#### Question

The diameter and a chord of a circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the centre of the circle?

#### Solution

AB is the diameter and AC is the chord.

Draw OL ⊥ AC

Since OL ⊥ AC and hence it bisects AC, O is the centre of the circle.

Therefore, OA = 10 cm and AL = 6 cm

Now, in Rt. ΔOLA,

`AO^2 = AL^2 + OL^2`

`⇒ 10 ^2 = 6^2 + OL^2`

`⇒ OL^2 = 100 - 36 = 64`

⇒ OL = 8 cm

Therefore, chord is at a distance of 8 cm from the centre of the circle.

Is there an error in this question or solution?

Solution The Diameter and a Chord of a Circle Have a Common End-point. If the Length of the Diameter is 20 Cm and the Length of the Chord is 12 Cm, How Far is the Chord from the Centre of the Circle? Concept: Chord Properties - a Straight Line Drawn from the Center of a Circle to Bisect a Chord Which is Not a Diameter is at Right Angles to the Chord.