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Geometry Puzzle: Find the Area of a Square Inside a Circle

Two squares, one red and the other yellow, meet at the center/centre of a circle. The red square is smaller than the yellow square. The yellow square is labeled/labelled twenty.

Two squares meet at the center/centre of a circle. The yellow square has area 20.

What is the area of the red square?

This puzzle was published on The Guardian.

Solution

Let rr be the radius of the circle.

As the area of the yellow square is 20, its side length is

20\sqrt{20}
252\sqrt{5}
Two squares, one red and the other yellow, meet at the center/centre of a circle. The red square is smaller than the yellow square. The yellow square is labeled/labelled twenty. The radius of the circle, that is the diagonal of the red square and the line segment that connects the center/centre of the circle and one of the vertex that is tangent to the circle of the yellow square, is labeled/labelled r. The side length of the yellow square is labeled/labelled two square root of five and half of the side length is labeled/labelled square root of five.

Using the Pythagorean/Pythagoras’ theorem in this square, we get:

r2=(5)2+(25)2r^2=\left(\sqrt{5}\right)^2+\left(2\sqrt{5}\right)^2
r2=5+20r^2=5+20
r2=25r^2=25
r=5r=5

The red square has its diagonal length 5.

Let xx be its side length.

Using the Pythagorean/Pythagoras’ theorem in this square, we get:

52=x2+x25^2=x^2+x^2
25=2x225=2x^2
x2=12.5x^2=12.5

So, the area of the red square is 12.5.

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