Consider a rectangle $A B C D$ where the lengths of the sides are $\overline{A B}=4$ and $\overline{B C}=8$. On the sides $B C$ and $A D$, points $M$ and $N$ are fixed, respectively, such that the quadrilateral $B M D N$ is a rhombus. Calculate the area of this rhombus.
Solution

Observe the following drawing:

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-27.jpg?height=380&width=662&top_left_y=1483&top_left_x=800)

Since $B M D N$ is a rhombus, the lengths of the segments $B M, M D, D N$, and $N B$ are all equal to the same value, say $x$. According to the problem, $\overline{A D}=8$, $\overline{A N}=8-x$.

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-27.jpg?height=383&width=660&top_left_y=2110&top_left_x=801)

We can now apply the Pythagorean Theorem to the triangle $B A N$:

$$
4^{2}+(8-x)^{2}=x^{2}
$$

Solving the equation, we get $x=5$. Finally, the area of the rhombus is $4 \times x = 4 \times 5 = 20$.