Three methods are shown below, but all of them use this idea.

If you draw lines from the centre of the hexagon to each of its vertices, 6 identical isosceles triangles are formed (with the green angles equal).

The blue angle is 360$^\circ\div$6 = 60$^\circ$, which means that the green angles must also be 60$^\circ$, and so the hexagon is split into 6 equilateral triangles - which have side length 1, the same as the hexagon.

We have seen that we can split the hexagon into 6 equilateral triangles with side length 1. We can try to do the same to the triangle.

Joining the midpoints of the sides, as shown on the right, we can split the triangle into 4 equilateral triangles with side length 2.

Doing the same to each of the smaller triangles formed gives equilateral triangles with side length 1. There are 4$\times$4 = 16 of these in the large equilateral triangle.

So the hexagon contains 6 equilateral triangles of side length 1, whilst the triangle contains 16. So their areas are in the ratio 6:16, which simplifies to 3:8.

**Scaling the hexagon up**

6 copies of the triangle will fit together to make a hexagon with side length 4.

So 6 copies of the triangle will make a copy of the hexagon which has been enlarged by a scale factor of 4.

That means its area has been enlarged by a scale factor of 4$^2$ = 16.

So 6 copies of the triangle have the same area as 16 copies of the hexagon.

So the ratio of the area of the hexagon to the area of the triangle is 6:16, which simplifies to 3:8

So the hexagon contains 6 equilateral triangles of side length 1, whilst the triangle contains 16. So their areas are in the ratio 6:16, which simplifies to 3:8.

6 copies of the triangle will fit together to make a hexagon with side length 4.

So 6 copies of the triangle will make a copy of the hexagon which has been enlarged by a scale factor of 4.

That means its area has been enlarged by a scale factor of 4$^2$ = 16.

So 6 copies of the triangle have the same area as 16 copies of the hexagon.

So the ratio of the area of the hexagon to the area of the triangle is 6:16, which simplifies to 3:8

6 equilateral triangles of side length 1 fit into the hexagon. These triangles are similar to the equilateral triangle of side length 4, but their sides are 4 times shorter.

This means that their area is 4$^2$ = 16 times smaller.

So, when scaled down by a factor of 16, the area of the triangle fits 6 times into the area of the hexagon. So the area of the hexagon is $\frac6{16}=\frac38$ of the area of the triangle.

So the ratio of the area of the hexagon to the area of the triangle is $\frac38$:1, which simplifies to 3:8.

This means that their area is 4$^2$ = 16 times smaller.

So, when scaled down by a factor of 16, the area of the triangle fits 6 times into the area of the hexagon. So the area of the hexagon is $\frac6{16}=\frac38$ of the area of the triangle.

So the ratio of the area of the hexagon to the area of the triangle is $\frac38$:1, which simplifies to 3:8.

You can find more short problems, arranged by curriculum topic, in our short problems collection.