A rectangle has one side that is $10 \mathrm{~cm}$ long. How long is the other side of the rectangle if a $10 \mathrm{~cm} \times 1 \mathrm{~cm}$ rectangle fits exactly diagonally inside it?
# Solution.

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In the rectangle $ABCD$, $AB=10$. The inscribed rectangle touches the side $AB$ at point $E$, the side $BC$ at point $F$, and the side $DA$ at point $G$. In the right triangle $AGE$, $GE=1$, the segment $AE$ is denoted by $x$, and the angle $\angle AGE$ is denoted by $\alpha$. In the right triangle $EBF$, $\angle FEB = \alpha$ (angles with perpendicular legs), $EB=10-x$, and $BF=y$. In the triangle $AGE$, $GA=\sqrt{1-x^{2}}$, and $\sin \alpha=x$. In the triangle $EBF$, $\cos \alpha=\frac{10-x}{10}$. We know that

$$
x^{2}+\left(\frac{10-x}{10}\right)^{2}=1
$$

Rearranging the equation, we get $101 x^{2}=20 x$. Since $x \neq 0$, we can simplify, and thus $\sin \alpha=x=\frac{20}{101} \approx 0.198$. From the triangle $EBF$, $\frac{y}{10}=\sin \alpha \approx 0.198$, hence $y \approx 1.98$. The length of the desired side is

$$
CB=y+CF=y+GA=y+\sqrt{1-x^{2}} \approx 1.98+0.98=2.96 \text{ cm}
$$