Problem 9. In trapezoid $A B C D$, segments $B C$ and $A D$ are the bases. $B C=9.5 ; A D=20 ; A B=5$; $C D=8.5$. Find its area.
Solution. To solve the problem, it is necessary to find the length of the height of this trapezoid. Place the trapezoid $ABCD$ relative to a rectangular coordinate system as shown in Figure 8. Then $A(0; 0)$, $D(20; 0)$, $B(x; y)$, $C(x+9.5; y)$. According to the problem's conditions,

$$
\begin{gathered}
AB^2 = (x-0)^2 + (y-0)^2 = 25 \\
CD^2 = (x+9.5-20)^2 + (y-0)^2 = 72.25
\end{gathered}
$$

After simplification, the system of equations (1) and (2) becomes:

$$
\left\{\begin{array}{l}
x^2 + y^2 = 25 \\
(x-10.5)^2 + y^2 = 72.25
\end{array}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-060.jpg?height=491&width=531&top_left_y=320&top_left_x=340)

Fig. 7
Fig. 8

![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-060.jpg?height=266&width=443&top_left_y=238&top_left_x=1252)

Fig. 9

After subtracting equation (3) from equation (4) term by term, we get:

$$
(x-10.5)^2 - x^2 = 47.25
$$

From this, $x=3$. Since the trapezoid $ABCD$ is located in the upper half-plane (Fig. 8), then $y=4$, i.e., the height of the trapezoid $ABCD$ is 4.

Thus, $S_{ABCD} = 0.5(9.5 + 20) \cdot 4 = 59$.