Rotating a square by 30 degrees what is the area of the common region

Question

A square ABCD, with sides of 10cm, is rotated by 30 degrees keeping its center fixed to result into another square, PQRS. What is the area of the region common to the two squares? I’ve seen this question with 45 degrees but I wanted to know what the answer would be with 30 degrees

Answer 1

A picture is worth a thousand words so I'll skip a lot of details.

By looking at the side $AB=a$ you get:

$$y+x+y\sqrt{3}=a$$

$$x+y(1+\sqrt{3})=a\tag{1}$$

By looking at the side $A'D'$ you get:

$$\frac x2+2y+x\frac{\sqrt3}{2}=a$$

$$x\frac{1+\sqrt3}{2}+2y=a\tag{2}$$

By solving (1) and (2) for $x,y$ you get:

$$x=\frac a3(3-\sqrt3)$$

$$y=\frac a6(3-\sqrt3)$$

So the common area must be:

$$A=a^2-4\cdot \frac12 y \cdot y\sqrt3=\frac{2a^2}{3}(3-\sqrt3)\approx0.845\ a^2$$

Answer 2

Alternatively and faster: you can preliminary show that the triangles carved in and out by the rotation are all congruent. In fact

• $\triangle OAA'$ is isosceles, implying (by angle chasing) that also $\triangle AQA'$ is isosceles.
• Thus $AQ \cong A'Q$, which gives $\triangle APQ \cong\triangle A'QR$ (ASA criterion).
• Proceed similarly on $\triangle OA'B$ (which is equilateral), to demonstrate that $\triangle A'QR\cong\triangle BRS$.

Now, if $\overline{AB} = a$, and $\ell$ is the hypothenuse of, say, $\triangle APQ$, we have $$ \ell + \frac{\ell}2 + \frac{\sqrt 3\ell}2 = a $$ $$ \ell= \frac2{3+ \sqrt 3}a. $$