The hexagon can cross itself, but must use only the six noncollinear vertices.
2007-07-06 06:56:02 · 4 answers · asked by Juli D 1 in Science & Mathematics ➔ Mathematics
Pick a vertex. How many vertices are left for you to choose from to attach this first side? 5. From this vertex, how many vertices are left to choose form to attach the next side? 4. Etc... all the way down to 1. This gives 5x4x3x2x1 = 120. However, since you are forming a cycle (starting at one vertex and ending at the same vertex), we must divide by an appropriate number. I'll leave it to you to figure out this number. But, that's basically it.
If you are having a hard time seeing this, try doing it with 4 vertices first.
2007-07-06 07:11:30 · answer #1 · answered by Doug 2 · 0⤊ 0⤋
Fix a starting vertex. The next vertex can be chosen in 5 ways, the one after that in 4, the next in 3 ways, then 2. Counting the starting vertex, that's 5 vertices so far. Now draw a line from the last one chosen to the start, closing the hexagon.
So there are 5 * 4 * 3 * 2 = 120 possible hexagons.
2007-07-06 12:54:22 · answer #2 · answered by jw 2 · 0⤊ 0⤋
Connect the six vertices of a regular hexagon to its center, creating 6 equilateral triangles. assuming the side of a hexagon is x cm, and because the triangles are equilateral, hence the height of the equilateral h = sqrt(x+x/4) = sqrt(x) ------- 2 hence the area of one triangle is x * sqrt(x)/4 therefore the area of the hexagon (6 equilateral triangles) is 6x*sqrt(x)/4 = 3x*sqrt(x) -------------- 2 so if the side(x) is 1 cm, area = 3*sqrt3/2
2016-05-19 23:06:42 · answer #3 · answered by Anonymous · 0⤊ 0⤋
how are the points arranged ???????????
2007-07-06 07:12:05 · answer #4 · answered by ag_iitkgp 7 · 0⤊ 0⤋