## Task 3 - 180923

For a right-angled triangle $A B C$ with the right angle at $B$ and $\angle B A C=60^{\circ}$, the length $r$ of the circumradius is given.

Calculate the perimeter and area of this triangle as well as the length of the height perpendicular to its hypotenuse!
After the inversion of Thales' theorem, $B$ lies on the semicircle over $A C$. Let $M$ be the midpoint of $A C$, then the circle around $M$ with radius $M A = M B = M C = r$ is the circumcircle of triangle $A B C$: Thus, $A C = 2 r$.

The isosceles triangle $A B M$ has, according to the given condition, an angle of $60^{\circ}$, making it equilateral. Therefore, $A B = r$.

By the Pythagorean theorem, one obtains $C B = r \sqrt{3}$. Thus, the perimeter $u = 3 r + r \sqrt{3} = r(3 + \sqrt{3})$ and the area

$$
I = \frac{1}{2} A B \cdot B C = \frac{1}{2} r^{2} \sqrt{3}
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1006.jpg?height=428&width=420&top_left_y=246&top_left_x=818)

Since the area can also be calculated using the formula $I = \frac{1}{2} A C \cdot h = r \cdot h$, with $h$ as the length of the height on the hypotenuse $A C$, it follows that

$$
h = \frac{I}{r} = \frac{1}{2} r \sqrt{3}
$$