We are asked to determine the perimeter of the given figure. The perimeter is found by adding the length of all the sides of the figure. Taking each square as a unit we divide the figure as follows:
we only need to determine the lengths of the exterior lines of the figure. For figure 1 we have a triangle that has base and height equal to 3 units, therefore, the length of the hypotenuse is:
[tex]h=3\sqrt[]{2}[/tex]
Figure 2 is an equal triangle as figure 1. For figure 3 we have 2 units on top and 2 units at the bottom, and for figure 4 we have a circle. The length of the arc of the circle is given by:
[tex]S=\pi r[/tex]
Where "r" is the radius. The radius of the circle is 3 units, therefore, the length is:
[tex]S=3\pi[/tex]
For this figure we must also add 2 units for the top and 2 units at the bottom. Therefore, the length of each side of the figure is:
[tex]P=3\sqrt[]{2}+2+2+2+3\pi+2+2+2+3\sqrt[]{2}[/tex]
Solving the operations:
[tex]P=29.9[/tex]
Therfore the perimeter is 29.9 units.
To determine the area we will add the areas of each figure. Figure 1 is a triangle and its area is:
[tex]A_1=\frac{bh}{2}[/tex]
Replacing the values:
[tex]A_1=\frac{(3)(3)}{2}=\frac{9}{2}[/tex]
Since figure 2 is an equal triangle we have:
[tex]A_2=\frac{9}{2}[/tex]
For figure 3 the area is the area of a square, that is:
[tex]A_3=2\times2=4[/tex]
The area of figure 4 is the area of half a circle, that is:
[tex]A_4=\frac{\pi r^2}{2}[/tex]
Replacing:
[tex]A_4=\frac{\pi(3)^2}{2}=\frac{9}{2}\pi[/tex]
Adding each of the areas:
[tex]\begin{gathered} A=A_1+A_2+A_3+A_4 \\ A=\frac{9}{2}+\frac{9}{2}+4+\frac{9}{2}\pi \end{gathered}[/tex]
Solving the operations:
[tex]A=27.1[/tex]
Therefore, the area is 27.1 square units.
