What is the area of this figure? Enter your answer in the box. in² A parallelogram with a right triangle created inside it with a short leg length of 16 inches and a long leg length of 20 inches. A triangle attached to the top of the parallelogram has a short leg measurement of 8 inches.

I constructed the figure with the information that you provide, please check the picture.
Since there is a right triangle inside the parallelogram, the parallelogram is a rectangle with [tex]length=20in[/tex] and [tex]width=16in[/tex]; remember that the area of a rectangle is (length)(width), so we can substitute our values to find its area:
[tex]A=(20in)(16in)[/tex]
[tex]A=320in^{2} [/tex]
On the other hand, the triangle attached to the top of the parallelogram has [tex]base=20in[/tex], and [tex]altitude=8in[/tex]; since the area of a triangle is [tex] \frac{1}{2} (base)(altitude)[/tex], we can substitute our values to get:
[tex]A= \frac{1}{2} (20in)(8in)[/tex]
[tex]A=80in^{2} [/tex]

Now the only thing lefts is add the area of the parallelogram and the triangle to get our total area:
[tex]totalA=320in^{2} +80in^{2} =400in x^{2} [/tex]

We can conclude that the area of the figure is [tex]400in^{2} [/tex].