Question

The figure shows a circle with centre O, in which the diameter AB bisects the chord CD at point E. If CE = ED = 8 cm and EB = 4 cm. Find the radius of the circle.
[3 Marks]

Answer 1

Join OD. Let radius of the circle be '$r$' cm.

OB = OD = $r$ cm
[1 Mark]

EB = 4 cm (given)

$\therefore$ OE = ($r$ - 4) cm

In $△OED$, $\angle OED={90}^o$

Using Pythagoras' theorem in $△OED$, we get

$O{D}^2=O{E}^2+E{D}^2$
[1 Mark]

$\Rightarrow$ $r^2=(r-4{)}^2+8^2$

$\Rightarrow$ $r^2=(r^2-8r+16)+64$

$\Rightarrow$ $r=10$ cm
[1 Mark]