This is the great rhombicosidodecahedron.
It has 62 faces and 120 vertices.
How many edges does it have?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Did not remember that formula, but the visual helped to show that every vertex is a junction of 3 edges, each being shared by 2 faces, so 120*3/2=180 is the answer I am looking for.
Log in to reply
I reminded myself of Euler's formula by thinking of a cube and remembering there's a 2 involved somewhere, but I like your approach Saya.
In case anybody tried it, there are 12 decagons, 20 hexagons and 30 squares.
This makes the number of edges 2 1 2 × 1 0 + 2 0 × 6 + 3 0 × 4 = 1 8 0 faces, dividing by 2 because every edge is the meeting of two faces..
Interesting that 1 2 × 1 0 = 2 0 × 6 = 3 0 × 4 = 1 2 0 . Is there a reason for that?
Log in to reply
Interesting observation, Paul... This seems to be true for three other Archimedean solids as well:
Since Euler's Polyhedral Formula is F+V=E+2 where F=Faces, V=Vertices, and E=Edges, we have that 62+120=E+2.
This can simplify to: 62+120=E+2 182=E+2 E=182-2 E=180
Therefore, this rhombicosidodecahedron has 180 edges.
Problem Loading...
Note Loading...
Set Loading...
Define the following:
Using Euler's formula:
V − E + F = 2
Or
E = V + F − 2 = 1 2 0 + 6 2 − 2 = 1 8 0