The figures presented in parts (a) and (b) are attracted in GeoGebra and also those displayed in (c) and also (d) are attracted on isometric graph paper.

You are watching: How to divide a hexagon into 4 equal parts

any kind of line with opposite vertices divides the hexagon into two congruent pieces:

*

each one is a trapezoid and a reflection across the heat takes one trapezoid come the other.

There are plenty of ways to construct exciting partitions.Any line v the facility of the hexagon (which is at the intersection of any kind of two diagonals) divides the hexagon right into two congruent pieces.

*

A 180o rotation around the facility shows this congruence because the line will certainly go come itself and each vertex will be mapped come the peak opposite it. Since line segments space taken to heat segments, every the vertices and also sides the the two halves will certainly "match up" under this rotation, showing they room congruent.

Or begin with a line with the center, and "cut out" any kind of piece of one of the halves and "add it ago in" by rotating that 180o about the center of the hexagon. Below are three examples:

*
*
*

In every case, the figures were constructed so that a 180o rotation about the center will "line castle up, one on height of the other."

Here is one way:

*

We recognize they room congruent due to the fact that if we think about the 2 lines displayed below:

*

The yellow an ar is the image of the red an ar when us reflect throughout $ell_1$, for this reason those two areas are congruent.The green an ar is the image of the blue region when we reflect across $ell_1$, therefore those two areas are congruent.The yellow region is the image of the green region when us reflect across $ell_2$, therefore those two regions are congruent.So all 4 regions are congruent.

Here is one way:

*

We can present they are congruent making use of reflections around the 2 lines displayed below:

*

All 4 regions top top the left next of $ell_1$ space congruent to their mirror photos on the right. Us can present the purple and orange areas are congruent by reflecting throughout $ell_2$. We can get the rest of the pair-wise congruencies by rotating a lot of of 60 degrees around point $P$. So every eight areas are congruent.

Solution:Solutions come the bonus questions

Bonus 1: here is one way:

*

Bonus 2: right here is one way:

*

Note the the regions don"t need to be bounded by simple closed curves.

See more: 2002 Honda Civic Vacuum Hose Diagram S? Vacuum Hose Diagrams

Bonus 3: here is one way:

*

Note the the regions don"t should be bounded by basic closed curves.


*

Typeset may 4, 2016 at 18:58:52. Licensed by Illustrative mathematics under a an imaginative Commons Attribution-NonCommercial-ShareAlike 4.0 worldwide License.