To answer this question, we have to find the are of the triangle, the rectangle, and the half of the circle (semicircle). Then we can proceed as follows:
It is given by the formula:
[tex]\begin{gathered} A_{\text{triangle}}=\frac{b\cdot h}{2} \\ b=3m \\ h=5m \end{gathered}[/tex]From the figure, we can see the values of the base, b = 3m, and the height of the triangle, h = 5m. Then we have:
[tex]\begin{gathered} A_{\text{triangle}}=\frac{b\cdot h}{2} \\ A_{\text{triangle}}=\frac{3m\cdot5m}{2}=\frac{15m^2}{2} \\ A_{\text{triangle}}=7.5m^2 \end{gathered}[/tex]Therefore, the area of the triangle is equal to 7.5 square meters.
We know that the area of the rectangle is given by:
[tex]\begin{gathered} A_{\text{rectangle}}=b\cdot h \\ b=6m \\ h=5m \\ A_{\text{rectangle}}=6m\cdot5m=30m^2 \\ A_{\text{rectangle}}=30m^2 \end{gathered}[/tex]To find this area, we needed to multiply the base, b = 6m, times the height, h = 5m.
Therefore, the area of the rectangle is 30 square meters.
We have that the area of a circle is given by:
[tex]A_{\text{circle}}=\pi r^2[/tex]Where
• π = 3.14 (we will use this value for π)
,• r is the radius of the circle
However, we need the area of the semicircle - half of the circle:
[tex]A_{\text{semicircle}}=\frac{1}{2}\pi r^2[/tex]We also need the value for the radius. The radius is half the diameter of the circle. In this case, the diameter of the circle is d = 6m. Therefore:
[tex]\begin{gathered} r=\frac{d}{2}\Rightarrow r=\frac{6m}{2}=3m \\ r=3m \end{gathered}[/tex]Then, we have:
[tex]\begin{gathered} A_{\text{semicircle}}=\frac{1}{2}\pi r^2 \\ A_{\text{semicircle}}=\frac{1}{2}\cdot3.14\cdot(3m)^2 \\ A_{\text{semicircle}}=\frac{1}{2}\cdot3.14\cdot9m^2 \\ A_{\text{semicircle}}=14.13m^2 \end{gathered}[/tex]Therefore, the area of the semicircle is 14.13 square meters.
Now, the area of the figure is the sum of the areas of the triangle, rectangle, and semicircle:
[tex]\begin{gathered} A_{\text{fig}}=A_{\text{triangle}}+A_{\text{rectangle}}+A_{\text{semicircle}} \\ A_{\text{fig}}=7.5m^2+30m^2+14.13m^2 \\ A_{\text{fig}}=51.63m^2 \end{gathered}[/tex]Therefore, in summary, the area of the figure is (to the nearest hundredth):
[tex]A=51.63m^2[/tex]