Answer:
The remaining area = 4.201 m²
Explanation:
Given that
AB=D= 4 m (R=2 m)
The area of AB semicircle
[tex]A=\pi R^2/2[/tex]
A=2π
Thee area of small semicircle
a=5π/6 + 2√3/4 m²
a=5π/6 + √3/2 m²
So the remaining area = A- a
= 2π - (5π/6 + √3/2) m²
The remaining area = 2π - (5π/6 + √3/2) m²