Its not a triangle. It's not a sector. It's a Trianglector!

Reference this diagram .

I will refer to the area obtained by removing the triangle portion of a sector as a segment . Note that the desired area is the area of the segment of the large quarter circle containing the desired area, minus the red and green regions. The red region is a segment of a sector of measure 60 degrees (since $EB = AB = AE$ which implies that $\triangle EAB$ is equilateral), and the green region is a sector of measure 15 degrees. The larger segment is a portion of a sector of 90 degrees.

Area of the Red Region

The area of the red region is $\frac{1}{6}(\pi) - \frac{\sqrt{3}}{4}$ . (A 60-degree sector with radius 1 minus the area of an equilateral triangle with radius 1.)

Area of the Green Region

The area of the green region is $\frac{1}{24}(\pi)$ . (A 15-degree sector with radius 1.)

Area of the Quarter-Circle Segment

The area of the quarter circle segment is $\frac{1}{4}(\pi) - \frac{1}{2}$ . (A 90-degree sector with radius 1 minus the area of an isosceles right triangle with leg length 1.)

Area of the Desired Region

This is equal to the area of the quarter-circle segment minus the sum of the areas of the red region and the blue region.

$\frac{1}{4}(\pi) - \frac{1}{2} - (\frac{1}{6}(\pi) - \frac{\sqrt{3}}{4} + \frac{1}{24}(\pi))$

I'll save you the simplification and tell you that the desired answer comes out to $\frac{\pi - 12 + 6\sqrt{3}}{24}$ , from which the desired answer is $1^2 + {(-12)}^2 + 6^2 + 3^2 + 24^2 = \boxed{766}$ .

There's lots of ways to divvy up this diagram to get at the black area. I first found the area of the triangle that shares the vertices of the "trianglector", using Heron's formula. Then I subtracted a 15° circle segment and added a 30° circle segment. As a check, we know that 2 circle sectors are involved, having areas π/12 and π/24, the difference being π/24, so right away we have a very good idea that a = 1 and e = 24, which is confirmed when all of this is worked out.

Tanishq Aggarwal probably has the more elegant solution, involving fewer computations. Very nice diagram.

First, calculate the area of triangle between DC and the intersection of the two quartercircles.

Where the two quarter circles meet, extend two lines to point A and B to make a equilateral triangle. The height of the equilateral triangle is sqrt(3/4) and the area is (sqrt3)/4.

Next, calculate the area between the equilateral triangle and the 30/360 circle . The area is is pi/2. Since theres two on each sides, multiply by two= pi/6.

Now that you have the parts, you can calculate the area of the triangle between DC and the intersection of the 2 quartercircles. Area= 1-(sqrt3)/4-(pi)/6

Finally, to calculate the blackened area, take area ACD( 1/2 sq unit)- quartercircle( 1/8(pi)- (1-(sqrt3)/4-(pi)/6) ( the area you just found above) Simplify =>1/2-1/8pi-1+sqrt3/4+pi/5 =>(-12+pi+6*sqrt3)/24 =>a=1, b=6, c=3, d=12 AND e=24