Divide the shape into two areas as above. Because angles in a semicircle are right angles, the length marked ‘y’ is an extension of one of the ‘x’ edges, and exactly meets the bottom corner of the semicircle.
y is the hypotenuse of an isosceles right triangle whose legs are x, and so therefore y = x*sqrt(2).
Using Pythagoras on the large triangle we get that:
x^2+(x+y)^2 = 4
x^2+(x+x(sqrt(2))^2 = 4
x^2 + x^2 + 2(sqrt(2))x^2 + 2x^2 = 4
x^2 (4+2(sqrt(2)) = 4
x^2 = 4/(4+2sqrt(2)) = 2/(2+sqrt(2))
multiplying the right hand side by (2-sqrt(2))/(2-sqrt(2)) to eliminate the square root in the denominator:
x^2 = (4-2sqrt(2))/(4-2) = 2-sqrt(2)
The area of the two triangles will be respectively x(x+y)/2 and x^2/2, so the combined area is (2x^2+xy)/2. Since y=xsqrt(2), the area is: (x^2)*(2+sqrt(2)/2)
Replace x^2 with the known value of 2-sqrt(2):
(2-sqrt(2))*(2+sqrt(2))/2
(4-2)/2
1
Therefore the area of the shape is simply 1.
