I get (D).
Observe the following (I'm omitting some intermetiate calculations that apply to 3-4-5 triangles, 45-45-90 triangles, and sines)
Aagc = Aaob - Aaod - Aagb
From the problem description you get that
<AOB = 90 since the semicircle is a quarter of the total circle
Segment AC = 1 and segment CO = 3 from the ratios given in the problem.
Given this you can quickly see that AOD is a 3-4-5 triangle and thatn AOB is a 45-45-90 triangle
Aaob = .5 * 4 * 4 = 8
Aod = .5 * 3 * 4 = 6
Now we only need to find the Aagb. Since AOB is 45-45-90 we know that segment AB is of length 4*root(2). Thus we are left with the problem of finding the height of AGB. That is tricky...
Manufacture a new point, D' which is perpendicular to segment AB and intersects D. Also make a point G' that bisects segment AB. Note that AGG' and ADD' are similar triangles.
Since D'BD is 45 degrees, both D'D and D'B must be root(2)/2 (use sine(45)=1/root(2))
Now to utilize the similar triangles note the following ratio will hold:
GG' / DD' = AG' / AD'
where GG' is the desired value. Filling in numbers:
GG' / (root(2)/2) = 2 * root(2) / (4*root(2) - root(2) / 2)
Cross multiplying we get
GG' = 4*root(2) / 14
Going back to the original formula
Aagc = Aaob - Aaod - Aagb
Aagc = 8 - 6 - .5 * (4*root(2)) * (4*root(2) / 14))
Simplifying you get 6/7
...
That being said if I had seen the problem and knew that I had to do those calculations I would have skipped it.
