Question

The figure below shows a circle centered at O and of radius 5 cm. AB and AC are two chords such that AB = AC = 6 cm. AP is perpendicular to BC. Find the length of the chord BC. [4 Marks]

Answer 1

In $△APC$,

$\Rightarrow A{C}^2=P{C}^2+A{P}^2$
$\Rightarrow P{C}^2=A{C}^2-A{P}^2$
$\Rightarrow P{C}^2=6^2-A{P}^2$
$\Rightarrow P{C}^2=36-A{P}^2$ ---(i) [1 Mark]

In $△POC$,
$PO=AO-AP$
$PO=5-AP$

$\Rightarrow P{C}^2=O{C}^2-O{P}^2$
$\Rightarrow P{C}^2=25-(5-AP{)}^2$​

$\Rightarrow P{C}^2=25-(25+A{P}^2-10AP)$

$\Rightarrow P{C}^2=10AP-A{P}^2$ ------- (ii) [1 Mark]

Substitute (i) in (ii)

$36-A{P}^2=10AP-A{P}^2$

$\Rightarrow AP=3.6cm$

$P{C}^2=36-A{P}^2$
$P{C}^2=36-12.96$
$P{C}^2=23.04$
$PC=4.8cm$ [1 Mark]

Since a perpendicular from the centre of a circle to a chord bisects the chord,
$BC=2\times PC$
$=2\times 4.8$
$=9.6cm$ [1 Mark]

$\therefore$ The length of chord $BC$ is $9.6cm$.