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Geometry Puzzle: Find the Area of the Square

Two concentric circles, a small one and a big one. A square is inscribed into the big circle and the small circle is inscribed into the square.

An annulus has area 4π4\pi.

What is the area of the square?

Solution

Let rr and RR be the radii of the small and big circles, respectively.

Two concentric circles, a small one and a big one. A square is inscribed into the big circle and the small circle is inscribed into the square. The radius of the small circle, from the center/centre to the right side of the square, is labeled/labelled r and the radius of the big circle, from the center/centre and the right top vertex of the square, is labeled/labelled R.

As the area of the annulus is 4π4\pi, we get

π(R2−r2)=4π\pi\left(R^2-r^2\right)=4\pi
R2−r2=4R^2-r^2=4
r2=R2−4(1)r^2=R^2-4\;\;\;\left(1\right)

Now, consider the isosceles right-angled triangle inside the square. It has sides with lengths rr, rr and RR.

Using the Pythagorean/Pythagoras’ theorem, we have:

R2=r2+r2R^2=r^2+r^2
R2=2r2(2)R^2=2r^2\;\;\;\left(2\right)

Substituting the value of (2) in (1) we get:

r2=2r2−4r^2=2r^2-4
r2=4r^2=4
r=2r=2

Then, as the side of the square has length 2r2r,

2r=42r=4
(2r)2=16\left(2r\right)^2=16

So, the area of the square is 16.

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