Notes
four squares and a triangle ii solution

Solution to the Four Squares and a Triangle II Puzzle

Solution by Area of a Square and Area of a Triangle

With the points labelled as above, let $a$ be the side length of the smaller squares and $b$ of the larger. Then the total area of all four squares is $2 a^2 + 2 b^2$.

The rectangle $A B C D$ has width $2a + b$ and height $2 b - a$, so its area is $(2 a + b)(2 b - a) = 2 b^2 + 3 a b - 2 a^2$.

Triangle $A C D$ has half this area. Then triangle $A G F$ has height $b$ and width $a$ so has area $\frac{1}{2} a b$. Triangle $F E C$ has height $b - a$ and width $b + a$ so has area $\frac{1}{2} (b^2 - a^2)$. Finally, rectangle $D G F E$ has width $a$ and height $b - a$ so has area $a b - a^2$. Putting this together means that the area of the orange triangle is:

and this simplifies all the way to $\frac{1}{2}(a^2 + b^2)$.

As it is given that the orange triangle has area $4$, it is therefore the case that the four squares together have area $4 \times 4 = 16$.

Solution by Invariance Principle

The relative sizes of the squares can be changed, so putting all four squares the same size shows that area of the orange triangle is the same as that of one square (or, rather, of half of two squares) so the area of all four is $4 \times 4 = 16$.