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or it is the greater acute angle in a right angled triangle, 3 one side of which is one third of the hypoténuse. (149) Cor. 2.-The volume of a regular tetraedron may now be determined numerically. Let the edge be the unit. The radius

1 O A of the circle which circumscribes the base will be

This

✓3 line, the edge AP, (which is = 1,) and the perpendicular OP, form a right angled triangle. Hence the perpendicular is 1 2

N3
1

The area of the base is
3
3

4 third of the product of this and the perpendicular is then V3 2 1

1 4 * X 3

which expresses the proportion 3

6.V2' of the tetraedron to a cube constructed with an equal edge.

V2

; one

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A

B

PROPOSITION XXIX. (150) To construct a regular solid with triangular

faces, and whose solid angles are formed by

four plane angles. Construct a square A B C D, and through its centre O draw a perpendicular, producing it on both sides of the plane of the square. From the points ABCD inflect on this perpendicular, at both sides of the planes, lines equal to the side of the square. It is evident that those which are on the same side of the plane will meet the perpendicular at the same point. Let these points be P and P'. Two pyramids will thus be constructed on opposite sides of the square, and the lateral faces of each of these pyramids will be the equilateral triangles constructed on the sides of the square. These two pyramids united form a regular solid with eight triangular sides, and of which the square is a diagonal plane.

This solid is therefore the regular octaedron. (151) Cor. 1.-To determine the inclination of the planes of each pair of intersecting faces, let them be considered as two sides of a solid angle of the pyramid whose base is the square. This solid angle is then formed by three plane angles, two of which are 60°, and the third a right angle. In the formula used in (148), let a = 90°, b = c = 60°, and it becomes

1 cos A.

= 0,...cos AS

3

1

Hence A is an obtuse angle whose cosine is or it is the

3' external angle to the greater obtuse angle in a right angled triangle, one side of which is a third of the hypotenuse. (152) Cor. 2.-Hence it appears that the angle under the adjacent planes of the regular octaedron, is the supplement of the angle under the adjacent planes of the regular tetraedron. (153) Cor. 3.-If three faces of the octraedron whose bases form the sides of the same face, (such as ADP, BCP, A P'B.) be produced through those sides until they form a solid angle, the produced parts will form a regular tetraedron with the face through whose sides they are produced. (154) Cor. 4.-Each pair of faces of the octaedron (A PB, DPC,) which are constructed on opposite sides, A B, DC of the square, and also on opposite sides of its plane are parallel. For the alternate angles which their planes form with that of the square are equal. (155) Cor. 5.—If the planes of three faces which are terminated in the sides of any one face A B P be produced until they form a solid angle, and also until they meet the plane of the face DCP which is parallel to A B P produced, they will with it form a regular tetraedron circumscribing the octaedron. Each face of this tetraedron will be divided into four equal equilateral triangles, by the face of the octaedron by whose production it is formed.

Hence it follows that the whole surface of the tetraedron is equal to sixteen times one of the faces of the octaedron, and these to double the whole surface of the octaedron.

It appears, therefore, that if the four corners be cut from a regular tetraedron by planes through the points of bisection of every three conterminous edges, the remaining figure will be a regular octaedron. Since each pyramid which is thus cut off is similar to the whole, and the edges are in the proportion of 1:2, the volumes are as 1 : 8. Therefore each of the four pyramids is equal to an eighth of the original pyramid, and to a fourth of the octaedron which remains after their removal. (156) Hence it appears, that the volume of a regular octaedron whose edge is 1, is half the volume of a regular tetraedron whose edge is 2. But by (149) the volume of a tetraedron whose edge is l is 1

And since a similar solid whose edge is 2 has eight times the volume (137), it follows that the volume of a tetraedron

8 2 N 2 whose edge is 2 is

Hence the regular octae6 12 3

6.72

or

N 2 dron whose edge is 1 is The volumes of an octaedron and

3 cube constructed on the same edge are therefore as v2: 3.

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PROPOSITION XXX.

P

(157) To construct a regular solid with triangular

faces and whose solid angles are formed by

five plane angles. Let a regular pentagon A B C D E be constructed, and through its centre let a perpendicular to its plane be drawn. From the points A, B, C, D, E, respectively, let right lines equal to the side of the pentagon be inflected on this perpendicular. Since the side of a regular pentagon is greater than the radius of its circumscribing circle, these lines will meet the perpendicular below the plane of the pentagon; and since the lines so inflected are equal, they will meet the perpendicular at the same point P so as to form a regular pentagonal pyramid. The solid angle P at the vertex of this pyramid will be then formed of five plane angles, each of which is two thirds of a right angle. Two of the plane angles which form each solid angle at the base of the pyramid, have evidently the same inclination as any two of the plane angles which form the solid angle P, being, in fact, the same planes. Hence the solid angles A, B, C, &c. at the base may be considered parts of solid angles equal to P formed by five plane angles, the part included by three of the plane angles being cut off by the plane angle of the base of the pyramid.

