The base of a straight pyramid is a square. The slant height of the pyramid is $27 \mathrm{~cm}$; the height is $3 \mathrm{~cm}$ greater than the side of the base. What is the surface area and volume of the pyramid?
If the base edge of the pyramid is $a$, the side edge is $l$, and its height is $m$, then its surface area is

$$
F=a^{2}+4 \cdot \frac{a}{2} \sqrt{l^{2}-\frac{a^{2}}{4}}=a^{2}+a \sqrt{4 l^{2}-a^{2}}
$$

and its volume is

$$
K=\frac{a^{2}}{3}(a+3)
$$

But

$$
l^{2}=\left(\frac{a}{\sqrt{2}}\right)^{2}+m^{2}=\frac{a^{2}}{2}+(a+3)^{2}
$$

or

$$
\begin{gathered}
\frac{a^{2}}{2}+(a+3)^{2}-27^{2}=0 \\
a^{2}+4 a-480=0
\end{gathered}
$$

from which the positive value of $a$ is $20 \mathrm{~cm}$, so $m=23 \mathrm{~cm}, \mathrm{~s}$ thus

$$
F=1403.19 \mathrm{~cm}^{2}
$$

and

$$
K=13066.7 \mathrm{~cm}^{3}
$$

(Erno Elischer, Budapest.)