Answer:
13/32
Explanation:
You want the area of the shaded portion of the unit square shown.
Circumcenter
Points B, C, E are shown as equidistant from point F, so will lie on a circle centered at F. The center of that circle is at the point of coincidence of the perpendicular bisectors of BE, BC, and CE.
Without loss of generality, we can let line EF lie on the x-axis such that E is at the origin. Chord EB of the circle has a rise of 1/2 for a run of 1, so a slope of 1/2. Its midpoint is (1, 1/2)/2 = (1/2, 1/4). The perpendicular line through this point will have slope -2, so its equation can be written ...
y -1/4 = -2(x -1/2)
y = -2x +5/4
Then the x-intercept (point F) will have coordinates (0, 5/8):
0 = -2x +5/4 . . . . . y=0 on the x-axis
2x = 5/4
x = 5/8
Trapezoid
Trapezoid EFCD will have upper base 5/8, lower base 1, and height 1/2. Its area is ...
A = 1/2(b1 +b2)h
A = (1/2)(5/8 +1)(1/2) = (1/4)(13/8) = 13/32
The shaded area is 13/32.
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Additional comment
The point-slope equation of a line through (h, k) with slope m is ...
y -k = m(x -h)