Bob is building a tunnel greenhouse. The greenhouse is 50 feet long and has semi-circular ends with a radius of 10 feet. We need to find the total surface area of plastic sheeting required to cover the top and both ends of the frame.

1. Problem Description

Bob is building a tunnel greenhouse. The greenhouse is 50 feet long and has semi-circular ends with a radius of 10 feet. We need to find the total surface area of plastic sheeting required to cover the top and both ends of the frame.

2. Solution Steps

The greenhouse consists of a cylinder cut in half lengthwise, with two half-circles as ends. The plastic sheeting will cover the curved surface of the half-cylinder and the two ends, which together make a complete circle.

First, let's find the area of the two ends. The two ends combined form a complete circle with radius $r = 10$ feet. The area of a circle is given by:

$A_{circle} = \pi r^2$

$A_{circle} = \pi (10)^2 = 100\pi$ square feet.

Next, we need to calculate the area of the curved part. This part is half of a cylinder with radius $r=10$ and length $h=50$. The surface area of a cylinder is given by:

$A_{cylinder} = 2\pi r h$

However, we only need half of the lateral surface area of the cylinder. The area of the half-cylinder (the curved part of the greenhouse) is given by:

$A_{half-cylinder} = \frac{1}{2} (2\pi r h) = \pi r h$

$A_{half-cylinder} = \pi (10)(50) = 500\pi$ square feet.

The total area of plastic sheeting needed is the sum of the area of the two ends (the circle) and the area of the half-cylinder:

$A_{total} = A_{circle} + A_{half-cylinder} = 100\pi + 500\pi = 600\pi$

Now, we need to calculate the numerical value and round it to the nearest tenth:

$A_{total} = 600\pi \approx 600 \times 3.14159 = 1884.954$

Rounding to the nearest tenth, we have $1885.0$ square feet.

3. Final Answer

1885.0 square feet