What’s the area of the polygon?

Answer:

[tex]108 in^{2}[/tex]

Step-by-step explanation:

Separate it into 2 parts:

The first would be that the length and width of the smaller rectangle would be, 7 in. and 6 in.

7 x 6 = 42 in^2

The second would be that the length and width of the larger rectangle would be, 11 in. and 6 in.

11 x 6 = 66 in^2

Total:

66 + 42 = 108 in^2

Answer:

108 [tex]in ^2[/tex]

Step-by-step explanation:

This polygon consists of two rectangles, one in the top-right hand corner and another in the bottom. To divide these two rectangles, imagine a line extended from the central horizontal line in the polygon. As I work the problem, remember that the formula for the area of a rectangle is [tex]A=l * h[/tex], where A is the area of the rectangle, l the length of the rectangle (one of the sides), and h is the height of the rectangle (the length of a side perpendicular to l).

Starting with the top-right rectangle:

We already know one of the sides of the rectangle is 4 in, so let's imagine this to be h, because it is the height of the rectangle. Finding l is more complicated. We know the length of the larger rectangle is 12 in (more on that later) and the 6 in length is the part of the length that is not a part of the top-right rectangle. This, since the vertical line all the way to the right is straight, the length is the difference between these two lengths, or:

[tex]l=12-6=6[/tex] in

Now that we know l and h, we can evaluate the area of the first rectangle, which is

[tex]A= 6 *4= 24[/tex] in^2

Now for the bottom rectangle:

We know both l and h for this rectangle, which are 12 in and 7 in respectively. Using the area formula for a rectangle above:

[tex]A=12*7=84[/tex] in^2

Since the polygon contains both rectangles, its area is the sum of the areas of the two rectangles. Thus,

[tex]A=24+84=108[/tex] in^2

QED

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