The area of the figure is the sum of the area of the two semi-circles and that of the rectangle.
Given a rectangle of length, l, and width, w, the area of the rectangle is given as:
[tex]A_r=l\times w[/tex]
The area of a semicircle with radius r has an area that is half that of a circle.
Hence,
[tex]A_S=\frac{\pi r^2}{2}_{}[/tex]
Substitute the values l=90 and w=58 into the area of a rectangle:
[tex]A_r=90\times58=5220[/tex]Since the semicircles are two, hence the area would be:[tex]2A_s=2(\frac{\pi r^2}{2})=\pi r^2[/tex]
The radius (r) is half the length of the diameter (d):
[tex]r=\frac{d}{2}[/tex]
Substitute d=58 which is also the width of the rectangle:
[tex]r=\frac{58}{2}=29[/tex]
Substitute r=29 into the area of two semicircles:
[tex]\begin{gathered} 2A_{s=}\pi r^2;r=29,\pi=3.14 \\ \Rightarrow3.14(29)^2=2640.74 \end{gathered}[/tex]
Add the areas 2As and Ar to get the area of the figure:
[tex]2A_s+A_r=2640.74+5220=7860.74m^2[/tex]
