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 AD.

• A. 60°
• B. 70°
• C. 80°
• D. 90°

Answer 1

The correct option is B

70°

Given OE = 9 cm, OA = 15 cm

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

Applying Pythagoras' theorem,

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

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

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

$\Rightarrow AE=12cm$

OD = 15 cm and OE = 9 cm (radius)

$\therefore$ DE = 15 - 9 = 6 cm

Consider $△AED$, $\angle AED={90}^o$

Applying Pythagoras theorem

$A{D}^2=A{E}^2+E{D}^2$

$A{D}^2={12}^2+6^2$

$A{D}^2=144+36=180$

$\Rightarrow AD=6\sqrt{5}cm$