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

Given: CE = ED = 8 cm and EB = 4 cm


Construction: Join OC (OC is radius)


Let AB = 2x (diameter)


OB = OC = x (radius)


In Δ COE


CO2 = CE2 + OE2


x2 = 82 + (OB – EB)2


x2 = 82 + (x – 4)2


x2 = 82 + x2 + 42 – 2(x)(4)


x2 = 64+ x2 + 16 – 8x


8x = 64+ 16 + x2 – x2


8x = 80


x = = 10


∴ Radius = x = 10 cm

10