warriors837, PLEASE find fault with this method.

KLAMKIN AREA METHOD:

**(i)** Area of BDC + Area of BDE + Area of BAE = (1/2)(Base x Height)

x + y + x = (1/2)(2)(2) = 2, then

2x + y = 2 - - -

**eq'n 1** **(ii)** Area of BDE + Area of BAE = Area of BAD

x + y = (1/2)(r^2)(theta) = (1/2)(2^2)(pi/4) = pi/2, then

x + y = pi/2,

y = (pi/2 - x) - - -

**eq'n 2** substitute** eq'n 1 **in** eq'n 2.** 2x + (pi/2 - x) = 2, then
2x - x = 2 - pi/2

x = 2 - pi/2

But we are looking for the unshaded portion of the figure,

thus, Area of unshaded portion = Area = 2x = 2(2 - pi/2) = 4 - pi = 0.8584 square units. **i hope it is correct . . . .**