Although the mathematics behind it comes from a very interesting field called "tessellation or tiling" but I can provide some intuition from my little understading.
1. When we are given a task of tiling then if we can somehow arrange that shape around a point such that the shapes cummulatively complete $360^{\circ}$ with their internal angles then that shape can fill the plane completely without leaving any empty space.
2. Let's understand this with the example of Rhombus:
A rhombus has the following properties:
- Opposite angles are always equal.
- $\alpha + \beta = 180^{\circ}$ where, $\alpha$ and $\beta$ are adjacent angles of rhombus.
So, if we form a group of rhombuses around a point such that there is always an angle $\alpha$ if there is an angle $\beta$ in the group and vice versa then they will cummulatively make $180^{\circ}$ and after $2$ such pairs we will have $360^{\circ}$ and that means there will be no empty space left on the plane. (fig. 1 and 2 in img 1 above)
But there is a catch, if we have a rhombus which has angles $\alpha$ and $\beta$ as factors of $360^{\circ}$ then we can take only one type of angle and achive the same results. (fig. 3 and 4 in img 1 above)
for example let's assume $\alpha = 60^{\circ}$ and $\beta = 120^{\circ}$ then $6 \times 60^{\circ} = 360^{\circ}$ and $3 \times 120^{\circ} = 360^{\circ}$.
3. Now, let's see why other shapes can't make tiling patterns:
a) Circles's story is crystal clear from the fig. 1 in img 2 above itself. There is no way we can remove the somewhat triangular looking space in center whenever $3$ circles meet around a point.
b) Regular pentagon has internal angle $= 108^{\circ}$ and if we try to find number of pentagons that we need around a point so that no empty space is left:
$n \times 108^{\circ} = 360^{\circ}$
$n = \frac{360^{\circ}}{108^{\circ}}$
$n \approx 3.33$
So, if we take $n = 3$ means only $3$ pentagons around a point (like fig. 2 in img 2 above) then $3 \times 108^{\circ} = 324^{\circ}$, which means there will be empty space left and if we take $n = 4$ means $4$ pentagons around a point (like fig. 3 in img 2 above) then $4 \times 108^{\circ} = 432^{\circ}$, which means there will be overlap b/w them.
c) Regular octagon has internal angle $= 135^{\circ}$ and if we try to find number of pentagons that we need around a point so that no empty space is left:
$n \times 135^{\circ} = 360^{\circ}$
$n = \frac{360^{\circ}}{135^{\circ}}$
$n \approx 2.66$
So, if we take $n = 2$ means only $2$ octagons around a point (like fig. 4 in img 2 above) then $2 \times 135^{\circ} = 270^{\circ}$, which means there will be empty space left and if we take $n = 3$ means $3$ octagons around a point (like fig. 5 in img 2 above) then $3 \times 135^{\circ} = 405^{\circ}$, which means there will be overlap b/w them.
