Find the surface area of a rectangular pyramid with these measurements: l= 8 cm, w= 4 cm, and h= 2 cm. Express your answer as a decimal rounded to the nearest hundredth.(1 point)

Answer 1

To find the surface area of a rectangular pyramid, we need to calculate the area of the base and the area of the triangular faces.

1. Calculate the area of the base: The base of the pyramid is a rectangle with length \( l = 8 \) cm and width \( w = 4 \) cm. \[ \text{Area of the base} = l \times w = 8 \times 4 = 32 \text{ cm}^2 \]

2. Calculate the area of the triangular faces: The pyramid has four triangular faces. We will first calculate the area of one triangular face and then multiply by 4.

   Each triangular face has a base along one edge of the rectangular base and an altitude which can be found by using the Pythagorean theorem.

   For the triangles with base along the length (8 cm): The height of these triangles can be found using: \[ h' = \sqrt{h^2 + \left(\frac{l}{2}\right)^2} = \sqrt{2^2 + \left(\frac{8}{2}\right)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \text{ cm} \] The area of one triangular face (base = 8 cm): \[ \text{Area}_{\text{length face}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 2\sqrt{5} = 8\sqrt{5} \text{ cm}^2 \]

   For the triangles with base along the width (4 cm): Using a similar method, the height of these triangles is: \[ h' = \sqrt{h^2 + \left(\frac{w}{2}\right)^2} = \sqrt{2^2 + \left(\frac{4}{2}\right)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \text{ cm} \] The area of one triangular face (base = 4 cm): \[ \text{Area}_{\text{width face}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 2\sqrt{2} = 4\sqrt{2} \text{ cm}^2 \]

3. Total area of the triangular faces: There are 2 length triangles and 2 width triangles: \[ \text{Total area}_{\text{triangular faces}} = 2 \times 8\sqrt{5} + 2 \times 4\sqrt{2} = 16\sqrt{5} + 8\sqrt{2} \text{ cm}^2 \]

4. Total surface area: Adding the base area and the area of the triangular faces: \[ \text{Surface Area} = \text{Area of base} + \text{Total area}_{\text{triangular faces}} = 32 + 16\sqrt{5} + 8\sqrt{2} \text{ cm}^2 \] We can use the approximations \( \sqrt{5} \approx 2.236 \) and \( \sqrt{2} \approx 1.414 \): \[ 16\sqrt{5} \approx 16 \times 2.236 \approx 35.776 \] \[ 8\sqrt{2} \approx 8 \times 1.414 \approx 11.312 \] Thus, \[ \text{Surface Area} \approx 32 + 35.776 + 11.312 \approx 79.088 \text{ cm}^2 \]

Finally, rounding to the nearest hundredth, the surface area of the rectangular pyramid is approximately: \[ \boxed{79.09 \text{ cm}^2} \]