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

1 answer

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:

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

  1. 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:

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

  1. 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.