3) [tex]55.2 in^2[/tex]
4) [tex]42 yd^2[/tex]
Step-by-step explanation:
3)
The regular hexagon can be seen as consisting of 6 identical triangles, so its area is equal to six times the area of one triangle:
[tex]A=6A_T[/tex]
The area of one triangle can be written as:
[tex]A_T=\frac{1}{2}bh[/tex]
where:
[tex]b=4.6 in[/tex] is the base of the triangle
[tex]h=4 in[/tex] is the height
Substituting,
[tex]A_T=\frac{1}{2}(4.6)(4)=9.2 in^2[/tex]
And so, the area of the regular hexagon is:
[tex]A=6A_T=6(9.2)=55.2 in^2[/tex]
4)
Here we have a complex figure consisting of several regular figures.
We observe that the figure consists of 2 parallelograms, on top and on bottom, so the total area of the figure is the sum of the areas of the two parallelograms:
[tex]A=2A_p[/tex]
where [tex]A_p[/tex] is the area of one parallelogram, which is given by
[tex]A_p = bh[/tex]
where:
b = 7 yd is the base of the parallelogram
h = 3 yd is the height of the parallelogram
Therefore, the area of the parallelogram is
[tex]A_p=(7)(3)=21 yd^2[/tex]
And therefore, the area of the figure is:
[tex]A=2A_p=2(21)=42 yd^2[/tex]
