Question

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.

Answer 1

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 $length=20in$ and $width=16in$; remember that the area of a rectangle is (length)(width), so we can substitute our values to find its area:
$A=(20in)(16in)$
$A=320in^{2}$
On the other hand, the triangle attached to the top of the parallelogram has $base=20in$, and $altitude=8in$; since the area of a triangle is $\frac{1}{2} (base)(altitude)$, we can substitute our values to get:
$A= \frac{1}{2} (20in)(8in)$
$A=80in^{2}$

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

We can conclude that the area of the figure is $400in^{2}$.

Answer 2

Answer:    400in^2  

Step-by-step explanation:    Just had a test.

Is this the pic.

Hope this Helps.