Construct a square of given area Given are three squares of side lengths $a$, $b$ and $c$ with $a>b>c$. Construct (with compass and straightedge) a square of the area $a^2-b^2-c^2$!
I have thought about "cutting" the two smaller squares into pieces and arranging them inside the big square in such a way that the space left is a square, too. 
Do you have any ideas?
 A: We can measure $a$ and $b$ in units of $c$.  You can use similar triangles to get segments of length $(\frac ac)^2$ and $(\frac bc)^2$, then can subtract to get $(\frac ac)^2-(\frac bc)^2-1$, then follow this question to get $\sqrt{(\frac ac)^2-(\frac bc)^2-1}$
A: Hint:
It is an application of the law of cosines. This figure is from wikipedia, and (changing $c$ and $a$ with respect to your question) shows that $c^2-a^2-b^2=-2ab\cos \gamma$

The two parallelograms of area $-2ab\cos \gamma$ can be transformed in a rectangle of sides $b$ and $d=|2a\cos \gamma|$, and this rectangle can be transformed in a square with a classical construction (that you can find here).
A: I have actually found out myself now and it is a lot easier than the answer given below.
Start with the square of the area $a^2$. Construct a semicircle towards its outside on one of its edges, say $AB$ and then construct a circle with radius $b$ and with centre $B$. Point $C$ should be the point of intersection of the circle and the semicircle. Then ABC is rectangular and Pythagoras tells us that $$|AC|^2=a^2-b^2.$$
Repeat this process now putting the square of area $a^2-b^2$ on the hypotenuse and the one of area $c$ on one of the legs. Then the square on the other leg has area $a^2-b^2-c^2$, as desired.
