1. Let a right triangle $ABC$ (right angle at vertex $C$) be given with the data $c=8 \text{ cm}$ and $v_{c}=\sqrt{7}$ $\text{cm} (a>b)$. Accurately calculate the area of the triangle, the radius of the circumscribed circle, and the lengths of the legs.

(10 points)
1. The area of triangle $ABC$ is $S=\frac{c \cdot v_{c}}{2}=\frac{8 \sqrt{7}}{2}=4 \sqrt{7} \mathrm{~cm}^{2}$. The radius of the circumcircle of the triangle is $R=\frac{c}{2}=4 \mathrm{~cm} . \mathrm{Z} x$ and $c-x$ denote the perpendicular projections of the legs onto the hypotenuse $c$. Using the altitude theorem $v_{c}^{2}=x(c-x)$. We get the quadratic equation $x^{2}-8 x+7=0$. The solutions to the equation are $x_{1}=7$ and $x_{2}=1$. Using the Pythagorean theorem for the lengths of the legs $a$ and $b$ we get $a^{2}=7^{2}+(\sqrt{7})^{2}$ and $b^{2}=1^{2}+(\sqrt{7})^{2}$. We calculate the lengths of the legs $a=\sqrt{56}=2 \sqrt{14} \mathrm{~cm}$ and $b=\sqrt{8}=2 \sqrt{2} \mathrm{~cm}$.

The area of the triangle $S=\frac{c \cdot v_{c}}{2}=\frac{8 \sqrt{7}}{2}=4 \sqrt{7} \mathbf{c m}^{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

![](https://cdn.mathpix.com/cropped/2024_06_07_3b81dd7f2d1678befe89g-14.jpg?height=60&width=1731&top_left_y=1832&top_left_x=174)

![](https://cdn.mathpix.com/cropped/2024_06_07_3b81dd7f2d1678befe89g-14.jpg?height=57&width=1727&top_left_y=1879&top_left_x=176)

Writing the quadratic equation $x^{2}-8 x+7=0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

The solutions to the quadratic equation are $x_{1}=7, x_{2}=1 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point

Using the Pythagorean theorem $a^{2}=7^{2}+(\sqrt{7})^{2}, b^{2}=1^{2}+(\sqrt{7})^{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ 1 1 point

Calculating the lengths of the legs $a=\sqrt{56}=2 \sqrt{14} \mathbf{~ m}, b=\sqrt{8}=2 \sqrt{2} \mathbf{~ c m} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \mathbf{1} \mathbf{1}$ point