4A. The base of a regular quadrilateral pyramid is a rhombus with side $a=2 \mathrm{~cm}$, composed of two equilateral triangles. The shorter edge of the pyramid has a length $b=2 \mathrm{~cm}$. Determine the area and volume of the pyramid.
Solution. Since the base of the pyramid is a rhombus composed of two equilateral triangles with side $a=2 \mathrm{~cm}$, we have

$$
B=2 \frac{a^{2} \sqrt{3}}{4}=\frac{a^{2} \sqrt{3}}{2}
$$

To determine the lateral surface area of the pyramid, we need to determine the height $h=\overline{S H_{1}}$. Triangle ACS is equilateral with side $a$, so its height is $H=\frac{a \sqrt{3}}{2}$. Since $\triangle A E O$ is half of an equilateral triangle, its height $h_{1}$ is half of the height of $\triangle A E C$, i.e., $h_{1}=\frac{a \sqrt{3}}{4}$. But $\Delta H_{1} O S$ is a right triangle, so

$$
h=\sqrt{H^{2}+h_{1}^{2}}=\sqrt{\frac{3}{4} a^{2}+\frac{3}{16} a^{2}}=\frac{a \sqrt{15}}{4}
$$

Thus, the surface area of the pyramid is

$$
P=B+4 \cdot \frac{a h}{2}=\frac{a^{2} \sqrt{3}}{2}+2 \frac{a^{2} \sqrt{15}}{4}=2 \sqrt{3}(1+\sqrt{5})
$$

and the volume of the pyramid is

$$
V=\frac{1}{3} B H=\frac{1}{3} \frac{a^{2} \sqrt{3}}{2} \frac{a \sqrt{3}}{2}=\frac{a^{3}}{4}=2
$$

NOTE. In a store, there were 6 crates of apples with weights of $15 \mathrm{~kg}, 18 \mathrm{~kg}$, $19 \mathrm{~kg}, 20 \mathrm{~kg}$, and $31 \mathrm{~kg}$. Two customers bought 5 crates, such that one bought twice as many apples as the other. Which crate was left unsold?