three diedrals contained by and 9, 10 and 11, 11 and 12 are those which belong to such a polyedral. Z.

The vertex C will fall at h, and a like polyedral will be formed at that pt., and so of all the others.

.. the polyedrals are all equal, and the solid is a regular icosaedron.

Q. E. F.

654. SCH. Models of the regular polyedrons may be easily constructed as follows:

Draw the following diagrams on cardboard, and cut them out. Then cut half way through the board in the dividing lines, and bring the edges together so as to form the respective polyedrons.

TETRAEDRON

HEXAEDRON

OCTAEDRON

DODECAEDRON

ICOSAEDRON

GENERAL PROPERTIES OF POLYEDRONS.

EULER'S THEOREM.†

Proposition. 27. Theorem.*

655. In any polyedron the number of edges increased by 2 is equal to the whole number of vertices and faces.‡

Hyp. Let E, F, and V denote the number of edges, faces, and vertices respectively of the polyedron S-ABCD. E+2= V + F.

To prove

Proof. Beginning with one face ABCD, we have E = V.

If we annex to this a second face SAB, by applying one of its edges, as

S

B

AB, to the corresponding edge of the first face, we form a surface having one edge AB, and two vertices A and B common to both faces; therefore, the whole number of edges is now one more than the whole number of vertices.

... for 2 faces ABCD and SAB, E = V + 1.

If we annex a third face SBC, adjacent to both ABCD and SAB, we form a new surface having two edges SB and BC, and three vertices S, B, C in common with the preceding surface; therefore the increase in the number of edges is again one more than the increase in the number of vertices.

.. for 3 faces, E = V + 2.

In like manner, for 4 faces, E V + 3.

And so on until every face but one has been annexed. .. for (F1) faces, E = V + F — 2.

But annexing the last face adds no edges nor vertices. .. for F faces, EV+F-2,

ΟΙ

E+2V+F.

+ The proof of this theorem was first published by Euler (1752).
This proof is due to Cauchy.

Q.E.D.

EXERCISES,

1. How many edges has the regular tetraedron ? 2. How many edges has the regular hexaedron ?

3. The frustum of a regular four-sided pyramid is 8 feet high, and the sides of its bases are 3 feet and 5 feet: find the height of an equivalent regular pyramid whose base is 10 feet square.

Proposition 28. Theorem.*

656. The sum of the face-angles of any polyedron is equal to four right angles taken as many times as the poly

edral has vertices less two.

Hyp. Let E, F, and N denote the number of edges, faces, and vertices respectively, and S the sum of the faces of any polyedron.

To prove

S = 4rt. s (V − 2).

Proof. Since each edge is common to two faces,

... the whole number of sides of the faces considered as independent polygons is 2E.

If we form an exterior at each vertex of every polygon, the sum of the interior and exterior s at each vertex is 2 rt. s.

And since the whole number of vertices in the faces is 2E,

.. the sum of all the interior and exteriors of the faces is 2 rt. s × 2E, or 4 rt. ≤s × E.

Since the sum of the exteriors of each face is 4 rt. LS, (151) .. the sum of the ext. Zs of the F faces is 4 rt. Zs X F. ... S4 rt. Zs X F /s

But

4 rt. s X E.

EXERCISES.

THEOREMS.

1. Show that a lateral edge of a right prism is equal to the altitude.

2. Show that the lateral faces of right prisms are rectangles.

3. Show that every section of a prism made by a plane parallel to the lateral edges is a parallelogram.

4. The lateral areas of right prisms of equal altitudes are as the perimeters of their bases.

5. The opposite faces of a parallelopiped are equal and parallel.

6. If the four diagonals of a quadrangular prism pass through a common point, the prism is a parallelopiped. 7. If any two non-parallel diagonal planes of a prism are perpendicular to the base, the prism is a right prism.

See (546).

8. Any straight line drawn through the centre of a parallelopiped, terminating in a pair of faces, is bisected at that point.

9. The volume of a triangular prism is equal to the product of the area of a lateral face by one-half the perpendicular distance of that face from the opposite edge.

10. In a cube the square of a diagonal is three times the square of an edge.

11. In any parallelopiped, the sum of the squares of the twelve edges is equal to the sum of the squares of its four diagonals.

12. In any quadrangular prism, the sum of the squares of the twelve edges is equal to the sum of the squares of its four diagonals plus eight times the square of the line joining the common middle points of the diagonals taken two and two.

13. A plane passing through a triangular pyramid,

parallel to one side of the base and to the opposite lateral edge, intersects its faces in a parallelogram.

14. The lateral surface of a pyramid is greater than the base.

Project the vertex on the base, etc.

15. The four middle points of two pairs of opposite edges of a tetraedron are in one plane, and at the vertices of a parallelogram.

(617).

16. The three lines joining the middle points of the three pairs of opposite edges of a tetraedron intersect in a point which bisects them all.

17. In a tetraedron, the planes passed through the three lateral edges and the middle points of the edges of the base intersect in a straight line.

The intersections of the planes with the base are medials of the base; .* . etc.

18. The lines joining each vertex of a tetraedron with the point of intersection of the medial lines of the opposite face all meet in a point, which divides each line in the ratio 1 4.

NOTE. This point is the centre of gravity of the tetraedron.

19. The plane which bisects a diedral angle of a tetraedron divides the opposite edge into segments which are proportional to the areas of the adjacent faces.

See (303), (369).

20. The volume of a truncated triangular prism is equal to the product of the area of its lower base by the perpendicular upon the lower base let fall from the intersection of the medial lines of the upper base.

21. The volume of a truncated parallelopiped is equal to the product of the area of its lower base by the perpendicular from the centre of the upper base upon the lower base.

22. The volume of a truncated parallelopiped is equal to the product of a right section by one-fourth the sum of its four lateral edges.

See (640).

23. Any plane passed through the centre of a parallelopiped divides it into two equivalent solids (640).