Question

PLEASE HELP! Urgent!!!

Answer 1

Answer:

$\frac{9}{2}\pi +35$

Step-by-step explanation:

The area of trapezoid is given by the formula:

$A=\frac{1}{2}(b_1+b_2)h$

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:

$A=\frac{1}{2}(b_1+b_2)h\\A=\frac{1}{2}(6+8)*5\\A=35$

Now, area of semicircle is:

$A=\frac{\pi r^2}{2}$

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:

$A=\frac{\pi r^2}{2}\\A=\frac{\pi (3)^2}{2}\\A=\frac{9}{2} \pi$

Total area of the figure is $\frac{9}{2}\pi +35$