Day 17: Overlapping Squares

It is clear that each square is of area $16$ . Then:

So the blue area is $16-(16-A+16-B)=A+B-16=27-16 = \boxed{11}$

Area of each square = $4^2 = 16 \text{ unit}^2$

Area of green regions + white regions = $16 \times 2 = 32 \text{ unit}^2$

$\therefore$ Area of white regions = $32 - 27 = 5 \text{ unit}^2$

$\Rightarrow$ Area of blue region = $16 - 5 =11 \text{ unit}^2$

Area of union = 27 + 16 = 43 = Blue + 32

Blue = 43 - 32 = 11

Answer: $\boxed{11}$

$green~area=2(4^2)-shaded~area=27$

$shaded~area=32-27=5$

$blue~area=4^2-shaded~area=16-5=\boxed{11}$

We are given that the side length of these identical squares are 4 and the area of the green squares is 27. We must first consider the area of the three squares.

The formula for the area of a square states that:

$A = B*H$

Where B is base and H is height. Therefore we can see that the area of the squares is shown below:

$A = B*H$ = $4*4$ = $16$

We also know that the area of the two green squares is 27. This means that the white overlapping area is not part of this value of the green area. Since the total area of one of the squares is $16$ , we know that the area of the two green and white squares is $2*16$ or $32$ . We can now subtract the area of the green and white squares from the green area to find the area of the white overlapping area, which is $32 - 27$ or $5$ . Since we have already established that all $3$ of the squares are of equal side length, therefore equal area, the area of the blue square is 16. If we then subtract the two values, $16-5$ , we get the final value of $11$ .