Question

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Answer 1

3) $55.2 in^2$

4) $42 yd^2$

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:

$A=6A_T$

The area of one triangle can be written as:

$A_T=\frac{1}{2}bh$

where:

$b=4.6 in$ is the base of the triangle

$h=4 in$ is the height

Substituting,

$A_T=\frac{1}{2}(4.6)(4)=9.2 in^2$

And so, the area of the regular hexagon is:

$A=6A_T=6(9.2)=55.2 in^2$

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:

$A=2A_p$

where $A_p$ is the area of one parallelogram, which is given by

$A_p = bh$

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

$A_p=(7)(3)=21 yd^2$

And therefore, the area of the figure is:

$A=2A_p=2(21)=42 yd^2$