## 3. Divided Square

A larger square is divided into four congruent rectangles and one smaller square as shown in the figure. If the perimeter of the larger square is $40 \mathrm{~cm}$ greater than the perimeter of the smaller square, and the perimeter of the smaller square is $8 \mathrm{~cm}$ greater than the perimeter of each of the four congruent rectangles, what is the difference in area between the larger and smaller square?

![](https://cdn.mathpix.com/cropped/2024_05_30_bfe8eca8742272ceb120g-2.jpg?height=357&width=369&top_left_y=958&top_left_x=1506)

## Result: $\quad 380$
Solution. Let $a$ be the length of the side of the smaller square and $b$ the length of the shorter side of the rectangle.

![](https://cdn.mathpix.com/cropped/2024_05_30_bfe8eca8742272ceb120g-2.jpg?height=525&width=619&top_left_y=1482&top_left_x=156)

Then the longer side of the rectangle has a length of $a+b$, and the side of the larger square has a length of $a+2b$.

The perimeter of the larger square is $o_{V}=4(a+2b)=4a+8b$, and the perimeter of the smaller square is $o_{M}=4a$.

The perimeter of each rectangle is $o=2(a+b+b)=$ $2(a+2b)=2a+4b$.

The difference in the perimeters of the larger and smaller squares is $40 \text{ cm}$, so we have

$$
\text { or } \quad 8b=40 \text {. }
$$

The length of the shorter side of the rectangle is $b=5 \text{ cm}$.

The difference in the perimeters of the smaller square and the rectangle is $8 \text{ cm}$, so we have

$$
4a-2a-4b=8 \quad \text { or } \quad 2a-20=8, \text { so } a=14
$$

The length of the side of the smaller square is $14 \text{ cm}$, and the side of the larger square is $24 \text{ cm}$.

The area of the larger square is $P_{V}=24^{2} \text{ cm}^{2}=576 \text{ cm}^{2}$, the area of the smaller square is

$P_{M}=14^{2} \text{ cm}^{2}=196 \text{ cm}^{2}$, and the difference in their areas is $380 \text{ cm}^{2}$.