Two side-by-side squares are inscribed in a semicircle. The diameter of the semicircle is 16. What is the sum of the two squares' areas? Two side-by-side squares are inscribed in a semicircle. The diameter of the semicircle is $16$. What is the sum of the two squares' areas?
This is a bonus question from my online math class that I've been trying to solve but haven't gotten very far. I named $CD$ $x$ and $HD$ $y$, then drew $OB$ and $OF$, then did the Pythagorean theorem to try to get $x^2+y^2$ but I didn't have much luck with that.

 A: Let $OD=c$ and the side lengths of the two squares $a$ and $b$. From right triangles OAB and OHF,
$$r^2=a^2+(a-c)^2;\>\>\>\>\> r^2=b^2+(b+c)^2$$
Eliminate $c$ to get,
$$a-\sqrt{r^2-a^2}=\sqrt{r^2-b^2}-b$$
Square both sides,
$$a\sqrt{ r^2-a^2} =b\sqrt{r^2-b^2}$$
Square again and rearrange,
$$r^2(a^2-b^2)=a^4-b^4$$
Thus, the sum of the two areas is
$$a^2+b^2=r^2=64$$
A: A somewhat devious way is to extract the fact that the answer doesn't depend on where D is, so place it at the origin!
Then it's simply:
$$ x = y $$
$$ \therefore x^2 + y^2 = 2 x^2 = 8^2 $$ 
A: 
Let $V$ be the common vertex of our squares which lies on the diameter of the semicircle. We may assume that the coordinates of $V$ are $(x,0)$. The upper-right-corner $A$ is located at the intersection (the one with a positive ordinate) of $x^2+y^2=64$ and $y=x-v$. The upper-left-corner $B$ is located at the intersection (always the one with a positive ordinate) of $y=v-x$ and $x^2+y^2=64$. It follows that the ordinate of $A$ is $\frac{1}{2}\left(-v+\sqrt{128-v^2}\right)$ and the ordinate of $B$ is $\frac{1}{2}\left(v+\sqrt{128-v^2}\right)$. By summing the squares of these numbers we get that the total area of our squares is 
$$ \frac{1}{4}(v^2+128-v^2+v^2+128-v^2) = 64,$$
i.e. the area of a square built on a radius.

In order to produce an elementary proof, we just have to show that the length of $AB=\sqrt{2}\sqrt{AA'^2+BB'^2}$ does not depend on the position of $V$ on the diameter. But this is trivial since the symmetric of $B$ with respect to the diameter "sees" the $AB$-chord under an angle of $45^\circ$, such that $\widehat{AOB}$ always is a right angle (and $AOVB$ is a cyclic quadrilateral).
A: You can use the Pythagorean theorem twice:

A: The base  $CA$ is partitioned into $(a,b)$ segments for the common squares corner and $(b,a) $ for locating center of circle.
To appreciate this direct analytical geometry is enough:
Equation of perpendicular bisector from the sketch:
$$\dfrac{ y-(a+b)/2}{x-(a+b)/2}= \dfrac{ (a+b)}{(a-b)}$$
At y=o, x=? Solving

$$ x_O=b $$
EDIT1:
So in the construction all you need to find the center of circle is marking off segment length $b$ with a compass centered at $C$ and complete the circle through square outer vertices $(F,B)$ of required circle radius $R$ centered at $O$.
Shown case is chosen $ (a=3,b=4)$ from familiar Pythagorean triplet with diameter of circle $10$ instead of $16$
$$ R^2= a^2+b^2 = R^2= 5^2 $$
The squares area sum equals area of square made on a side of radius $R$
A: *

*We know the line between the semi circle and D exactly bisects each square, and thus FDB forms a right angle.

*We know that for some angle FDM, the y coordinate at F is $sin \theta$

*The right angle FDB then requires that the y coordinate for B is $sin (\theta + \pi/2) = cos \theta$

*The area of each square is the square of those y coordinates, and thus the sum is $(r sin \theta)^2 + (r cos \theta)^2$

*Given the identity $sin^2 \theta + cos^2 \theta = 1$, we can simplify the result to $r^2 = 64$.


Incidentally if you combine this approach for the unit circle with the Pythagorean approach in other answers, you have a neat proof of the trig identity we used. It also demonstrates that all such paths between points on a circle which form a right angle have equal length.
A simpler form is to note #1 as above, then:


*All valid combinations of F and B form a right angle triangle with the same length hypotenuse. 

*The square of that hypotenuse is always the same as the required area by substituting for the squares' diagonals, and thus the result is constant.

*If the result is constant it must equal the easy symmetric case and thus equals $r^2$.

A: The sum of areas is not always 64.
Example : consider the square whose vertice are $O$, center of the semi-circle, $M_1 = (a, 0)$, $M_2=(a, a)$ and $M_3 = (0, a)$ with $a=4.\sqrt 2$. This square is inscribed in the semi-circle at $M_2$, its area is $a^2 = 32$.
Then consider the square whose vertice are $M_1$, $M_5=(a+b, 0)$, $M_6 = (a+b, b)$ and $M_7 = (a, b)$ with $b =2.\sqrt 6 - 2.\sqrt 2$. This square is inscribed in the semi-circle at $M_6$, its area is $b^2 = 32 -8.\sqrt 3$. 
So the sum of the two area is $64-8.\sqrt 3$, different from 64.
As Phil H noticed, "All the solutions given (and indeed the question, I suppose) assume that length OD <= length OA". I find it very important to give demonstrations that give a response to the explicit question, implicit assumption are a way to non-sense. 
Every answer given is wrong until you change the question : for example you could assert that the point of contact between the semi-circle and the square cannot be on the line $CD$.
A: If we embed the bigger square with side $A$ into a coordinate system first quadrant and the smaller square with side $a$ adjacent to it we can easily see that the following four triangles are congruent:
$$((0,0),(a,0),(0,A))$$
$$((a,0),(A+a,0),(A+a,a))$$
$$((0,A),(A,A),(A,A+a))$$
$$((A,a),(A+a,a),(A,A+a)).$$ From this, it also follows that $((a,0),(0,A),(A,A+a),(A+a,a))$ has equal diagonals $((a,0),(A,A+a))$ and $((0,A),(A+a,a)).$
Hence, they form a new square with area $r^2$. As the coset of the new square, with the sum of the bigger and smaller squares extended with exactly two of these congruent triangles, gives exactly the new square on one hand and the sum of the bigger and smaller squares on the other hand hence $r^2=A^2+a^2$ follows.
