Question

In a circle with centre \( P \). chord \( A B \) is drawn of length of \( 23 cm . \operatorname{seg} P Q \) is perpendicular to chord \( A B \), then find I (\(QB)\).

Answer 1

P is the centre of the circle.

AB = 23 cm

seg PQ ⊥ chord AB

Perpendicular drawn from the centre of the circle to the chord bisects the chord.

So, the length of l(QB) = \(\frac{1}{2}\) l(AB)

l(QB) = \(\frac{1}{2}\) ×23   ...[∵ l(AB) = 23 cm]

l(QB) = 11.5 cm