(a) Each of these triangles has three right angles, for an angle sum of 270
degrees.
(b) The total angle sum is 8
·
270 = 2160 degrees, which is 720 degrees more than the
total angle sum of eight ordinary plane triangles, 8
·
180 = 1440 degrees. The difference of
720 degrees is the total angle defect of a polyhedron, as expected.
6.
Repeat problem 5 for either the tetrahedron or the dodecahedron. (B+S 4.6.33 or 34)
Solution for the tetrahedron.
(a) Each triangle has three angles. Three of these angles
meet at the center of each face, so each angle is 360
/
3 = 120 degrees.
The angle sum of
each triangle is 360 degrees. (b) The sum of the angles of all four triangles is 4
·
360 = 1440
degrees, which is 720 degrees more than the 720degree total for four plane triangles.
Solution for the dodecahedron.
(a) Each triangle has three angles. Five of these angles
meet at the center of each face, so each angle is 360
/
5 = 72 degrees. The angle sum of each
triangle is 216 degrees. (b) The sum of the angles of all twenty triangles is 20
·
216 = 4320
degrees, which is 720 degrees more than the 3600degree total for twenty plane triangles.
3
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 Summer '09
 Lugo
 Math, Polyhedron, 120 degrees, Art Gallery Theorem, 72 degrees, Angle sum

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