To calculate the surface area of a triangular prism, we need to determine the areas of its two triangular bases and three rectangular lateral faces.
1. Let's calculate the area of the triangular bases:
Given:
- Width (base of the triangle) \( W = 27.9 , \text{cm} \)
- Height of the triangle (h from the base to the top vertex) = \( H = 5.1 , \text{cm} \)
The area \( A \) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Substituting in the values:
\[ A = \frac{1}{2} \times 27.9 , \text{cm} \times 5.1 , \text{cm} \]
\[ A = \frac{1}{2} \times 142.29 , \text{cm}^2 \]
\[ A \approx 71.145 , \text{cm}^2 \]
Since there are two triangular bases:
\[ \text{Total area of triangular bases} = 2 \times 71.145 \approx 142.29 , \text{cm}^2 \]
2. Now calculate the area of the rectangular lateral faces:
There are three rectangular faces:
- Two rectangles with dimensions \( L \times W \)
- One rectangle with dimensions \( L \times \text{Slant Height} \)
Using the dimensions provided:
- Length \( L = 30.5 , \text{cm} \)
- Slant height \( \text{Slant H} = 28.4 , \text{cm} \)
Calculating the areas of the three rectangles:
-
Area of the first rectangle: \[ A_1 = L \times W = 30.5 , \text{cm} \times 27.9 , \text{cm} = 851.95 , \text{cm}^2 \]
-
Area of the second rectangle (same dimensions as the first): \[ A_2 = L \times W = 851.95 , \text{cm}^2 \]
-
Area of the third rectangle (with slant height): \[ A_3 = L \times \text{Slant H} = 30.5 , \text{cm} \times 28.4 , \text{cm} = 865.42 , \text{cm}^2 \]
3. Total surface area:
Now we can sum all areas together to find the total surface area \( S.A. \):
\[ S.A. = \text{Area of triangular bases} + A_1 + A_2 + A_3 \] \[ S.A. \approx 142.29 , \text{cm}^2 + 851.95 , \text{cm}^2 + 851.95 , \text{cm}^2 + 865.42 , \text{cm}^2 \] \[ S.A. \approx 1711.61 , \text{cm}^2 \]
4. Round to the nearest whole number:
\[ S.A. \approx 1712 , \text{cm}^2 \]
Therefore, the amount of wrapping paper needed to wrap the toy is approximately 1712 cm².