3B. Given a right triangle СВА with legs of length 3 and 4. Squares BCSR and $A B Q P$ are constructed outside the triangle on the smaller leg and the hypotenuse, respectively. Calculate the area of triangle $B R Q$.
Solution. By the Pythagorean theorem, the length of the hypotenuse is 5.

Let $\measuredangle A B C=\beta$ and $\quad \measuredangle Q B R=\theta$ (see diagram). From

$$
\measuredangle A B Q=\measuredangle C B R=90^{\circ},
$$

we have

$$
\beta+\theta=180^{\circ} \text { i.e. } \theta=180^{\circ}-\beta \text {. }
$$

Then

$\sin \theta=\sin \left(180^{\circ}-\beta\right)=\sin \beta=\frac{4}{5}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_e56627a2e774dea33268g-08.jpg?height=483&width=646&top_left_y=1306&top_left_x=816)

Finally, the area of triangle $Q B R$ is $P=\frac{1}{2} \overline{B Q} \cdot \overline{B R} \sin \theta=6$.

## IV year