Question

In a circle of radius 5 cm and centre O, AB and AC are two chords such that AB = AC = 6 cm. AO is the perpendicular bisector to BC. Find the length of the chord BC.

• A. 5.6 cm
• B. 9.6 cm
• C. 10.6 cm
• D. 11.6 cm

Answer 1

The correct option is B

9.6 cm

Given, AO is the perpendicular bisector of chord BC i.e, PC = PB. Join O and C


In $△ACP$,
$A{C}^2=P{C}^2+A{P}^2$
$\Rightarrow P{C}^2=A{C}^2-A{P}^2$
$\Rightarrow P{C}^2=(6{)}^2-A{P}^2$
$\Rightarrow P{C}^2=36-A{P}^2\dots \left(1\right)$

In $△POC$,
$P{C}^2=O{C}^{-}O{P}^2$
$\Rightarrow P{C}^2=O{C}^2-(AO-AP{)}^2$
$\Rightarrow P{C}^2=(5{)}^2-(5-AP{)}^2$
$\Rightarrow P{C}^2=25-(25+A{P}^2-10AP)$
$\Rightarrow P{C}^2=10AP-A{P}^2\dots \left(2\right)$

From (1) and (2),
$36-A{P}^2=10AP-A{P}^2$
$\Rightarrow AP=3.6cm\dots \left(3\right)$

Substituting (3) in (1),
$P{C}^2=36-(3.6{)}^2$
$\Rightarrow P{C}^2=36-12.96$
$\Rightarrow P{C}^2=23.04$
$\Rightarrow PC=4.8$

Hence, length of the chord $BC=2PC=9.6cm$