Question

In the given figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm. The radius of the circle is

(a) 10 cm

(b) 12 cm

(c) 6 cm

(d) 8 cm

Answer 1

(a) 10 cm

Let the radius of the circle be r cm.
Let OD = OB = r cm.
Then OE = (r - 4) cm and ED = 8 cm
Now, in right ΔOED, we have:
$O{D}^2=O{E}^2+E{D}^2$
⇒ $(r{)}^2=(r-4{)}^2+8^2$
⇒ $r^2=r^2+16-8r+64$
⇒ 8r = 80
⇒ r = 10 cm
Hence, the required radius of the circle is 10 cm.