Tied Cans

Relevant wiki: Length and Area - Composite Figures

We draw the squareABCD . ∵ $r= 3\text{ cm}$ , all circles are the same.
The area of the square ABCD is $(12\text{ cm})^2 = 144\text{ cm}^2$ .

We then draw the square DEFG ∵ The area of the square DEFG $(6\text{ cm})^2 = 36\text{ cm}^2$ .

The area of each of these circles is $\pi \times 3^2 = 9\pi \text{ cm}^2$ .

The green area is $\dfrac14 \left( \text{ area of square DEFG} - \text{ area of of one of these circles} \right) = \dfrac{36-9\pi}4 \approx 1.913 \text{ cm}^2$ .

The yellow area is the area of the square minus the area of all the circles - 4 times the green area,

$144 - (9\pi\times 4) - (1.913 \times4) = 23.176 \text{ cm}^2.$