Pythagoras:

OA² = OM² + AM²

5² = OM² + 3²

25 = OM² + 9

OM² = 16

⇒ `OM = sqrt(16)`

⇒ OM = 4 cm

Distance from centre to chord AB = 4 cm.