Question

The figure below shows a circle centered at O and of radius $6cm$. AB and AC are two chords such that $AB=AC=6cm$. If AP is perpendicular to BC, find the length of the chord BC.

• A. $\sqrt{3}$ cm
• B. $3\sqrt{3}$ cm
• C. $6\sqrt{3}$ cm
• D. $8\sqrt{3}$ cm

Answer 1

The correct option is C $6\sqrt{3}$ cm


In $△APC$
$\because$ AP is perpendicular to BC
$\Rightarrow A{C}^2=P{C}^2+A{P}^2$
$\Rightarrow P{C}^2=A{C}^2-A{P}^2$
$\Rightarrow (6{)}^2-A{P}^2$
$\Rightarrow 36-A{P}^2$ .....(i)
In $△POC$
$\Rightarrow P{C}^2=O{C}^2-O{P}^2$
$=36-(6-AP{)}^2$
[$\because PO=AO-AP=6-AP$]
$P{C}^2=36-(36+A{P}^2-12AP)$ ......(ii)
Compare (i) and (ii)
$36-A{P}^2=12AP-A{P}^2$
$\Rightarrow AP=3cm$
$P{C}^2=36-A{P}^2$
$=36-9=27$
$PC=\sqrt{27}=3\sqrt{3}cm$
Since a perpendicular from the centre of a circle to a chord bisects the cord
$BC=2\times PC=2\times 3\sqrt{3}cm=6\sqrt{3}cm$