From this diagram, we can see that:

- the length of the yellow edge multiplied by the length of the blue edge must be 12
- the length of the red edge multiplied by the length of the blue edge must be 18
- the length of the yellow edge multiplied by the length of the red edge must be 24.

If the yellow edge is 6 cm long and the blue edge is 2 cm long, then the red edge must be 9 cm long, because 2$\times$9 = 18. However, then the orange face would have area 6$\times$9 = 54 square centimetres, which is not right.

If the yellow edge is 2 cm long and the blue edge is 6 cm long, then the red edge must be 3 cm long, because 6$\times$3 = 18. However, then the orange face would have area 2$\times$3 = 6 square centimetres, which is not right.

If the yellow edge is 4 cm long and the blue edge is 3 cm long, then the red edge must be 6 cm long, because 3$\times$6 = 18. Then the orange face would have area 4$\times$6 = 24 square centimetres, which is right.

So the edges of the cuboid are 3 cm, 4 cm and 6 cm long, which means the volume of the cuboid is 3$\times$4$\times$6 = 72 cm$^3$.

If we call the lengths of the edges $a$, $b$ and $c$, then $ab=12$, $ac=18$ and $bc=24$.

From here, we can use a trial approach to work out $a$, $b$ and $c$ similar to the method described above, or we can solve these equations, as shown below.

Since $ab=12$, we can say that $a=\dfrac{12}{b}$.

Substituting into $ac=18$, we get $\dfrac{12}{b}\times c=18$. Rearranging, $12c=18b$, so $c=\dfrac{18b}{12}=\dfrac{3b}{2}$.

Substituting into $bc=24$ now gives $b\times\dfrac{3b}{2}=24$, so $\dfrac{3b^{2}}{2}=24$, so $b^{2}=16$, so $b=4$.

So $a=\dfrac{12}{4}=3$ and $c=\dfrac{3\times4}{2}=6$.

So the volume, given by $abc$, is equal to $3\times4\times6=72$ cm$^3$.

As in the picture above, we can call the lengths of the edges $a$, $b$ and $c$. So $ab=12$, $ac=18$ and $bc=24$.

We want the volume of the cuboid, $V=abc$.

Notice that $ab\times ac\times bc=a^2b^2c^2=(abc)^2=V^2$.

So $12\times18\times24=V^2$.

So $V=\sqrt{12\times18\times24}=72$.

So the volume is $72$ cm$^3$.

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