The base of a certain straight pyramid is a regular pentagon around which a circle with a radius of $6 \mathrm{dm}$ can be inscribed, and its lateral faces are equilateral triangles. What is the total surface area and volume of this pyramid?
First, let's calculate the area of the base.

We can achieve this by dividing it into 5 congruent isosceles triangles and taking 5 times the area of one such triangle.

The area of one such triangle is

$$
S=\frac{a r^{\prime}}{2}
$$

where $a$ is the side of the pentagon, and $r^{\prime}$ is the radius of the inscribed circle of the pentagon.

But

$$
r^{\prime 2}=r^{2}-\frac{a^{2}}{4}
$$

where $r$ is the radius of the circumscribed circle of the pentagon.

The value of $a$ is

$$
a=\frac{r}{2} \sqrt{10-2 \sqrt{5}}
$$

and

$$
\frac{a^{2}}{4}=\frac{r^{2}}{16}(10-2 \sqrt{5})
$$

thus

$$
\begin{aligned}
r^{\prime 2}= & \frac{r^{2}}{16} 16-\frac{r^{2}}{16}(10-2 \sqrt{5}) \\
& =\frac{r^{2}}{16}(6+2 \sqrt{5}) \\
& =\frac{r^{2}}{16}(1+\sqrt{5})^{2}
\end{aligned}
$$

and therefore

$$
r^{\prime}=\frac{r}{4}(1+\sqrt{5})
$$

From this,

$$
S=\frac{r^{2}}{16} \sqrt{10-2 \sqrt{5}}(1+\sqrt{5})
$$

The area of the side is:

$$
\begin{gathered}
S^{\prime}=\frac{a^{2}}{4} \sqrt{3} \\
=\frac{r^{2}}{16}(10-2 \sqrt{5}) \sqrt{3}
\end{gathered}
$$

The total surface area is therefore

$$
\begin{aligned}
5\left(S+S^{\prime}\right) & =5\left\{\frac{r^{2}}{16} \sqrt{10-2 \sqrt{5}}(1+\sqrt{5})+\frac{r^{2}}{16}(10-2 \sqrt{5}) \sqrt{3}\right\}= \\
& =\frac{5}{16} r^{2} \sqrt{10-2 \sqrt{5}}\{1+\sqrt{5}+\sqrt{30-6 \sqrt{5}}\} \\
= & \frac{45}{4} \sqrt{10-2 \sqrt{5}}\{1+\sqrt{5}+\sqrt{30-6 \sqrt{5}}\} \mathrm{dm}^{2}
\end{aligned}
$$

I get the volume by multiplying the area of the base by one-third of the height. But the height $h$ is a leg of a right triangle whose hypotenuse is $a$ and the other leg is $r$; that is, $h$ is nothing other than the side of a decagon inscribed in a circle of radius $r$.

Thus,

$$
h=\frac{r}{2}(\sqrt{5}-1)
$$

The sought volume is:

$$
\begin{gathered}
V=\frac{r}{6}(\sqrt{5}-1) 5 \frac{r^{2}}{16} \sqrt{10-2 \sqrt{5}}(\sqrt{5}+1) \\
=\frac{5}{96} r^{3} 4 \sqrt{10-2 \sqrt{5}} \\
=\frac{5}{24} r^{3} \sqrt{10-2 \sqrt{5}} \\
=\frac{5}{24} 216 \sqrt{10-2 \sqrt{5}} \mathrm{dm}^{3} \\
V=45 \sqrt{10-2 \sqrt{5}} \mathrm{dm}^{3}
\end{gathered}
$$

(Gergely Jorga, 8th grade student, Arad Main Real School).