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