Answer:
![\frac{9}{2}\pi +35](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B2%7D%5Cpi%20%2B35)
Step-by-step explanation:
The area of trapezoid is given by the formula:
![A=\frac{1}{2}(b_1+b_2)h](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%28b_1%2Bb_2%29h)
Where
A is the area
b_1 is base 1
b_2 is base 2, and
h is the height
Looking at the figure, base 1 is the left line which goes from y = 3 to y = 3, so 6 units. Also, base 2 is the right line which goes from y = -3 to y =5, so 8 units.
The height is the horizontal distance in the middle, which goes from x = -2 to x = 3, so 5 units. Hence area of trapezoid is:
![A=\frac{1}{2}(b_1+b_2)h\\A=\frac{1}{2}(6+8)*5\\A=35](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%28b_1%2Bb_2%29h%5C%5CA%3D%5Cfrac%7B1%7D%7B2%7D%286%2B8%29%2A5%5C%5CA%3D35)
Now, area of semicircle is:
![A=\frac{\pi r^2}{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20r%5E2%7D%7B2%7D)
Where
A is the area
r is the radius
Looking at the figure, the diameter (twice radius) goes from y = -3 to y = 3, so 6 units. But radius is half of that, so 3 units. Hence area of semicircle is:
![A=\frac{\pi r^2}{2}\\A=\frac{\pi (3)^2}{2}\\A=\frac{9}{2} \pi](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20r%5E2%7D%7B2%7D%5C%5CA%3D%5Cfrac%7B%5Cpi%20%283%29%5E2%7D%7B2%7D%5C%5CA%3D%5Cfrac%7B9%7D%7B2%7D%20%5Cpi)
Total area of the figure is ![\frac{9}{2}\pi +35](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B2%7D%5Cpi%20%2B35)