From a regular quadrilateral pyramid, we cut off four small tetrahedra with planes that bisect the edges originating from each vertex of the base. What is the surface area and volume of the remaining solid if each edge of the pyramid is one unit long?
The surface area of the quadrilateral pyramid consists of one unit-sided square and 4 identical unit-sided equilateral triangles; $F_{g}=1+\sqrt{3}$.

The volume of the pyramid is: $V_{g}=\frac{A_{t} \cdot m}{3}$. The base is a square, so $m=\sqrt{1-\left(\frac{\sqrt{2}}{2}\right)^{2}}=\frac{\sqrt{2}}{2}, V_{g}=\frac{\sqrt{2}}{6}$.

Since the pyramid is regular, the small tetrahedra that are cut off are congruent. Let's determine the volume of one small tetrahedron. The base of the tetrahedron is an isosceles right triangle with legs of $\frac{1}{2}$, and its area is: $\frac{1}{8}$. The height of the tetrahedron is half the height of the pyramid: $\frac{\sqrt{2}}{4}$.

The combined volume of the four small tetrahedra is: $\frac{4 \cdot \frac{1}{8} \cdot \frac{\sqrt{2}}{4}}{3}=\frac{\sqrt{2}}{24}$. The volume of the remaining solid is:

$$
\frac{\sqrt{2}}{6}-\frac{\sqrt{2}}{24}=\frac{\sqrt{2}}{8} \text { cubic units. }
$$

The surface area of one small tetrahedron consists of one isosceles right triangle with legs of $\frac{1}{2}$, two equilateral triangles with sides of $\frac{1}{2}$, and one triangle with sides $\frac{1}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2}$, which, by the converse of the Pythagorean theorem, is congruent to the first-mentioned isosceles right triangle. Thus,

$$
t_{1}=\frac{1}{8}, \quad t_{2}=\frac{\left(\frac{1}{2}\right)^{2} \sqrt{3}}{4}=\frac{\sqrt{3}}{16}
$$

We need to subtract from the surface area of the pyramid the total area of the three faces of the small tetrahedra at their apexes, and add the total area of the fourth faces.

The surface area of the remaining solid is:

$$
F=1+\sqrt{3}-\left(4 \cdot \frac{1}{8}+8 \cdot \frac{\sqrt{3}}{16}\right)+4 \cdot \frac{1}{8}=\frac{2+\sqrt{3}}{2} \approx 1.866 \text { square units. }
$$

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