7. Suppose a large piece of jungle gym equipment has the shape of a truncated cone. It needs to be painted, but you need to know how much paint to get so that you can cover its entire surface. The piece of equipment has big radius 3 feet, little radius 2 feet, height 5 feet, and slant height 6 feet. What is the surface area of the piece of equipment? $$ 123.98 \text { square inches } $$ $$ \approx 111.07 $$ square inches $$ \approx 135.09 $$ square inches

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We begin by recalling that the total surface area of a truncated cone (including both circular bases) is given by

$$
\text{Surface Area} = \pi (R+r)l + \pi R^2 + \pi r^2,
$$

where 
- $$R$$ is the larger radius,
- $$r$$ is the smaller radius,
- $$l$$ is the slant height.

For this problem, we are given:
- $$R = 3$$ feet,
- $$r = 2$$ feet,
- $$l = 6$$ feet.

The lateral (side) surface area is calculated first:

$$
\pi (R+r) l = \pi (3+2) \cdot 6 = \pi \cdot 5 \cdot 6 = 30\pi.
$$

Next, we calculate the areas of the two bases:
- The area of the larger base is

$$
\pi R^2 = \pi (3^2) = 9\pi,
$$

- The area of the smaller base is

$$
\pi r^2 = \pi (2^2) = 4\pi.
$$

Now, adding these contributions, the total surface area is

$$
30\pi + 9\pi + 4\pi = 43\pi.
$$

To obtain a numerical value, we use $$\pi \approx 3.1416$$:

$$
43\pi \approx 43 \times 3.1416 \approx 135.09.
$$

Thus, the surface area of the piece of equipment is

$$
\boxed{43\pi\ \text{square feet} \approx 135.09\ \text{square feet}}.
$$
The surface area of the truncated cone is approximately 135.09 square feet.

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