4. Given a right triangle $ABC$. From the vertex $A$ of the right angle, the median $AM$ is drawn. The height $AD$ of triangle $ACM$ bisects the opposite side.

a) Determine the angles of triangle $ABC$.
b) Calculate the area and volume of the solid obtained by rotating triangle $ABC$ around the side $AB$, if the length of the segment $AD$ is $k \sqrt{3}$.
Solution. a) In a right-angled triangle, the median to the hypotenuse is half the hypotenuse, so $A M = C M$ (see diagram on the right). Furthermore, since $D M = C D, A D = A D$, and $\measuredangle A D M = \measuredangle A D C = 90^{\circ}$, we conclude that triangles $A D M$ and $A D C$ are congruent. Therefore,

![](https://cdn.mathpix.com/cropped/2024_06_05_6f06ce47627480a4d953g-4.jpg?height=327&width=310&top_left_y=418&top_left_x=1172)
$A M = A C$. Thus, triangle $A M C$ has equal sides, making it equilateral. From this, we conclude that $\measuredangle A C B = 60^{\circ}, \measuredangle A B C = 30^{\circ}$, and by assumption $\measuredangle B A C = 90^{\circ}$.

b) Since $A D = k \sqrt{3}$, from $A D = \frac{A C \sqrt{3}}{2}$ it follows that $A C = 2 k$. Therefore, $B C = 4 k$ and $A B = 2 k \sqrt{3}$. The resulting solid is a cone with radius $A C$, height $A B$, and slant height $B C$. Therefore,

$$
P = \pi \cdot (2 k)^{2} + \pi \cdot 2 k \cdot 4 k = 12 k^{2} \pi \text{ and } V = \frac{1}{3} \pi \cdot (2 k)^{2} \cdot 2 k \sqrt{3} = \frac{8 \sqrt{3} k^{3} \pi}{3}
$$