## Task A-3.8. (10 points)

A right trapezoid is given, in which a circle can be inscribed. If the distances from the center of the inscribed circle to the ends of the longer leg are $15 \mathrm{~cm}$ and $20 \mathrm{~cm}$, what is the area of the trapezoid?
## Solution.

Let $A B C D$ be the trapezoid under consideration, with $\varangle D A B=\varangle C D A=90^{\circ}$ and $|A B|>|C D|$. Let $O$ be the center of the inscribed circle of the trapezoid, $r$ its radius, and $P, Q, R$ and $S$ the points of tangency of the circle with the sides $\overline{A B}, \overline{B C}, \overline{C D}$ and $\overline{A D}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_30_5dbbe9edfad2d8808b54g-21.jpg?height=605&width=805&top_left_y=1622&top_left_x=634)

Then $|O B|=20,|O C|=15$.

Notice that $|P B|=|B Q|,|Q C|=|C R|$ and $\varangle A B O=\varangle O B C, \varangle B C O=\varangle O C D$.

Triangles $P O B$ and $Q O B$ are congruent, as are triangles $C Q O$ and $C R O$.

(2 points)

Quadrilaterals $A P O S$ and $S O R D$ are squares with side length $r$.

(1 point)

Therefore,

$$
\begin{aligned}
P & =P(A P O S)+P(S O R D)+P(O P B)+P(O Q B)+P(O Q C)+P(O R C) \\
& =2 r^{2}+2[P(O Q B)+P(O Q C)] \\
& =2 r^{2}+2 P(O B C)
\end{aligned}
$$

Let $\alpha=\varangle A B O=\varangle O B C$.

Then $\varangle B C O=\varangle O C D=90^{\circ}-\alpha$, since $\varangle A B C+\varangle B C D=180^{\circ}$.

By examining the angles of triangle $B C O$, we conclude that $\varangle B O C=90^{\circ}$.

Thus, triangle $O B C$ is a right triangle.

(2 points)

Using the Pythagorean theorem, we can calculate the length of side $\overline{B C}$ of triangle $O B C$:

$$
|B C|=\sqrt{|O B|^{2}+|O C|^{2}}=\sqrt{20^{2}+15^{2}}=25
$$

The radius of the inscribed circle $\overline{O Q}$ is the height of the right triangle $O B C$, so

$$
r=\frac{|O B||O C|}{|B C|}=\frac{20 \cdot 15}{25}=12
$$

Now we can calculate the desired area:

$$
P=2 r^{2}+|O B| \cdot|O C|=2 \cdot 12^{2}+20 \cdot 15=588
$$

## Variations.

## Method of calculating the area.

The area of the trapezoid is $P=\frac{1}{2}(|A B|+|C D|) \cdot|A D|$.

Since this trapezoid is tangential, we have $|A B|+|C D|=|B C|+|D A|$.

As in the above solution, we determine $|B C|=25$ and $r=12$,

so $|A D|=24,|A B|+|B C|=|B C|+|D A|=49$ and $P=\frac{1}{2} \cdot 49 \cdot 24=588$.

We can also determine $|A B|=12+16=28,|C D|=9+12=21$.

## Trigonometry.

Knowing that $O B C$ is a right triangle, we can use trigonometry.

For the angle $\alpha$ from the above solution, we have $\operatorname{tg} \alpha=\frac{|O C|}{|O B|}=\frac{3}{4}$.

It follows that $\sin \alpha=\frac{3}{5}$ and $\cos \alpha=\frac{4}{5}$.

We have $r=|O B| \sin \alpha=12$ and $|B C|=|O B| \cos \alpha+|O C| \sin \alpha=16+9=25$.

## SCHOOL (CITY) MATHEMATICS COMPETITION

## 4th grade - high school - A variant

February 4, 2010.

IF A STUDENT HAS A DIFFERENT METHOD OF SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO EVALUATE AND GRADE THAT METHOD APPROPRIATELY.