4. The base of the oblique prism is a kite, which has a shorter diagonal of length e. The internal angles of the kite at the endpoints of the longer diagonal measure $90^{\circ}$ and $60^{\circ}$. The height of the prism is equal to the longer diagonal of the kite. Express the volume of the prism in terms of e. The result should be exact.
4. The volume of the prism is $V=\frac{e \cdot f}{2} \cdot f$. The longer diagonal is divided into segments $f_{1}$ and $f_{2}$. Both segments are expressed in terms of the length $e$ of the shorter diagonal. The right triangle $D B C$ is isosceles, so $f_{2}=\frac{e}{2}$. Triangle $B D A$ is equilateral, so $f_{1}=\frac{e \sqrt{3}}{2}$, which is the height of the equilateral triangle. Therefore, the length of the diagonal $f$ is $\frac{e \sqrt{3}}{2}+\frac{e}{2}=\frac{e}{2}(\sqrt{3}+1)$. The volume of the prism is

![](https://cdn.mathpix.com/cropped/2024_06_07_40efadbcfacda9b65264g-11.jpg?height=460&width=599&top_left_y=661&top_left_x=1248)
$\frac{e \cdot \frac{e}{2}(\sqrt{3}+1)}{2} \cdot \frac{e}{2}(\sqrt{3}+1)=\frac{e^{3}(2+\sqrt{3})}{4}$.

Volume $V=\frac{e \cdot f}{2} \cdot f$ expressed. 1 point

$f_{1}=\frac{e \sqrt{3}}{2}$ expressed. 1 point

$$
f_{2}=\frac{e}{2}
$$

$f=\frac{e \sqrt{3}}{2}+\frac{e}{2}=\frac{e}{2}(\sqrt{3}+1)$ expressed. 1 point

$V=\frac{e \cdot \frac{e}{2}(\sqrt{3}+1)}{2} \cdot \frac{e}{2}(\sqrt{3}+1)=\frac{e^{3}(3+2 \sqrt{3}+1)}{8}$ calculated. 1 point

Result $V=\frac{e^{3}(2+\sqrt{3})}{4}$.