## Task B-3.5.

A trapez with mutually perpendicular diagonals has bases of length $a=12$ and $c=4$, and the extensions of the legs of the trapezoid intersect at an angle $\alpha$. If $\cos \alpha=\frac{4}{5}$, calculate the area of this trapezoid.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_aa38f3c63b8013e55583g-16.jpg?height=806&width=691&top_left_y=1733&top_left_x=614)

Draw a parallel to side $BC$ through vertex $D$ (or to side $AD$ through vertex $C$). Let the intersection of this parallel and the larger base be $E$. Then we have

$$
|DE| = b, \quad |AE| = a - c = 8, \quad \angle ADE = \alpha
$$

From triangle $AED$, using the cosine rule, we get the relationship between the lengths of the legs of the trapezoid $b$ and $d$:

$$
\begin{gathered}
(a - c)^2 = b^2 + d^2 - 2bd \cos \alpha \\
64 = b^2 + d^2 - 2bd \cdot \frac{4}{5}
\end{gathered}
$$

If the intersection of the diagonals, point $S$, divides the diagonals into segments of lengths $k, l, m, n$ as shown in the figure, then by the Pythagorean theorem, we have: $a^2 = k^2 + n^2$ and $c^2 = l^2 + m^2$. Adding these equations, we get

$$
a^2 + c^2 = k^2 + l^2 + m^2 + n^2
$$

Similarly, from $b^2 = l^2 + n^2$ and $d^2 = k^2 + m^2$, we get

$$
b^2 + d^2 = k^2 + l^2 + m^2 + n^2
$$

Thus, $b^2 + d^2 = a^2 + c^2 = 160$, and from equation $(*)$ we get

$$
bd = \frac{160 - 64}{\frac{8}{5}} = 60
$$

From the area formula for triangle $AED$, $P = \frac{(a - c)h}{2} = \frac{bd \sin \alpha}{2}$, and

$$
\sin \alpha = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}
$$

we get

$$
h = \frac{bd \sin \alpha}{a - c} = \frac{60 \cdot \frac{3}{5}}{8} = \frac{9}{2}
$$

Finally, the area of the trapezoid is $P = \frac{a + c}{2} h = \frac{16}{2} \cdot \frac{9}{2} = 36$.

## NATIONAL MATHEMATICS COMPETITION

## 4th grade - high school - B variant

## May 11, 2022.