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. The dimensions of the rectangular prism are a length of 30.5cm, a base of 5.1cm, a slope of 28.4cm and a rise of 27.9cm. Don't forget to round to the nearest whole number.

Answer 1

To find the amount of wrapping paper needed to wrap a toy in the shape of a triangular prism, we first need to calculate the surface area of the prism. The surface area of a triangular prism can be calculated using the formula:

\[
\text{Surface Area} = \text{Base Area} + \text{Lateral Area}
\]

### Step 1: Calculate the Base Area of the Triangle
The base of the triangle can be represented as follows:
- Base \( b = 5.1 \, \text{cm} \)
- Height \( h = 27.9 \, \text{cm} \) (this represents the rise of the triangle).

The area \( A \) of the triangular base is given by:

\[
A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5.1 \times 27.9
\]

Calculating the area:

\[
A = \frac{1}{2} \times 5.1 \times 27.9 = \frac{1}{2} \times 142.29 \approx 71.145 \, \text{cm}^2
\]

### Step 2: Calculate the Lateral Area
The lateral area consists of three rectangles:

1. Two rectangles corresponding to the triangular base edges (these have height equal to the length of the prism):
- Side 1 (slope edge): \( 28.4 \, \text{cm} \)
- Side 2 (base edge): \( 5.1 \, \text{cm} \)
- Side 3 (rise edge): \( 27.9 \, \text{cm} \)

We calculate the area of these rectangles as follows:

- Area of rectangle along the slope (side 1):
\[
\text{Area}_1 = \text{length} \times \text{slope} = 30.5 \times 28.4
\]

- Area of rectangle along the base (side 2):
\[
\text{Area}_2 = \text{length} \times \text{base} = 30.5 \times 5.1
\]

- Area of rectangle along the rise (side 3):
\[
\text{Area}_3 = \text{length} \times \text{rise} = 30.5 \times 27.9
\]

Calculating these areas:

1. \( \text{Area}_1 = 30.5 \times 28.4 \approx 865.42 \, \text{cm}^2 \)
2. \( \text{Area}_2 = 30.5 \times 5.1 \approx 155.55 \, \text{cm}^2 \)
3. \( \text{Area}_3 = 30.5 \times 27.9 \approx 851.95 \, \text{cm}^2 \)

Summing the areas:

\[
\text{Lateral Area} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3
\]
\[
\text{Lateral Area} \approx 865.42 + 155.55 + 851.95 \approx 1872.92 \, \text{cm}^2
\]

### Step 3: Calculate the Total Surface Area
Now, we add the base area and the lateral area:

\[
\text{Total Surface Area} = \text{Base Area} + \text{Lateral Area}
\]
\[
\text{Total Surface Area} \approx 71.145 + 1872.92 \approx 1944.065 \, \text{cm}^2
\]

### Step 4: Round to the Nearest Whole Number
Finally, rounding the total surface area to the nearest whole number:

\[
\text{Total Surface Area} \approx 1944 \, \text{cm}^2
\]

Therefore, the amount of wrapping paper needed to wrap the toy is approximately **1944 cm²**.