6. Given the circle $k_{1}(A ; 4 \text{~cm})$, its point $B$ and the circle $k_{2}(B ; 2 \text{~cm})$. Point $C$ is the midpoint of segment $A B$ and point $K$ is the midpoint of segment $A C$. Calculate the area of the right triangle KLM, whose vertex L is one of the intersection points of circles $k_{1}, k_{2}$ and whose hypotenuse $K M$ lies on the line $A B$.
SOLUTION. First of all, note that due to the axial symmetry with respect to the line $A B$, it does not matter which of the two intersections of the circles $k_{1}$ and $k_{2}$ we choose as point $L$.

The area of the triangle $K L M$ will be expressed not by the lengths of its legs $K L$ and $L M$, but by the lengths of its hypotenuse $K M$ and the corresponding height $L D$ (Fig. 3 on the left), using the formula ${ }^{3}$

$$
S_{K L M}=\frac{|K M| \cdot|L D|}{2} .
$$

To determine the distances of point $D$ from points $B$ and $L$, consider the midpoint $S$ of the segment $B L$ (Fig. 3 on the right). Triangles $A S B$ and $L D B$ are both right-angled with a common acute angle at vertex $B$. Therefore, they are similar by the AA similarity criterion, so the ratio of their sides is (we are calculating with lengths without units, so according to the problem, $|A B|=4,|B L|=2$, and thus $|B S|=|B L| / 2=1)$

$$
\frac{|B D|}{|B S|}=\frac{|B L|}{|B A|}=\frac{2}{4}, \quad \text { hence } \quad|B D|=\frac{1}{2}|B S|=\frac{1}{2}
$$[^1]![](https://cdn.mathpix.com/cropped/2024_04_17_222317d063c1e447ab0bg-8.jpg?height=742&width=1330&top_left_y=237&top_left_x=364)

Fig. 3

From the Pythagorean theorem for triangle $L D B$, we get ${ }^{4}$

$$
|L D|=\sqrt{|B L|^{2}-|B D|^{2}}=\sqrt{4-\frac{1}{4}}=\frac{\sqrt{15}}{2} .
$$

From the equality $|B D|=1 / 2$, we also derive the length of the segment $K D$ of the hypotenuse $K M$ of the right-angled triangle $K L M$, namely $|K D|=|A B|-|A K|-|B D|=4-1-1 / 2=5 / 2$. The length of the other segment $D M$ is now determined from the Euclidean theorem of the height, according to which $|L D|^{2}=|K D| \cdot|D M|$. We get $|D M|=|L D|^{2} /|K D|=(15 / 4) /(5 / 2)=3 / 2$, so the entire hypotenuse $K M$ has a length of $|K M|=|K D|+|D M|=5 / 2+3 / 2=4$. By substituting into the formula at the beginning of the solution, we arrive at the result

$$
S_{K L M}=\frac{|K M| \cdot|L D|}{2}=\frac{4 \cdot \frac{\sqrt{15}}{2}}{2}=\sqrt{15}
$$

Answer. The area of triangle $K L M$ is $\sqrt{15} \mathrm{~cm}^{2}$.