To solve for the volume of the oblique pyramid with a square base and given parameters, we will follow these steps meticulously:
1. Calculate the area of the square base:
- The edge length of the square base is given as \(2 \text{ cm}\).
- The area \(A_{\text{base}}\) of a square with side length \(a\) is given by \(a^2\).
- Therefore, \(A_{\text{base}} = 2 \text{ cm} \times 2 \text{ cm} = 4 \text{ cm}^2\).
2. Determine the height of the pyramid:
- The pyramid's height can be calculated using trigonometry from the given angle \( \angle BAC = 45^\circ \).
- Since \(\angle BAC = 45^\circ\) and the square base has a length of \(2 \text{ cm}\), we can use the tangent function.
- For height \(h\), it can be computed using the formula \( h = \frac{\text{opposite side}}{\tan(\text{angle})} \).
3. Plug in the values:
- The opposite side in this case is half the edge length of the square base, i.e., \(\frac{2}{2} = 1 \text{ cm}\).
- The angle is \(45^\circ\).
- Thus, \( h = \frac{1 \text{ cm}}{\tan(45^\circ)} \).
- Knowing that \(\tan(45^\circ) = 1\):
- \( h = \frac{1 \text{ cm}}{1} = 1 \text{ cm} \).
4. Calculate the volume of the pyramid:
- The volume \(V\) of a pyramid is given by \( \frac{1}{3} A_{\text{base}} \times h \).
- Substitute the area of the base and the height:
[tex]\[ V = \frac{1}{3} \times 4 \text{ cm}^2 \times 1 \text{ cm} \][/tex]
[tex]\[ V = \frac{1}{3} \times 4 \text{ cm}^3 \][/tex]
[tex]\[ V = \frac{4}{3} \text{ cm}^3 \][/tex]
5. Simplify the volume:
- \(\frac{4}{3}\) in decimal form is approximately \(1.333\).
6. Match the volume to the given choices:
- The volume is approximately \(1.333 \text{ cm}^3\).
- None of the given options exactly match this volume.
Given the choices:
- \(2.4 \text{ cm}^3\)
- \(3.6 \text{ cm}^3\)
- \(4.8 \text{ cm}^3\)
- \(7.2 \text{ cm}^3\)
None of these options match our calculated volume of roughly [tex]\(1.333 \text{ cm}^3\)[/tex]. This discrepancy means there may be a mistake in the understanding of the problem or the provided choices may be incorrect.
