Question

Find the surface area of a rectangular pyramid with these measurements: = 5.5 yd., = 4 yd.,
and = 6 yd. (1 point)
l𝑙 w𝑤
hℎ
83.19 yd.
2
83.19 yd.
2
255.06 yd.
2
255.06 yd.
2
56.79 yd.
2
56.79 yd.
2
242.27 yd.
2

Answer 1

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

The measurements given are:
- Length \( l = 5.5 \) yd
- Width \( w = 4 \) yd
- Height \( h = 6 \) yd

### Step 1: Area of the Base
The base of the pyramid is a rectangle, so the area of the base \( A_{\text{base}} \) is calculated as:
\[
A_{\text{base}} = l \times w = 5.5 \times 4 = 22 \text{ yd}^2
\]

### Step 2: Area of the Triangular Faces
There are 4 triangular faces. Two of them have a base equal to the length \( l \), and the other two have a base equal to the width \( w \).

#### Area of Triangular Faces on Length Side
The area of one triangular face with base \( l \) is given by the formula:
\[
A_{\text{triangular}} = \frac{1}{2} \times \text{base} \times \text{height}
\]
The height of the triangle can be found using the Pythagorean theorem. The slant height \( s \) of the pyramid can be calculated as:
\[
s = \sqrt{\left(\frac{l}{2}\right)^2 + h^2} = \sqrt{\left(\frac{5.5}{2}\right)^2 + 6^2} = \sqrt{(2.75)^2 + 6^2} = \sqrt{7.5625 + 36} = \sqrt{43.5625} \approx 6.6 \text{ yd}
\]

Now, the area of one triangular face (with base \( l \)):
\[
A_{\text{triangular for l}} = \frac{1}{2} \times l \times s = \frac{1}{2} \times 5.5 \times 6.6 \approx 18.15 \text{ yd}^2
\]
Since there are 2 such triangular faces:
\[
A_{\text{triangular for l total}} = 2 \times 18.15 \approx 36.30 \text{ yd}^2
\]

#### Area of Triangular Faces on Width Side
Using the same method for the faces with base \( w \):
\[
s = \sqrt{\left(\frac{w}{2}\right)^2 + h^2} = \sqrt{\left(\frac{4}{2}\right)^2 + 6^2} = \sqrt{(2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} \approx 6.32 \text{ yd}
\]

Now, the area of one triangular face (with base \( w \)):
\[
A_{\text{triangular for w}} = \frac{1}{2} \times w \times s = \frac{1}{2} \times 4 \times 6.32 \approx 12.64 \text{ yd}^2
\]
Since there are 2 such triangular faces:
\[
A_{\text{triangular for w total}} = 2 \times 12.64 \approx 25.28 \text{ yd}^2
\]

### Step 3: Total Surface Area
Now we can sum up the areas to find the total surface area \( A_{\text{total}} \):
\[
A_{\text{total}} = A_{\text{base}} + A_{\text{triangular for l total}} + A_{\text{triangular for w total}}
\]
\[
A_{\text{total}} = 22 + 36.30 + 25.28 \approx 83.58 \text{ yd}^2
\]

### Conclusion
The closest approximate answer for the surface area of the rectangular pyramid, judging by the options provided, would be **83.19 yd²**.