Mr. Teacher asked Adam and Eve to calculate the perimeter of a trapezoid, whose longer base measured $30 \mathrm{~cm}$, height $24 \mathrm{~cm}$, and the non-parallel sides $25 \mathrm{~cm}$ and $30 \mathrm{~cm}$. Adam got a different perimeter than Eve, yet Mr. Teacher praised both for their correct solutions.

Determine the results of Adam and Eve.

(L. Hozová)
A trapez with given base $(AB)$, height $(AF)$, and legs $(AD$ and $BC)$ is not uniquely determined; the following possibilities may occur:
![](https://cdn.mathpix.com/cropped/2024_04_17_e6d56f2031023e9db813g-5.jpg?height=844&width=620&top_left_y=617&top_left_x=318)
![](https://cdn.mathpix.com/cropped/2024_04_17_e6d56f2031023e9db813g-5.jpg?height=840&width=694&top_left_y=618&top_left_x=1049)

We understand all these trapezoids as arising from the rectangle $ABEF$ by adding or, respectively, removing right triangles $AFD$ and $BEC$. Depending on the lengths of the given segments, the quadrilateral $ABCD$ can generally become non-convex. † This will become clear once we calculate the unknown lengths of the segments.

In the following calculations, we do not write units (everywhere in cm) and we substitute the values from the problem: $|AB|=30$, $|AF|=24$, $|AD|=25$, and $|BC|=30$. According to the Pythagorean theorem in triangles $AFD$ and $BEC$, we calculate the lengths of the remaining legs:

$$
\begin{aligned}
& |FD|=\sqrt{|AD|^2-|AF|^2}=\sqrt{25^2-24^2}=7 \\
& |EC|=\sqrt{|BC|^2-|BE|^2}=\sqrt{30^2-24^2}=18
\end{aligned}
$$

In the first, respectively, in the second case, we get

$$
\begin{aligned}
& |CD|=|AB|-|FD|-|EC|=30-7-18=5, \\
& |CD|=|AB|+|FD|-|EC|=30+7-18=19.
\end{aligned}
$$

In both cases, the resulting difference is positive and less than 30, so it is a trapezoid whose longer base is $AB$. In the remaining two cases, the longer base is $CD$, so we do not need to consider these cases. [^0]

In the first, respectively, in the second case, the perimeter of the trapezoid $|AB|+|BC|+|CD|+|DA|$ is

$$
30+30+5+25=90, \quad \text{resp.} \quad 30+30+19+25=104
$$

and these are the results of Adam and Eve.

Note. When calculating the lengths of the legs $FD$ and $EC$ by hand, it is advantageous to use the following transformations:

$$
\begin{aligned}
& \sqrt{25^2-24^2}=\sqrt{(25-24)(25+24)}=\sqrt{49}=7 \\
& \sqrt{30^2-24^2}=\sqrt{(30-24)(30+24)}=\sqrt{6 \cdot 54}=\sqrt{18 \cdot 18}=18
\end{aligned}
$$

Evaluation. 2 points for analyzing the possibilities; 2 points for auxiliary calculations; 2 points based on the quality of the commentary.

[^0]:    † For example, in the first case, this situation would occur if $|AB|<|FD|+|EC|$.