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