Put me down for C as well.
We have triangle ABF which is a right triangle (it shares a straight line with another right triangle) and we have two measurements for it's sides which means we can find the third side via Pythagorean theorem:
9^2 + b^2 = 41^2
81 + b^2 = 1681
b^2 = 1600
b = 40Now we have another triangle ADF which is also right because it shares a vertical angle with right triangle BFE. We also have two sides of this triangle so we can find the third by the Pythagorean theorem:
40^2 + b^2 = 50^2
1600 + b^2 = 2500
b^2 = 900
b = 30
Now look at triangle FEB and FAD. They both share a right angle and two parallel lines. These triangles are similar. We can set up a ratio knowing two sides of FAD and one side and one unknown side of FEB to find that side:
(30/40) : (9/x)
30x = 360
x = 12
We can use another set of similar triangles to find the base for triangle BDC. Note that line FA is parallel to DC and that AD is parallel to CB. Also, both triangles have a right angle at points D and F. If you were to rotate FAD onto BDC (as done with FAD and FEB) you would see that their angle measures are the same and all of their corresponding sides parallel.
(30/50) : (39/x)
30x = 1950
x = 65
We need to find the measure for line DC to get the other base of FEDC. Triangle FEB is similar to DCB because they both share a right angle (at f and d respectively) and because they share two sets of parallel lines.
(9/12) : (39/x)
3/4 : 39/x
3x = 156
x=52
(We actually do not need to find the base of BC)
Now we have the two bases of FEDC and the height:
A = h * (b1 + b2) / 2
A = 30 * (64) / 2
A = 30 * 32
A = 960
C
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