On each side of the base of the pyramid let an equilateral triangle be constructed, so that its plane shall be inclined to the adjacent lateral face of the pyramid at the same angle as any two of the adjacent lateral faces; that is, so that the angle under the planes A B C' and AB P shall be equal to the angle under any two adjacent planes containing the angle P, and so that the same may be true of the planes BCD and BCP, CDE' and CDP, &c.

Hence it follows, that at each of the vertices A, B, C, &c. of the base of the pyramid there are four angles each two thirds of a right angle, and which are united at the same inclinations as four of the angles which form the solid angle P. It follows, therefore, that the angle CBD included between the conterminous sides (B C', B'D') of two equilateral triangles A B C', CBD', constructed upon conterminous sides of the pentagonal

base must be an angle of an equilateral triangle, so placed that if its plane be supposed to be drawn it will complete the solid angle B, and render it equal to P. The same conclusion is obviously applicable to each of the other angular points of the base.

We have thus a figure formed having a solid angle at P formed of five angles of equilateral triangles, having ten equilateral triangular faces, and a serrated edge or boundary A CBD'CE', &c., the planes of the angles being so disposed that if the gaps C'B D', D'CE', &c. be filled up, solid angles will be formed at A, B, C, &c, equal to P.

Let another figure in every respect equal and similar to this be formed, the corresponding points being marked by the small letters a, b, c,....a', b, c, &c. Let the point d be placed upon B, and the sides d'a, d'b, upon the equal sides B C', B D' of the equal angle C'BD'. It is evident that the points a and b will coincide with C and D' respectively. Thus the angle ad b inserted in C'B D' will complete the solid angle B, which will then be equal to P.

The plane of the angle D'BC has been already proved to be inclined to that of D'B C at the same angle as any two adjacent plane angles of P, and the same is true of the planes of the angles ac' b and cbd'. Since then the plane ac'b coincides with CBD', and the planes c'b d' and B D'C are equally inclined to that plane, the plane c'b d must coincide with B D' C. Since the line B D' coincides with db, and the angles BD' C and c'b d' are equal, and in the same plane, the point d' must coincide with C. In the same manner we may prove that the points c, e', &c. coincide with E' D, &c.; and we may prove that each of the solid angles at these points is equal to P, as we have already proved of the solid angle B.

Hence it appears, that by the union of the two shells formed of ten equilateral triangles in the manner already described, a regular solid with twenty triangular faces is formed.

This solid is called the regular icosaedron. 158) COR.- To determine the inclination of the planes of each pair of intersecting faces. The edges B A, B C, and B P form a solid angle, the two plane angles PBA and PBC being those whose inclination is sought. Applying the formula used in the former cases, let a, b, and c be the plane angles A B C, PBC, and P BA. Hence a = 108° (the angle of a regular pentagon,) b=c=60°. Hence,

3 1 cos A.

= 0,

4
4 cos 108°

4 cos 18° +1
...cos AS
3

3
The value of A may hence be found by the aid of tables.

cos 108°

PROPOSITION XXXI.

(159) To construct a regular solid with square faces.

This is, obviously, a rectangular parallelopiped whose base is a square, and whose altitude is equal to the side of the base.

The regular hexaedron is therefore the cube.

PROPOSITION XXXII.

(160) To construct a regular solid with pentagonal

faces.

DI

Let ABCDE be a regular pentagon. From the vertex A draw the line Aa equal to the side of the pentagon, and inclined to AB and AE at angles equal to the angle of the pentagon. The solid angle formed by the three lines which meet at that point is one of the angles of the required solid, formed by the three pentagonal angles a A B, a A E, and B A E. In the same manner, let the lines Bb, Cc, &c. be drawn from each of the angles of the pentagon forming solid angles of the same kind at the points B, C, D, &c. Let the pentagon, of which a A B b are three sides, be completed, and in the same manner let each of the other pentagons on the sides of the base ABCDE be completed. We shall thus have a shell with six regular and equal pentagonal faces, and a serrated edge, a C'b D'c, &c. The adjacent planes forming several pentagonal faces, are inclined each to each at the same angle; and it may be proved in the same manner as in (157), that if a plane be drawn through the angle C' b D', a solid angle will be formed at b equal to those at A, B, C, &c. As in (157), let another shell in every respect equal and similar to this be constructed, and let them be united at their serrated edges. It will follow by the reasoning used in the former case, that the several solid angles which will be formed at a, C', 6, D', &c. will be equal to those at A, B, C, &c.

Hence by the union of those two shells with six pentagonal faces, a regular solid with twelve pentagonal faces is formed.

This solid is called the regular dodecaedron. (161) COR.–To determine the inclination of the planes, we have

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