Question

A chord AB of a circle whose centre is O, is bisected at E by a diameter CD. OC= OD = 15 cm and OE = 9 cm. Find the length of BC

• A. $12cm$
• B. $12\sqrt{5}cm$
• C. $6\sqrt{10}cm$
• D. $10\sqrt{3}cm$

Answer 1

The correct option is B

$12\sqrt{5}cm$

Given OE = 9 cm, OA = 15 cm

Consider $△OAE$, $\angle OEA={90}^o$
[Since the segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord]

Applying Pythagoras' theorem,

$A{E}^2=O{A}^2-O{E}^2$

$A{E}^2={15}^2-9^2$

$A{E}^2=225-81=144$

$\Rightarrow AE=12cm$ and $BE=12cm$

$\therefore$ CE = 15 + 9 = 24 cm

Consider $△CEB$, $\angle CEB={90}^o$

Applying Pythagoras theorem

$B{C}^2=C{E}^2+E{B}^2$

$B{C}^2={24}^2+{12}^2$

$B{C}^2=576+144=720$

$\Rightarrow BC=12\sqrt{5}cm$