Question
Find the surface area of a rectangular pyramid with these measurements: l
= 5.5 yd., w
= 4 yd., and h
= 6 yd. (1 point)
Responses
255.06 yd.2
255 point 0 6 yards squared
83.19 yd.2
83 point 1 9 yards squared
242.27 yd.2
242 point 2 7 yards squared
56.79 yd.2
= 5.5 yd., w
= 4 yd., and h
= 6 yd. (1 point)
Responses
255.06 yd.2
255 point 0 6 yards squared
83.19 yd.2
83 point 1 9 yards squared
242.27 yd.2
242 point 2 7 yards squared
56.79 yd.2
Answers
GPT-4o mini
To find the surface area of a rectangular pyramid, you need to calculate the area of the base and the areas of the four triangular faces.
1. **Calculate the area of the base**:
The base is a rectangle, so its area \(A_{base}\) is given by:
\[
A_{base} = l \times w = 5.5 \, \text{yd} \times 4 \, \text{yd} = 22 \, \text{yd}^2
\]
2. **Calculate the area of the triangular faces**:
There are four triangular faces. We need to calculate the area of each triangular face using the formula for the area of a triangle:
\[
A_{triangle} = \frac{1}{2} \times \text{base} \times \text{height}
\]
For the two triangular faces with base \(l\):
\[
A_{triangle1} = \frac{1}{2} \times 5.5 \, \text{yd} \times h' \quad \text{(where } h' \text{ is the slant height)}
\]
For the two triangular faces with base \(w\):
\[
A_{triangle2} = \frac{1}{2} \times 4 \, \text{yd} \times h'
\]
**Find the slant height \(h'\)**.
To find the slant height \(h'\), we should use the Pythagorean theorem. The height \(h\) of the pyramid and half of the base's width form a right triangle:
\[
h' = \sqrt{h^2 + \left(\frac{l}{2}\right)^2} = \sqrt{6^2 + \left(\frac{5.5}{2}\right)^2} = \sqrt{36 + 6.0625} = \sqrt{42.0625} \approx 6.48 \, \text{yd}
\]
3. **Calculate the areas of the four triangular faces**:
For the triangles formed on the length side (2 triangles):
\[
A_{triangle1} = \frac{1}{2} \times 5.5 \times 6.48 \approx 17.83 \, \text{yd}^2 \quad \Rightarrow \text{Total from both triangles} \approx 35.66 \, \text{yd}^2
\]
For the triangles formed on the width side (2 triangles):
\[
A_{triangle2} = \frac{1}{2} \times 4 \times 6.48 \approx 12.96 \, \text{yd}^2 \quad \Rightarrow \text{Total from both triangles} \approx 25.92 \, \text{yd}^2
\]
4. **Total surface area**:
\[
\text{Surface Area} = A_{base} + A_{triangle1} + A_{triangle2}
\]
\[
\text{Surface Area} = 22 + 35.66 + 25.92 = 83.58 \, \text{yd}^2
\]
Thus, the closest answer to the calculated surface area is:
\[
83.19 \, \text{yd}^2
\]
So the correct response is **83.19 yd²**.
1. **Calculate the area of the base**:
The base is a rectangle, so its area \(A_{base}\) is given by:
\[
A_{base} = l \times w = 5.5 \, \text{yd} \times 4 \, \text{yd} = 22 \, \text{yd}^2
\]
2. **Calculate the area of the triangular faces**:
There are four triangular faces. We need to calculate the area of each triangular face using the formula for the area of a triangle:
\[
A_{triangle} = \frac{1}{2} \times \text{base} \times \text{height}
\]
For the two triangular faces with base \(l\):
\[
A_{triangle1} = \frac{1}{2} \times 5.5 \, \text{yd} \times h' \quad \text{(where } h' \text{ is the slant height)}
\]
For the two triangular faces with base \(w\):
\[
A_{triangle2} = \frac{1}{2} \times 4 \, \text{yd} \times h'
\]
**Find the slant height \(h'\)**.
To find the slant height \(h'\), we should use the Pythagorean theorem. The height \(h\) of the pyramid and half of the base's width form a right triangle:
\[
h' = \sqrt{h^2 + \left(\frac{l}{2}\right)^2} = \sqrt{6^2 + \left(\frac{5.5}{2}\right)^2} = \sqrt{36 + 6.0625} = \sqrt{42.0625} \approx 6.48 \, \text{yd}
\]
3. **Calculate the areas of the four triangular faces**:
For the triangles formed on the length side (2 triangles):
\[
A_{triangle1} = \frac{1}{2} \times 5.5 \times 6.48 \approx 17.83 \, \text{yd}^2 \quad \Rightarrow \text{Total from both triangles} \approx 35.66 \, \text{yd}^2
\]
For the triangles formed on the width side (2 triangles):
\[
A_{triangle2} = \frac{1}{2} \times 4 \times 6.48 \approx 12.96 \, \text{yd}^2 \quad \Rightarrow \text{Total from both triangles} \approx 25.92 \, \text{yd}^2
\]
4. **Total surface area**:
\[
\text{Surface Area} = A_{base} + A_{triangle1} + A_{triangle2}
\]
\[
\text{Surface Area} = 22 + 35.66 + 25.92 = 83.58 \, \text{yd}^2
\]
Thus, the closest answer to the calculated surface area is:
\[
83.19 \, \text{yd}^2
\]
So the correct response is **83.19 yd²**.