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
W =27.9
L = 30.5
H =5.1
and there's a side slant that's 28.4

Answer 1

To calculate the surface area of a triangular prism, we need to determine the areas of the two triangular bases and the three rectangular sides. Given the dimensions:

- **W** (width of the base triangle) = 27.9 cm
- **L** (length of the prism) = 30.5 cm
- **H** (height of the base triangle) = 5.1 cm
- **s** (slant height) = 28.4 cm

### Step 1: Area of the Triangular Base

The area \( A \) of a triangle can be calculated using the formula:

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

For our triangular base:

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

### Step 2: Total Area of Two Triangular Bases

Since there are two identical triangular bases:

\[
\text{Area of two bases} = 2A = 2 \times 71.145 = 142.29 \, \text{cm}^2
\]

### Step 3: Area of the Rectangular Sides

Now we calculate the areas of the three rectangular sides:

1. **Base Rectangle** (length × width):
\[
\text{Area} = L \times W = 30.5 \times 27.9 = 852.45 \, \text{cm}^2
\]

2. **Side Rectangle 1** (length × height):
\[
\text{Area} = L \times H = 30.5 \times 5.1 = 155.55 \, \text{cm}^2
\]

3. **Side Rectangle 2** (length × slant height):
\[
\text{Area} = L \times s = 30.5 \times 28.4 = 865.42 \, \text{cm}^2
\]

### Step 4: Total Area of the Rectangular Sides

Combining the areas of the three rectangular sides:

\[
\text{Total Area of Rectangles} = 852.45 + 155.55 + 865.42 = 1873.42 \, \text{cm}^2
\]

### Step 5: Total Surface Area

Now we can find the total surface area \( S.A. \):

\[
S.A. = \text{Area of two bases} + \text{Total Area of Rectangles}
\]
\[
S.A. = 142.29 + 1873.42 = 2015.71 \, \text{cm}^2
\]

### Final Step: Rounding

Rounding to the nearest whole number:

\[
\text{S.A.} \approx 2016 \, \text{cm}^2
\]

Therefore, the amount of wrapping paper needed to wrap the toy is approximately **2016 cm²**.