## Task A-1.6. (20 points)

Trapezoid $A B C D$ has a right angle at vertex $B$, and diagonal $\overline{B D}$ is perpendicular to side $A D$. The length of side $\overline{B C}$ is $5 \mathrm{~cm}$, and the length of diagonal $\overline{B D}$ is $13 \mathrm{~cm}$.

Calculate the area of trapezoid $A B C D$.
## Solution.

First method.

We know $|BC|=5 \text{ cm}, |BD|=13 \text{ cm}$.

From the right triangle $BCD$, it follows that $|CD|=\sqrt{13^2-5^2}=12 \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_59925bf190fc107f19b7g-04.jpg?height=389&width=805&top_left_y=842&top_left_x=634)

Let $|AB|=a, |AD|=d$. Let $N$ be the foot of the perpendicular from point $D$ to the base of the trapezoid. Then $|AN|=a-12$,

(2 points)

so from the right triangle $AND$ we get: $(a-12)^2 + 5^2 = d^2$.

(5 points) and from triangle $ABD$: $d^2 + 13^2 = a^2$.

Solving the obtained system, we get: $a=\frac{169}{12} \text{ cm}$,

so the area is $P=\frac{1}{2}\left(\frac{169}{12} + 12\right) \cdot 5, \quad P=\frac{1565}{24} \text{ cm}^2$.

Second method.

We know $|BC|=5 \text{ cm}, |BD|=13 \text{ cm}$.

From the right triangle $BCD$, it follows that $|CD|=\sqrt{13^2-5^2}=12 \text{ cm}$.

Triangles $ABD$ and $BDC$ are similar ($\varangle ABD = \varangle DBC, \varangle ADB = \varangle CBD = 90^\circ$).

Therefore, it holds that: $\frac{|AB|}{13} = \frac{13}{12}$,

from which we get $|AB| = \frac{169}{12} \text{ cm}$,

and finally: $P=\frac{1}{2}\left(\frac{169}{12} + 12\right) \cdot 5, \quad P=\frac{1565}{24} \text{ cm}^2 \quad \left(=65.208 \dot{3} \text{ cm}^2\right)$.