### Building Tetrahedra

Can you make a tetrahedron whose faces all have the same perimeter?

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load.

# Square in a Circle in a Square

##### Age 14 to 16 ShortChallenge Level
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.
You can find more short problems, arranged by curriculum topic, in our short problems collection.