Answer:
[tex]\frac{9}{2}\pi +35[/tex]
Step-by-step explanation:
The area of trapezoid is given by the formula:
[tex]A=\frac{1}{2}(b_1+b_2)h[/tex]
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:
[tex]A=\frac{1}{2}(b_1+b_2)h\\A=\frac{1}{2}(6+8)*5\\A=35[/tex]
Now, area of semicircle is:
[tex]A=\frac{\pi r^2}{2}[/tex]
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:
[tex]A=\frac{\pi r^2}{2}\\A=\frac{\pi (3)^2}{2}\\A=\frac{9}{2} \pi[/tex]
Total area of the figure is [tex]\frac{9}{2}\pi +35[/tex]
