Geometry Forum - Problem of the Week
Solutions - Tiling a Plane with a Hexagon, Nov. 21- 25, 1994
Annie says:Some excellent answers this week, but the first thing that got me excited was Ruth Carver's message accompanying her students' sketches. Make sure you read it.
Megan (who's doing a great job, and she's only in seventh grade!) and Sean submitted very nice answers - they state the questions, their answers, and their explanations. Very clear, very well organized, and no room for doubt. Kenneth Chan sent some good pictures as well. Not every one chose to tackle the tiling part of the problem, but Megan gave the clearest explanation of why it won't work.
Mount St. Joseph Academy
My students had a lot of fun constructing this problem on Sketchpad and learned a lot about constructions while doing it. Because of the short week, not all my students had a chance to complete it and one class worked in pairs but compiled their findings as a group. By the way, we had parent's night this past Tuesday and many parents remarked on how they are getting hooked on the Problem of the Week and look forward to it each week. Isn't that great--whole families talkin' math:-) Happy Thanksgiving! --Ruth CarverAnnie Chan
1) The hexagon is not equilateral since the longest side of the isoceles triangles and the (1 unit long) side of the square are not the same.2) The hexagon is equiangular because each of the angle in the hexagon consists of a right angle from the square and a 30ƒ angle from the isoceles triangle.
3) We can find the area of the hexagon. We use the area of triangle formula to find the area of the equilateral triangle which is root 3 divided by 4 unit¾. Then each of the square is 1 unit¾. For the remainder of it, which are 3 isosceles triangles, we use the area of triangle formula to find the third side which is root 3 divided by 4 units¾. The area of the whole hexagon is 3+root3 units¾.
Megan Guichard
1) Is the hexagon equilateral? No; assuming that the sides of the triangle have a length of one unit, Pythagorean's Theorem can be used to prove that three of the sides have lengths equal to the square root of three units, and the remaining three sides have lengths equal to one unit. Therefore, the hexagon is not equilateral.2) Is the hexagon equiangular? Yes; I drew a picture like the one described, and it is easy to see that each angle of the hexagon is 90 + 30 = 120 degrees.
3) Can we find the area of the hexagon if the sides of the triangle are each one unit in length? Yes; by finding the area of all the little triangles and squares that make up the hexagon, it is easy to see that the area of the hexagon is 3 times 1^2 + 8 times sqrt(3)/8 = 3 + sqrt(3).
Will the figure tesselate on a plane? No, it will not; if you put two sides of equal length together, as you must to prevent making a hole with a 60 degrees angle, then the third hexagon must have two sides of equal length adjacent to each other, which the hexagon doesn't have.
Sean Nichols
First of all, the hexagon given is not equilateral. The sides formed by the sides of the squares is equal to 1 (the same as the sides of the equilateral triangle in the center, which are given to be 1). But then, the other sides, formed by joining together the outside vertices of the squares, have a length of (sqrt 3). This can be found by dividing the triangle in half, along the altituide to this outside edge. Doing this produces two congruent "30-60-90" triangles. Using the formula for the lengths of the sides of a "30-60-90" triangle (sides are 1-2-(sqrt 3)), we can determine that the length of the outside edge of one of these triangles is (sqrt 3)/2, and therefore the total length of this edge is (sqrt 3).Next, we can determine that all the angles of this hexagon are congruent, at 120 degrees (90 degrees for the square + 30 degrees for the triangle between the squares). Therefore, this hexagon _is_ equiangular.
Finally, we can find the area of the hexagon. It is:
Area of center triangle
+ 3 * Area of square
+ 3 * Area of outside triangle.
Given the length of one of the sides of the center triangle to be 1, the
area of the center triangle is (sin*60)/2, the area of one square is 1,
and the area of one outside triangles is (sqrt 3)/4. So we have:
(sin*60)/2
+ 3
+ 3 * (sqrt 3)/4
=~ 4.732050808...
From this information, we can determine that this hexagon cannot
be perfectly tessellated, as although the angles are congruent,
and can be perfectly divided into 360, the sides are not either
of a congruent length, or even of lengths such that the ratio of
their sides is a rational number ((sqrt 3) is not rational).
Kenneth Chan
(a) D______________________E
/ \
/ \
/ A \
/ /\ \
/ / \ \
/ / \ \
I/ / \ \F
\ / \ /
\ / \ /
\ B/____________\C /
\ | | /
\ | | /
\ | | /
\ | | /
\ | | /
\ | | /
\ | | /
\|____________|/
H G
AB=BC=CA=DI
Consider triangle ABC and ADE,
AB=AD (given)
CA=EA (given)
But angle BAC is not equal to angle DAE.
Triangle ABC is not congruent to triangle ADE.
DE is not equal to BC and also DI.
Therefore, the hexagon is not equilateral.
(b) D______________________E
/ 30ƒ 30ƒ \
/90ƒ 90ƒ\
/ A \
/ /\ \
/ / \ \
/ / \ \
I/90ƒ / \ 90ƒ\F
\ / \ /
\ / \ /
\30ƒ B/____________\C 30ƒ/
\ | | /
\ | | /
\ | | /
\ | | /
\30ƒ| |30ƒ/
\ | | /
\ |90ƒ 90ƒ| /
\|____________|/
H G
From the figure above, the hexagon is equianglular.
(c) Area of hexagon
= area of equilateral triangle + area of 3 squares + area of 3
isosceles triangle
= ‡(1 unit)¾sin 60degrees + 3(1 unit)¾ + 3*‡(1 unit)¾sin 120degrees
= (3+root 3)unit¾
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