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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.

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 three rectangular sides.

Step 1: Area of the Triangular Bases

The triangular base is a right triangle, so we can use the formula for the area of a triangle:

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

Given:

  • Base = 27.9 cm
  • Height = 5.1 cm

Substituting the values:

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

Since there are two triangular bases:

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

Step 2: Area of the Rectangular Faces

Each rectangular face can be calculated using the formula for the area of a rectangle:

\[ \text{Area} = \text{length} \times \text{width} \]

The three rectangular faces are:

  1. Base × Length (27.9 cm × 30.5 cm)
  2. Height × Length (5.1 cm × 30.5 cm)
  3. Hypotenuse × Length (28.4 cm × 30.5 cm)

Calculating each:

  1. First rectangular face (Base × Length): \[ \text{Area} = 27.9 , \text{cm} \times 30.5 , \text{cm} = 852.45 , \text{cm}^2 \]

  2. Second rectangular face (Height × Length): \[ \text{Area} = 5.1 , \text{cm} \times 30.5 , \text{cm} = 155.55 , \text{cm}^2 \]

  3. Third rectangular face (Hypotenuse × Length): \[ \text{Area} = 28.4 , \text{cm} \times 30.5 , \text{cm} = 867.2 , \text{cm}^2 \]

Step 3: Total Surface Area

Now, we sum the areas of the triangular bases and the rectangular faces:

\[ \text{Total Surface Area} = \text{Area of triangular bases} + \text{Area of rectangular faces} \]

Calculating the total area of the rectangular faces:

\[ \text{Total Area of Rectangular Faces} = 852.45 , \text{cm}^2 + 155.55 , \text{cm}^2 + 867.2 , \text{cm}^2 = 1875.2 , \text{cm}^2 \]

Adding all parts together:

\[ \text{Total Surface Area} = 142.29 , \text{cm}^2 + 1875.2 , \text{cm}^2 = 2017.49 , \text{cm}^2 \]

Final Answer

Rounding to the nearest whole number:

\[ \text{Total Surface Area} \approx 2017 , \text{cm}^2 \]

Thus, the amount of wrapping paper needed is: \[ \text{S.A.} = 2017 , \text{cm}^2 \]

So the final answer is: S.A. = 2017 cm²