Sum of areas of squares

Here's my alternate approach:

Let $x$ be the distance from the center of the semi circle to the right edge of the yellow square. $\\$ The yellow square has side length $r-x$ , the green square has side length $r+x$ , and the blue square has side length $(r+x)-(r-x) = 2x$ . $\\$ The sum of the areas of the squares is then $A =(r-x)^2 + (r+x)^2 + (2x)^2 = 2r^2 + 6x^2$ .

Draw the radius of the semicircle to the vertex of the blue square on the semicircle, then the perpendicular down to the diameter of the semicircle. $\\$ The right triangle formed has height $r-x$ , base $x+2x = 3x$ , and hypotenuse $r$ . $\\$ By the Pythagorean Theorem: $(r-x)^2 + (3x)^2 = r^2 \Rightarrow x = \frac{r}{5}$ or $x = 0$ .

$x>0 \Rightarrow x = \frac{r}{5} = \frac{1}{2}$ . The sum of the areas is $A = 2\cdot(\frac{5}{2})^2 + 6\cdot(\frac{1}{2})^2 = 14$ .

(Note that $x = 0$ corresponds to the blue square disappearing, and the yellow and green squares each having side length $\frac{5}{2}$ . $\\$ In this case, the total area is $A = 12\frac{1}{2}$ ).

Let the side length of the green square be $a$ and that of the yellow square be $b$ , where $a > b$ . Then the side length of the blue square is $a-b$ and the diameter $AB=a+b=5$ . Let $C$ be the vertex of the blue square on the semicircle and $CD$ be perpendicular to $AB$ . Because $AB$ is a diameter, $\angle ACB = 90^\circ$ . Therefore $\triangle ABC$ and $\triangle BCD$ are similar. Then we have:

$\begin{aligned} \frac {BC}{AB} & = \frac {BD}{BC} \\ BC^2 & = AB \cdot BD \\ CD^2 + BD^2 & = AB \cdot BD \\ b^2 + (a+a-b)^2 & = (a+b)(a+a-b) \\ 4a^2 - 4ab + 2b^2 & = 2a^2 +ab - b^2 \\ 2a^2 - 5ab + 3b^2 & = 0 \\ (2a-3b)(a-b) & = 0 & \small \blue{\text{Since }a> b} \\ \implies 2a & = 3b \end{aligned}$

Since $a+b=5$ , $\implies a = 3$ , $b=2$ , and $a-b = 1$ , and the sum of areas of the three squares is $a^2 + b^2 + (a-b)^2 = 9+4+1 = \boxed{14}$ .

Why not $12.5$ ? In that case the blue and the yellow squares will merge into one square.

Link to the solution video