A trapezoid-shaped plot of land has parallel sides measuring 2100 meters and 1500 meters, and the non-parallel sides (legs) are 613 meters and 37 meters long. What is the area of the plot in square meters?
I. Solution. For the trapezoid $ABCD$ ($AB=2100, CD=1500, DA=37, BC=613$), draw perpendiculars from vertices $C$ and $D$ to the base $AB$, intersecting the base at points $E$ and $F$ respectively. We know that the area of the trapezoid is:

$$
T=\frac{AB+CD}{2} \cdot m
$$

The heights $EC = FD = m$ can be calculated using the Pythagorean theorem in the right triangles $AFD$ and $BCE$.

![](https://cdn.mathpix.com/cropped/2024_05_02_5aa3ee708c1b5c18e6edg-1.jpg?height=249&width=596&top_left_y=488&top_left_x=751)

Let $AF = x$, then $EB = (2100 - 1500) - x = 600 - x$, so
(1) $\quad 37^{2} = m^{2} + x^{2}$;
(2) $\quad 613^{2} = (600 - x)^{2} + m^{2}$.

From this,

$$
37^{2} - x^{2} = 613^{2} - 600^{2} + 1200x - x^{2}
$$

Performing the indicated operations and rearranging, we get $x = -12$ and $m = 35$. Since $x < 0$, point $A$ is an internal point of $EF$ (denoted as $A'$ in the first diagram).

The area of the trapezoid is:

$$
T = \frac{2100 + 1500}{2} \cdot 35 = 63000 \, \text{m}^2
$$

which is approximately $\frac{63000}{3.596} \approx 17519$ square fathoms (1 Viennese square fathom $\approx 3.596 \, \text{m}^2$).

II. Solution. Using the notation in the second diagram, first calculate the area of triangle $A'BC$ using Heron's formula $\left(s = \frac{K}{2} = 625 \, \text{m}\right)$:

$$
\begin{aligned}
T_{A'BC} & = \sqrt{s(s-a)(s-b)(s-c)} = \\
& = \sqrt{625 \cdot 12 \cdot 588 \cdot 25} = 10500 \, \text{m}^2
\end{aligned}
$$

From the area formula of the triangle:

$$
T_{A'BC} = \frac{600 \cdot m}{2} = 10500 \Rightarrow m = 35 \, \text{m}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_5aa3ee708c1b5c18e6edg-1.jpg?height=325&width=504&top_left_y=1865&top_left_x=797)

With $m$ known, we can calculate the area of the trapezoid as per the first solution: $T_{ABCD} \approx 17519$ square fathoms.