+38 
The figure is make up of 3 circles. The small circle has centre O and a radius of 6 cm. The big circle, has centre O and a radius of 10 cm. The diameter of the big circle cuts through the centre of the medium-sized circle and the small circle. The three regions formed are indicated as X, Y and Z
(a) Find the radius of the medium-sized circle.
(b) Find the area of region Z. Use a calculator to obtain the value of π. (Round off to nearest 2 decimal places).
(c) Express the area of the region Y as a ratio to the area of region X
+129852 I'm assuming here that the small circle touches the medium circle like so
(a) The radius of the medium circle = (6 + 10) / 2 = 8
(b) The area of the medium circle = pi * 8^2 = 64 pi
The area of Z = area of large circle - area of medium circle = (10^2 - 8^2) pi = 36pi ≈ 113.1 cm^2
(c) The area of the small circle = Y = pi * 6^2 = 36pi
X is the area between the medium circle and the small circle = (8.^2 - 6^2) pi
So Y / X = 36 / [ (8^2 - 6^2) ] = 36/28 = 9 / 7
+237 (a) D = 6 * 2 + (10 - 6)
= 12 + 4
= 16
r = 1/2D = 16 * 1/2 = 8
(b) Sz = π ((102) - π - (82))
= π(100 - 64) = 113.097 cm2 ≈ 113.10 cm2
(c) Sy = π * 62 = 36π
Sx = π * 82 - π * 62 = 64π - 36π = 28π
Y/X = SY/SX = 36π/28π = 9/7
