Certainly! Let's solve for the volume of an oblique pyramid that has a square base with an edge length of \(5 \text{ cm}\) and a height of \(7 \text{ cm}\).
### Step-by-Step Solution:
1. Calculate the area of the square base:
The area \(A\) of a square is given by the formula:
[tex]\[
A = \text{side}^2
\][/tex]
Here, the side length of the square base is \(5 \text{ cm}\).
[tex]\[
A = 5 \text{ cm} \times 5 \text{ cm} = 25 \text{ cm}^2
\][/tex]
2. Use the formula for the volume of a pyramid:
The volume \(V\) of a pyramid is given by the formula:
[tex]\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\][/tex]
We already know the base area \(A = 25 \text{ cm}^2\) and the height \(h = 7 \text{ cm}\).
[tex]\[
V = \frac{1}{3} \times 25 \text{ cm}^2 \times 7 \text{ cm}
\][/tex]
[tex]\[
V = \frac{1}{3} \times 175 \text{ cm}^3
\][/tex]
[tex]\[
V = 58.\overline{3} \text{ cm}^3
\][/tex]
This can also be written as:
[tex]\[
V = 58 \frac{1}{3} \text{ cm}^3
\][/tex]
3. Identify the correct answer from the given choices:
- \(11 \frac{2}{3} \text{ cm}^3\)
- \(43 \frac{3}{4} \text{ cm}^3\)
- \(58 \frac{1}{3} \text{ cm}^3\)
- \(87 \frac{1}{2} \text{ cm}^3\)
Based on our calculation, the volume of the pyramid is \(58 \frac{1}{3} \text{ cm}^3\), which matches one of the given options.
Therefore, the volume of the pyramid is:
[tex]\[
\boxed{58 \frac{1}{3} \text{ cm}^3}
\][/tex]
