A right-angled trapezoid $A B C D$ is given, with a right angle at vertex $B$ and parallel sides $A B$ and $C D$. The diagonals of the trapezoid are perpendicular to each other and have lengths $|A C|=12 \mathrm{~cm}$, $|B D|=9 \mathrm{~cm}$. Calculate the perimeter and area of this trapezoid.

( $M$. Krejčová)

Hint. The area of this trapezoid is the same as the area of a suitable right-angled triangle. First, find such a triangle, then determine the lengths of the sides of the trapezoid and its perimeter.
We draw a line through point $C$ parallel to the diagonal $B D$ and denote its intersection with line $A B$ as $B^{\prime}$, see the figure. Since lines $A B$ and $C D$ are also parallel, $B B^{\prime} C D$ is a parallelogram, and it follows that $\left|B^{\prime} C\right|=|B D|=9 \mathrm{~cm}$ and $\left|B^{\prime} B\right|=|C D|$.

![](https://cdn.mathpix.com/cropped/2024_04_17_db2558a90e80c9854485g-7.jpg?height=445&width=774&top_left_y=240&top_left_x=641)

The essence of this construction lies in the observation that triangles $A C D$ and $C B^{\prime} B$ have the same area (sides $C D$ and $B^{\prime} B$ are equal, and the heights of both triangles to these sides are the same). Therefore, the area of trapezoid $A B C D$ is the same as the area of triangle $A B^{\prime} C$, which we can easily determine: From the construction, it follows that triangle $A B^{\prime} C$ is a right triangle, and from the problem statement, we know both of its legs $|A C|=12 \mathrm{~cm}$ and $\left|B^{\prime} C\right|=9 \mathrm{~cm}$. The area of triangle $A B^{\prime} C$, and thus of the given trapezoid, is

$$
S=\frac{1}{2} \cdot 12 \cdot 9=54\left(\mathrm{~cm}^{2}\right)
$$

To determine the perimeter of the trapezoid, we need to know the lengths of all its sides.

a) Side $B C$ is the height to side $A B^{\prime}$ in the aforementioned right triangle. Using the Pythagorean theorem, we calculate the length of the hypotenuse in triangle $A B^{\prime} C$:

$$
\left|A B^{\prime}\right|=\sqrt{12^{2}+9^{2}}=15(\mathrm{~cm}) .
$$

From the knowledge of the area of this triangle, we determine its height $|B C|$:

$$
\begin{aligned}
\frac{1}{2} \cdot 15 \cdot|B C| & =54, \\
|B C| & =7.2(\mathrm{~cm}) .
\end{aligned}
$$

b) In the right triangle $A B C$, we know its hypotenuse and now also one leg; using the Pythagorean theorem, we determine the length of the other leg:

$$
|A B|=\sqrt{12^{2}-7.2^{2}}=9.6(\mathrm{~cm}) .
$$

c) Clearly, $\left|A B^{\prime}\right|=|A B|+\left|B B^{\prime}\right|$ and $\left|B B^{\prime}\right|=|C D|$, from which we can easily express the length of side $C D$:

$$
|C D|=\left|A B^{\prime}\right|-|A B|=15-9.6=5.4(\mathrm{~cm}) \text {. }
$$

d) Side $A D$ can be seen as the hypotenuse in the right triangle $A P D$, where $P$ is the foot of the perpendicular from point $D$ to side $A B$. The lengths of the legs in this triangle are $|P D|=|B C|=7.2 \mathrm{~cm}$ and $|A P|=|A B|-|C D|=9.6-5.4=4.2(\mathrm{~cm})$. According to the Pythagorean theorem, we calculate the length of the hypotenuse:

$$
|A D|=\sqrt{7.2^{2}+4.2^{2}} \doteq 8.3(\mathrm{~cm}) .
$$

The perimeter of the given trapezoid is thus approximately equal to

$$
o=|A B|+|B C|+|C D|+|D A| \doteq 9.6+7.2+5.4+8.3=30.5(\mathrm{~cm}) .
$$

Note. For interest and verification, we also provide the calculation of the area of the trapezoid using the usual formula:

$$
S=\frac{1}{2}(|A B|+|C D|) \cdot|B C|=\frac{1}{2}(9.6+5.4) \cdot 7.2=54\left(\mathrm{~cm}^{2}\right) .
$$

Notice that the considerations at the beginning of our solution actually justify the validity of this formula.

The formula for calculating the area of the given trapezoid can also be derived using the following figure. On it, the dashed lines represent a rectangle, each side of which passes through one of the vertices of the trapezoid and is parallel to one of its diagonals. The area of the rectangle is equal to the product $|A C| \cdot|B D|$. The area of the trapezoid is clearly half of this, i.e., $S=\frac{1}{2}|A C| \cdot|B D|$.

![](https://cdn.mathpix.com/cropped/2024_04_17_db2558a90e80c9854485g-8.jpg?height=703&width=731&top_left_y=1096&top_left_x=662)