1. What is the surface area of the cylinder in terms of pi? The diagram is not drawn to scale.

A) 990 pi in^2
B) 342 pi in^2
C) 378 pi in^2
D) 504 pi in^2

The radius is 9 and the height is 19.

when I solve this, no matter what the formula is, I always get 523, which means there's something I'm doing wrong. Any help would be really great. Thank you

a = 2πr(r+h) = 2π(9)(9 + 19) = 504π

So, D

How did you get 523?

I used the formula, 2 pi r^2+ pi dh.

It seems what you did was much much easier though.

Thank you so much by the way!

D) 504 pi in^2

thanks s

To find the surface area of a cylinder, we need to consider two parts: the curved surface area and the area of the two circular bases.

1. Curved Surface Area:
The curved surface area of a cylinder is calculated by multiplying the height (h) of the cylinder by the circumference of its base (2πr), where r is the radius of the cylinder. The formula for the curved surface area (CSA) is: CSA = 2πrh.

In this case, the radius (r) is given as 9 and the height (h) is given as 19. Plugging in these values into the formula, we get:
CSA = 2π(9)(19)
CSA = 342π in^2

2. Area of the Circular Bases:
The area of a circle is given by the formula A = πr^2, where r is the radius. Since a cylinder has two circular bases, we multiply the area of one base by 2 to get the total area of both bases.

In this case, the radius (r) is given as 9. Plugging in this value into the formula, we get:
Total area of the circular bases = 2(π)(9^2)
Total area of the circular bases = 2(π)(81)
Total area of the circular bases = 162π in^2

Now, to find the total surface area of the cylinder, we sum up the curved surface area and the area of the circular bases:
Total Surface Area = Curved Surface Area + Total area of the circular bases
Total Surface Area = 342π + 162π
Total Surface Area = 504π in^2

Therefore, the surface area of the cylinder in terms of pi is 504π square inches. Comparing this to the given answer choices, the correct option is D) 504π in^2.

your formula is correct, but your answer is not. Oh, well.