A symmetric trapezoid has non-parallel sides of length $5 \mathrm{~cm}$, and the difference in length between the parallel sides is $6 \mathrm{~cm}$. If the base of the trapezoid is reduced by $10 \mathrm{~cm}$, and the height is increased by $1 \mathrm{~cm}$, the area of the trapezoid remains unchanged. What are the lengths of the sides of the trapezoid? What is its height and area?
Let the parallel sides of the trapezoid be denoted by $x$ and $y$, such that $x>y$. Let the height be denoted by $z$. The problem requirements are:

$$
\begin{gathered}
x-y=6 \ldots \\
\frac{(x-10)+y}{2}(z+1)=\frac{x+y}{2} z \ldots
\end{gathered}
$$

From (2), it follows that

$$
x+y=10(z+1) \ldots
$$

Any of the non-parallel sides is the hypotenuse of a right triangle, whose legs are the height ($z$) and half the difference of the parallel sides. By the Pythagorean theorem:

$$
z^{2}+\left(\frac{x-y}{2}\right)^{2}=5^{2} \cdots
$$

Considering (1),

$$
z^{2}=5^{2}-3^{2}=16, \quad z= \pm 4
$$

Given the geometric meaning of $z$, it can only be positive, so $z=4$.

Now from (2a): $x+y=50$; and from (1) $x-y=6$, thus

$$
x=28, \quad y=22
$$

The area of the trapezoid is:

$$
\frac{28+22}{2} 4=100 \mathrm{~cm}^{2}
$$

János Almási (Érseki g. V. o. Bp. II.)