Why can't a regular pentagon (five equal sides) tessellate?

Question

Well this was a question in a maths test a sat but i'm not sure if I got it right. The question was "Why can't a regular pentagon tessellate but a regular hexagon can" or something along those lines. Then there was a picture of hexagons tessellating and one of their interior angles marked 120 degrees if this helps. Thanks heaps.

Answer 1

hexagons work because each angle is 120 degress, as you say, and 3 times 120 equals 360 degrees. So three hexagons will surround a point with no 'space" left over.

but the interior angle of a pentagon is108 degrees. three pentagons together only fill up 3 time 108, or 324 degrees. There is space left over. But four pentagons would overlap.

so 3 is not enough and 4 is too many. Pentagons cannot surround a point the way hexagons do.

Answer 2

Pentagon Tessellation

Answer 3

This Site Might Help You.

RE:
Why can't a regular pentagon (five equal sides) tessellate?
Well this was a question in a maths test a sat but i'm not sure if I got it right. The question was "Why can't a regular pentagon tessellate but a regular hexagon can" or something along those lines. Then there was a picture of hexagons tessellating and one of their interior...

Answer 4

I agree with the answers of others above concerning 2D tessellation, but it is interesting to note that regular pentagons CAN "tessellate" in three dimensions. Since three pentagons only use 324 degrees of "space" in the 2D plane, they must be draw toward each other in the third dimension to make the edges meet. It is possible to continue this tessellation using 12 regular pentagons, which forms a Regular Dodecahedron. Sometimes referred to as a D12 by gamers, these shapes are often used for dice where 12 (not 6) options are needed. The other regular polyhedrons are the tetrahedron (4 equilateral triangles), cube (6 squares), octahedron (8 equilateral triangles) and icosahedron (20 equilateral triangles). All of these can be viewed as 3D tessellations of their face polygons.

Answer 5

A tessellation or tiling of the plane is a collection of plane figures that fills the plane with no overlaps and no gaps.

The angles at the vertices which will form each repeat ot the tessellation must add up to 360 degrees where they touch at the center. This only works for 6 equilateral triangles, 4 squares or 3 regular hexagons. All the other polygons have vertex angles that are not factors of 360.

Answer 6

I think it would be something to do with the angle at which the sides of the figure meet.
If the inside angles of a regular hexagon are 120 degrees. then 3 x 120 degrees = 360 degrees which is a complete circle. Therefore they fit together with no gaps.

The angles in a regular pentagon don't work out so that they fill a complete circle, and so they wouldn't lie edge to edge and fill up all the gaps.

Does this help?
I'm not a great mathematician, but that's my best guess.

Answer 7

To form a tessellate, the measure of interior angle must be a factor of 360 degrees so that spaces in all directions can be filled completely.

For regular hexagons, each interior angle is 120 degrees, which is the factor of 360 degrees.

For regular pentagons, each interior angle is 108 degrees, which is not the factor of 360 degrees.

Answer 8

What Is A Regular Pentagon

Answer 9

To tessellate means to fall into mosaic. If you take your hexagon and run a line between opposing angles the resultant will be six equal shapes. It is impossible to obtain equal shapes with a pentagon.

Answer 10

For the best answers, search on this site https://shorturl.im/QNmjf

To make it easier, just add 2 letters on your pic. Call E the vertex vertically top of B and F the projection of E on the line AO. You just want area (REF) = area (APR). Now given that those 2 triangles have a common angle say, theta, at the point R, the areas of (REF) and (APR) are actually equal to (1/2)*RE*RF*(sin theta) and (1/2)*RA*RP*(sin theta). Hence the equality of areas is equivalent to RE*RF = RP*RA. Now this is equivalent to RF/RP = RA/RE, which is equal to (RF - RA) / (RP - RE) or FA / PE. So FA / RF = PE / RP, which by Thales is BE / RA. Finally FA*RA = RF*BE which allows to compute FA. Setting OB = RE = 1 and RA = FA + FR = x + sin(18) we have FA*RA = (x+sin 18)*x = FR*BE = 1/2, by standard pentagone stuff. So x^2 + (sin 18)*x -1/2 = 0 and x = (-sin(18) + sqrt(sin^2(18)+2)) / 2. Finally your total area is (cos 18)*(phi + x), where phi = (1+sqrt(5))/2 = EB. This gives Area = 2.0802612... To draw the point P, just take a point G same height as R so that RG=sqrt(2)/2. Take H the middle of RF and draw the circle centered at H of radius HG. This circle intersects (RO) at A.