3. A square is divided by two lines into four rectangles. What is the side length of the given square if the areas of three of the four rectangles are $48 \mathrm{~cm}^{2}, 96 \mathrm{~cm}^{2}$, and $144 \mathrm{~cm}^{2}$.

The intersection is such that the two largest of the mentioned three rectangles share only one common vertex.
Solution. Let's denote the sides of the rectangle as in the diagram. Then

$$
a d=144, b c=48 \text { and } b d=96
$$

and we need to determine $a d$. If we multiply the first and third equations, we get $a b c d=96 \cdot 144$, i.e.,

![](https://cdn.mathpix.com/cropped/2024_06_05_aa5de80eeceb04c02556g-2.jpg?height=301&width=274&top_left_y=359&top_left_x=1209)

$$
(a d)(b c)=96 \cdot 144
$$

However, $b c=48$, and substituting this into the previous equation, we find $a d=288$. Therefore, the area of the square is

$$
144+48+96+288=576 \mathrm{~cm}^{2}
$$

which means the length of its side is $24 \mathrm{~cm}$.