Pile them up

Label the three squares as $ABCD$ , $EFGH$ , and $IJKL$ as shown. Let the side lengths of square $EFGH$ and square $IJKL$ be $a$ and $b$ respectively, and $KM$ be perpendicular to $AB$ . By Pythagorean theorem ,

$\begin{aligned} AF^2 + FG^2 & = AG^2 \\ (AE+EF)^2 + FG^2 & = AG^2 \\ \left(\frac {10-a}2 + a \right)^2 + a^2 & = 10^2 \\ 5a^2 + 20a - 300 & = 0 \\ a^2 + 4a - 60 & = 0 \\ (a-6)(a+10) & = 0 & \small \blue{\text{Since }a > 0} \\ \implies a & = 6 \end{aligned}$

Similarly,

$\begin{aligned} AM^2 + KM^2 & = AK^2 \\ (AF - MF)^2 + (KJ+JM)^2 & = AK^2 \\ \left(8 - \frac {6-b}2 \right)^2 + (b+6)^2 & = 10^2 \\ 5b^2 + 68b - 156 & = 0 \\ (5b+78)(b-2) & = 0 & \small \blue{\text{Since }b > 0} \\ \implies b & = 2 \end{aligned}$

The sum of areas of square $EFGH$ and square $IJKL$ is $a^2 + b^2 = 6^2 + 2^2 = \boxed{40}$ .

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Generalization: Answering questions by @Vijay Simha and @Saya Suka .

Let the side length of the $n$ th stacked square be $a_n$ . Then $a_n$ is given by:

$\left(5 + \frac {a_n}2 \right)^2 + (a_1 + a_2 + a_3 + \cdots + a_n)^2 = 10^2$

And we have:

$\begin{array} {lll} n = 1: & \left(5 + \frac {a_1}2 \right)^2 + a_1^2 = 100 & \implies a_1 = 6 \\ n = 2: & \left(5 + \frac {a_2}2 \right)^2 + (6+a_2)^2 = 100 & \implies a_2 = 2 \\ n = 3: & \left(5 + \frac {a_3}2 \right)^2 + (8+a_3)^2 = 100 & \implies a_3 = \frac {8\sqrt{31}-42}5 \\ n = 4: & \left(5 + \frac {a_4}2 \right)^2 + \left(\frac {8\sqrt{31}-2}5+a_4\right)^2 = 100 & \implies a_4 = \frac {4\sqrt{1953 + 208 \sqrt{31}}-32\sqrt{31}-42}5 \\ \cdots \end{array}$

Let a be the side of the larger orange square and b the side of the small orange square, and place the whole diagram on a Cartesian graph so that the bottom center of the large square is at the origin.

Then the equation of the quarter circle with its center on the left side has an equation of $(x + 5)^2 + y^2 = 100$ .

The top right corner of the larger orange square is at $(\frac{1}{2}a, a)$ , so $(\frac{1}{2}a + 5)^2 + a^2 = 100$ , which solves to $a = 6$ for $a > 0$ .

The top right corner of the smaller orange square is at $(\frac{1}{2}b, a + b) = (\frac{1}{2}b, b + 6)$ , so $(\frac{1}{2}b + 5)^2 + (b + 6)^2 = 100$ , which solves to $b = 2$ for $b > 0$ .

Therefore, the total area of the two orange squares is $A = a^2 + b^2 = 6^2 + 2^2 = \boxed{40}$ .

Due to symmetry, both of the orange squares would have to be horizontally centered in the middle of the diagram (since the two centers are at the bottom vertices on their left and right each).

If we denote the 2 shorter lines that make up the common radius of the quarters together with the larger orange square's base as x each, then
10² = (10 – x)² + (10 – 2x)²
100 = 5x² – 60x + 200
0 = x² – 12x + 20
= (x – 2)(x – 10)
Since 0 < x < 10,
x = 2
for an area of big square of
= (10 – 2x)² = (10 – 2 × 2)² = 36

Now, with a small square with sides of 2y, we have an equation of
10² = (10/2 + y)² + (10 – 2×2 + 2y)²
= (5 + y)² + (6 + 2y)²
100 = 5y² + 34y + 61
0 = 5y² + 34y – 39
= (5y + 39)(y – 1)
Again, since y > 0,
y = 1
for an area of small square of
= (2y)² = (2 × 1)² = 2² = 4

Answer = 36 + 4 = 40

Let $r=10$ . For both red squares with sides $h_1,\:h_2$ , we can find a right triangle containing its top-left corner, and the bottom-right corner of the entire figure: $\begin{aligned} h_1:&& r^2&=h_1^2+\left(\frac{r}{2}+\frac{h_1}{2}\right)^2 &&&\Rightarrow &&&& h_1^2 + \frac{2r}{5}h_1 - \frac{3r^2}{5}&=0 &&&\Rightarrow &&&& h_1&=\frac{-1\pm4}{5}r &&& \Rightarrow &&&& h_1&=\frac{3r}{5}\\[.5em] h_2:&& r^2&=(h_1+h_2)^2+\left(\frac{r}{2}+\frac{h_2}{2}\right)^2 &&&\Rightarrow &&&& h_2^2 + \frac{34r}{25}h_1 - \frac{39r^2}{125}&=0 &&&\Rightarrow &&&& h_2&=\frac{-17\pm22}{25}r &&& \Rightarrow &&&& h_2&=\frac{r}{5} \end{aligned}$ The total read area is $h_1^2+h_2^2=\frac{2r^2}{5}=\boxed{40}$