## Task B-1.4.

In a circle $k$ with center at point $O$ and radius $6 \mathrm{~cm}$, two larger circles with centers $B$ and $D$, and two smaller circles with centers $A$ and $C$ are inscribed, as shown in the figure. These 4 circles are tangent to circle $k$. The larger circles touch each other at point $O$, and the two smaller circles touch the larger circles. Determine the area of quadrilateral $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_05_30_bcbf631b6726be7fa637g-03.jpg?height=454&width=460&top_left_y=1778&top_left_x=1255)
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_bcbf631b6726be7fa637g-04.jpg?height=659&width=657&top_left_y=293&top_left_x=631)

The two larger circles have equal radii $r=3 \text{ cm}$, and due to symmetry with respect to the line $B D$, the two smaller circles have equal radii $r_{1}$.

The quadrilateral $A B C D$ is a rhombus because $|A B|=|B C|=|C D|=|D A|=r+r_{1}$.

Since the diagonals of a rhombus bisect each other and intersect at right angles, triangle $A O D$ is a right triangle and the following holds:

$$
3^{2}+\left(6-r_{1}\right)^{2}=\left(3+r_{1}\right)^{2}
$$

From this, $r_{1}=2 \text{ cm}$.

The area of the rhombus is $P=\frac{4 \cdot r \cdot\left(6-r_{1}\right)}{2}=\frac{48}{2}=24 \text{ cm}^{2}$.