Question

A toy came in a container that is shaped like a triangular prism. How much wrapping paper is needed to wrap the toy? Round your answer to the nearest whole number. (1 point) S.A.= cm2
L =27.9 cm
W = 30.5 cm
H =5.1 cm
slant H 28.4 cm

Answer 1

To find the surface area of a triangular prism, we need to calculate the area of the two triangular bases and the area of the three rectangular sides (the lateral surface area).

1. **Area of the triangular base**:

The area \( A \) of a triangle is given by the formula:

\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]

In this case, we can use the width \( W \) as the base and the height \( H \) (the height of the triangle from base to the apex) for the triangular base:

\[
A = \frac{1}{2} \times 30.5 \, \text{cm} \times 5.1 \, \text{cm}
\]

Calculating this:

\[
A = \frac{1}{2} \times 30.5 \times 5.1 = 77.775 \, \text{cm}^2
\]

Since there are two triangular bases:

\[
\text{Total area of triangular bases} = 2 \times 77.775 = 155.55 \, \text{cm}^2
\]

2. **Area of the rectangular sides**:

Next, we compute the areas of the three rectangular sides. The three rectangles have the following dimensions:

- Rectangle 1 (base × length):
\[
\text{Area}_1 = \text{width} \times \text{length} = 30.5 \, \text{cm} \times 27.9 \, \text{cm} = 852.45 \, \text{cm}^2
\]

- Rectangle 2 (slant height × length):
\[
\text{Area}_2 = \text{slant height} \times \text{length} = 28.4 \, \text{cm} \times 27.9 \, \text{cm} = 793.56 \, \text{cm}^2
\]

- Rectangle 3 (height × length):
\[
\text{Area}_3 = \text{height} \times \text{length} = 5.1 \, \text{cm} \times 27.9 \, \text{cm} = 142.29 \, \text{cm}^2
\]

Now we add the areas of the three rectangular sides together:

\[
\text{Total area of rectangular sides} = 852.45 + 793.56 + 142.29 = 1788.30 \, \text{cm}^2
\]

3. **Total surface area**:

To find the total surface area \( S.A. \), we add the area of the triangular bases and the area of the rectangular sides:

\[
S.A. = \text{Total area of triangular bases} + \text{Total area of rectangular sides}
\]
\[
S.A. = 155.55 \, \text{cm}^2 + 1788.30 \, \text{cm}^2 = 1943.85 \, \text{cm}^2
\]

Finally, rounding this to the nearest whole number, the total surface area is:

\[
\boxed{1944} \, \text{cm}^2
\]

This is the amount of wrapping paper needed to wrap the toy.