Square in a circle in a square
What is the ratio of the areas of the squares in the diagram?
In the diagram, the circle touches every edge of the larger square and every corner of the smaller square.
Image
![Square in a circle in a square Square in a circle in a square](/sites/default/files/styles/large/public/thumbnails/content-id-12521-square%252520in%252520a%252520circle%252520in%252520a%252520square%2525201.png?itok=-vbNdCyY)
What is the ratio of the area of the smaller square to the area of the larger square?
This problem is adapted from the World Mathematics Championships
Cutting up the squares to compare their areas
Rotating the smaller square so that its corners touch the sides of the larger square, and then removing the circle, gives the images shown below.
It is clear from the image with the red dotted lines on it that the smaller square occupies half of the area of the larger square.
So the ratio of the area of the smaller square to the area of the larger square is 1:2.
Using relationships between the side lengths of the squares and the diameter of the circle
In the diagrams below, two diameters labelled d are drawn onto the circle, and then the squares are each shown individually with one of the diameters.
The area of the larger square is d$^2$, and the area of the smaller square is s$^2$, so we want the ratio s$^2$:d$^2$.
We can use Pythagoras' theorem to find a relationship between s and d, because the smaller square can be split into two right-angled triangles.
d$^2$ = s$^2$ + s$^2$, so d$^2$ = 2s$^2$.
Which means that the ratio s$^2$:d$^2$ is 1:2.
Using relationships between the side lengths of the squares and the radius of the circle
In the diagram below, two radii of the circle are shown, both labelled $\text{r}$.
The larger square has side length $2\text{r}$, so its area is $(2\text{r})^2=4\text{r}^2$.
The diagram below is helpful for finding $x$, which is half of the side length of the smaller square.
By Pythagoras' theorem, $\text{r}^2=x^2 + x^2$, so $\text{r}^2 = 2x^2$, so $x^2$ = $\frac{1}{2}\text{r}^2$.
That means $x$=$\frac{1}{\sqrt{2}}\text{r}$, so the side length of the smaller square is $\frac{2}{\sqrt{2}}\text{r}$ (which is equal to $\sqrt{2}\text{r}$).
So the area of the smaller square is $\left(\frac{2}{\sqrt2}\text{r}\right)^2=\frac{4}{2}\text{r}^2=2\text{r}^2$.
So the ratio of the area of the smaller square to the area of the larger square is $2\text{r}^2$:$4\text{r}^2$, which simplifies to 1:2.
Rotating the smaller square so that its corners touch the sides of the larger square, and then removing the circle, gives the images shown below.
Image
![Square in a circle in a square Square in a circle in a square](/sites/default/files/styles/large/public/thumbnails/content-id-12521-square%252520in%252520a%252520circle%252520in%252520a%252520square%2525207.png?itok=SEaquojh)
It is clear from the image with the red dotted lines on it that the smaller square occupies half of the area of the larger square.
So the ratio of the area of the smaller square to the area of the larger square is 1:2.
Using relationships between the side lengths of the squares and the diameter of the circle
In the diagrams below, two diameters labelled d are drawn onto the circle, and then the squares are each shown individually with one of the diameters.
Image
![Square in a circle in a square Square in a circle in a square](/sites/default/files/styles/large/public/thumbnails/content-id-12521-square%252520in%252520a%252520circle%252520in%252520a%252520square%2525202.png?itok=-lQ-WJDE)
Image
![Square in a circle in a square Square in a circle in a square](/sites/default/files/styles/large/public/thumbnails/content-id-12521-square%252520in%252520a%252520circle%252520in%252520a%252520square%2525203a.png?itok=uA7_MHp7)
Image
![Square in a circle in a square Square in a circle in a square](/sites/default/files/styles/large/public/thumbnails/content-id-12521-square%252520in%252520a%252520circle%252520in%252520a%252520square%2525203b.png?itok=0BMaY5qz)
The area of the larger square is d$^2$, and the area of the smaller square is s$^2$, so we want the ratio s$^2$:d$^2$.
We can use Pythagoras' theorem to find a relationship between s and d, because the smaller square can be split into two right-angled triangles.
d$^2$ = s$^2$ + s$^2$, so d$^2$ = 2s$^2$.
Which means that the ratio s$^2$:d$^2$ is 1:2.
Using relationships between the side lengths of the squares and the radius of the circle
In the diagram below, two radii of the circle are shown, both labelled $\text{r}$.
Image
![Square in a circle in a square Square in a circle in a square](/sites/default/files/styles/large/public/thumbnails/content-id-12521-square%252520in%252520a%252520circle%252520in%252520a%252520square%2525208.png?itok=4IBIgjtr)
The larger square has side length $2\text{r}$, so its area is $(2\text{r})^2=4\text{r}^2$.
The diagram below is helpful for finding $x$, which is half of the side length of the smaller square.
Image
![Square in a circle in a square Square in a circle in a square](/sites/default/files/styles/large/public/thumbnails/content-id-12521-square%252520in%252520a%252520circle%252520in%252520a%252520square%2525209.png?itok=ciSHVS61)
By Pythagoras' theorem, $\text{r}^2=x^2 + x^2$, so $\text{r}^2 = 2x^2$, so $x^2$ = $\frac{1}{2}\text{r}^2$.
That means $x$=$\frac{1}{\sqrt{2}}\text{r}$, so the side length of the smaller square is $\frac{2}{\sqrt{2}}\text{r}$ (which is equal to $\sqrt{2}\text{r}$).
So the area of the smaller square is $\left(\frac{2}{\sqrt2}\text{r}\right)^2=\frac{4}{2}\text{r}^2=2\text{r}^2$.
So the ratio of the area of the smaller square to the area of the larger square is $2\text{r}^2$:$4\text{r}^2$, which simplifies to 1:2.