What's the area of this square?

Question

There's is a square with inner rectangles as in the diagram.

If the sum of perimeters of these 6 rectangles is 120 cm, what is the area in $\text{cm}^2$ of the square?

Answer 1

You can align the perimeters and find out that the sum of the perimeters of the rectangles is $10$ times the base of the square. This means that the measure of the square's base is $12$ centimeters; therefore, area of the square is $144$ square centimeters.

Answer 2

Look at the top of the square and it is broken into two lengths. Call those $L_1$ and $L_2$. And we know $L_1 + L_2 = s$ the side of the square.

Look at the side to the left. It is broken into three heights. Call those $H_1, H_2, H_3$ and we know $H_1 + H_2 + H_3 = s$.

Look at the side to right. It is broken into three heights. Call those $h_1, h_2, h_3$ and we know that $h_1 + h_2 + h_3 = s$.

To add up the perimeters of the six smaller rectangles we get:

$2(L_1 + H_1) + 2(L_1+H_2) + 2(L_1 + H_3) + 2(L_2 + h_1) + 2(L_2 + h_2) + 2(L_3+h_3) =$

$6(L_1 + L_2) + 2(H_1 + H_2 + H_3) + 2(h_1 + h_2 + h_3)=$

$6s + 2s + 2s = 10s = 120 cm$

So $s = 12 cm$.

Okay, that was more detail than needed. All that we needed to notice was that in counting how many times the sides of the rectangles are counted to get the perimeter of the 6 rectangles. The horizontal sides are counted six times; once for the top and bottom and twice for each horizontal bit in the middle. Whereas the vertical sides are counted four times; once on the left and right and two times for the single vertical bit in the center.

(I made a mistake the first time I did this in that I both the horizantol and the vertical inside lines were counted the same number of times inside.)

....

where were we? Oh, yes.

$s = 12 cm$ so $A = s^2 = 144 cm^2$.

Answer 3

As was already mentioned in a comment taking into account lines which counted once and twice one obtains:

$$ 6a+4a=120\Rightarrow a=12 \Rightarrow a^2=144. $$