Two regular quadrilateral pyramids have a slant height of one unit on their side faces. The lengths of their side edges are 1.25 units and 1.33 units, respectively. Which one has the larger volume?
The base of a regular quadrilateral pyramid is a square, and its side faces are isosceles triangles. The slant height of the pyramid is the height of the isosceles triangle corresponding to its base.

The volume of the pyramid is given by \( V = \frac{a_{t} \cdot m}{3} \), where \( a_{t} \) is the area of the square base, and \( m \) is the height of the pyramid.

Let \( x \) be half of the base edge (side of the square). From the right triangle OFB:

\[
x^{2} = B O^{2} - O F^{2}
\]

From the right triangle \( O O_{1} F \):

\[
m = \sqrt{O F^{2} - O^{\prime} F^{2}}, \quad \text{where} \quad O F = 1, \quad O^{\prime} F^{2} = x^{2}
\]

Using these, the volume of the pyramid is:

\[
V = \frac{(2 x)^{2} \sqrt{1^{2} - x^{2}}}{3} = \frac{4 x^{2} \sqrt{1 - x^{2}}}{3}
\]

Let's write the volume for both given values of \( B O \). If \( B O = 1.25 \), then \( x^{2} = (1.25^{2} - 1^{2}) \),

\[
V_{1} = \frac{4 (1.25^{2} - 1^{2}) \sqrt{1 - (1.25^{2} - 1)}}{3} \approx 0.49615
\]

For \( B O = 1.33 \), \( x^{2} = (1.33^{2} - 1^{2}) \),

\[
V_{2} = \frac{4 (1.33^{2} - 1^{2}) \sqrt{1 - (1.33^{2} - 1)}}{3} \approx 0.4928
\]

We can see that in the second case, although the side edge is longer, the volume of the pyramid is actually smaller.

![](https://cdn.mathpix.com/cropped/2024_05_02_ec32db20c119d6caa77bg-1.jpg?height=426&width=436&top_left_y=1256&top_left_x=823)