Use the image to answer the question.

An illustration shows a 3 D triangular prism with the triangular face as a right triangle. The edges that are not visible are represented by dashed lines. The length and base of the triangular prism measures 30.5 centimeters and 27.9 centimeters. The perpendicular side of the triangular face measures 5.1 centimeter and the hypotenuse measures 28.4 centimeter.

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

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 lateral faces.

  1. Area of the triangular face: The triangular face is a right triangle. The area \( A \) of a right triangle can be calculated using the formula: \[ A = \frac{1}{2} \times \text{base} \times \text{height} \] Given:

    • Base \( = 27.9 , \text{cm} \)
    • Height \( = 5.1 , \text{cm} \)

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

    Since there are two triangular bases, we multiply by 2: \[ \text{Total area of the triangular bases} = 2 \times 71.145 = 142.29 , \text{cm}^2 \]

  2. Area of the rectangular faces: The three rectangular faces can be calculated using the lengths of the prism and the sides of the triangle.

    • The first rectangle has dimensions:

      • Length \( = 30.5 , \text{cm} \)
      • Width \( = 27.9 , \text{cm} \)

      \[ \text{Area} = 30.5 \times 27.9 = 852.45 , \text{cm}^2 \]

    • The second rectangle has dimensions:

      • Length \( = 30.5 , \text{cm} \)
      • Width \( = 5.1 , \text{cm} \)

      \[ \text{Area} = 30.5 \times 5.1 = 155.55 , \text{cm}^2 \]

    • The third rectangle has dimensions:

      • Length \( = 30.5 , \text{cm} \)
      • Width \( = 28.4 , \text{cm} \)

      \[ \text{Area} = 30.5 \times 28.4 = 864.2 , \text{cm}^2 \]

    Now, adding up the areas of the rectangular faces: \[ \text{Total area of rectangular faces} = 852.45 + 155.55 + 864.2 = 1872.2 , \text{cm}^2 \]

  3. Calculating the total surface area: Finally, we add the area of the triangular bases to the area of the rectangular faces: \[ \text{Total Surface Area} = 142.29 + 1872.2 = 2014.49 , \text{cm}^2 \]

Rounding to the nearest whole number, the total surface area required for wrapping paper is: \[ \text{S.A.} = 2014 , \text{cm}^2 \]

Therefore, the answer is: \[ \boxed{2014} , \text{cm}^2 \]