Question

In a circle of radius 5 cm, $AB$ and $AC$ are two chords such that $AB=AC=6cm.$ Find the length of the chord $BC$.

Answer 1

Consider the diagram:

It is given that $AC=AB$.

Join OA. Let it meet BC at P and let OA$\perp$BC.

Then, OP is the perpendicular bisector of BC.
(Since the segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord)
$△PAC$ is a right angled triangle.

$\Rightarrow {PC}^2={AC}^2-{AP}^2$

$\Rightarrow ,{PC}^2=6^2-{AP}^2$
$\Rightarrow {PC}^2=36-{AP}^2$ ... (i)

$\text{In}△POC$

${PC}^2={OC}^2-{OP}^2$

$\Rightarrow {PC}^2=25-{(5-AP)}^2$ [$PO=AO-AP$]

$\Rightarrow {PC}^2=25-(25+{AP}^2-10AP)$

$\Rightarrow {PC}^2=10AP-{AP}^2$ ... (ii)

Using (i) and (ii)

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

$\Rightarrow AP=3.6cm$

Using Pythagoras' theorem for $△APC$, we have

$\Rightarrow P{C}^2=A{C}^2-A{P}^2$

$\Rightarrow P{C}^2=6^2-{3.6}^2$

$\Rightarrow P{C}^2=36-12.96=23.04$

$\Rightarrow PC=\sqrt{23.04}=4.8cm$

Since, $BC=2\times PC$

$\Rightarrow BC=2\times 4.8=9.6cm$