D a If three re gular pentagons intersect at a vertex of a t

D. (a) If three re gular pentagons intersect at a vertex of a tessellation, the sum of the measures of the interior angles would be . If four regular pentagons intersected in a vertex of a tessellation, the sum of the measures of the interior angles about the vertex would be If five regular pentagons intersected in a vertex of a tessellation, the sum of the measures of the interior angles would be Can the plane be tessellated by regular pentagons? Explain! (b) Explain carefully why regular heptagons cannot tessellate the plane. (e) Explain carefully why regular octagons cannot tessellate the plane. (d) Explain carefully why a regular n-gon will not tessellate the plane if n > 6.

Solution

a) 3*108=324

4*108=432

5*108=540

A regular pentagon does not tessellate.

In order for a regular polygon to tessellate vertex-to-vertex, the interior angle of your polygon must divide 360 degrees evenly. Since 108 does not divide 360 evenly, the regular pentagon does not tessellate this way.

Trying to place one of the vertices on an edge somewhere instead of on the vertex does not work for similar reasons, the angles don’t match up.

b),c),d)

No, A regular n-gon (n>6) has angles that measure (n-2)(180)/n, in this case (5)(180)/7 = 900/7 = 128.57.and 135 A polygon will tessellate if the angles are a divisor of 360. 128.57and 135 is not a divisor of 360.

The only regular polygons that tessellate are Equilateral triangles, each angle 60 degrees, as 60 is a divisor of 360. A regular quadrilateral (square), each angle is 90 degrees, as 90 is a divisor of 360. And a regular Hexagon (6 sides), each angle is 120 degrees, and 120 is a divisor of 360.

However, the n-gon can tile the hyperbolic plane; so you could decorate a ball with tiled n-gon when>6