The figure consists of a semi-circle and a triangle.
Hence,
[tex]\text{ the area of the figure = area of the triangle }+\text{ area of the semi-circle}[/tex]Given a triangle with base b and perpendicular height, h, then the area, A, of the triangle is given by
[tex]A=\frac{bh}{2}[/tex]In this case,
b = 3km, and h = 4km
therefore,
[tex]\text{ area of the triangle = }\frac{3\times4}{2}=3\times2=6\operatorname{km}^2[/tex]Given a semi-circle with diameter, d, the area, say S, of the semi-circle, is given by
[tex]S=\frac{\pi d^2}{8}[/tex][tex]\begin{gathered} \text{taking,} \\ \pi\approx3.142 \end{gathered}[/tex]We must have that,
[tex]S=\frac{3.142d^2}{8}[/tex]In our case,
d = 4km
therefore,
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