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PYRAMIDS AND CONES.

DEFINITIONS.

28. A Pyramid is a polyedron bounded by a polygon called the base, and by triangular planes meeting at a common point called the vertex.

29. A pyramid is called triangular, quadrangular, pentagonal, according as its base is a triangle, a quadrangle, or a pentagon; and so on.

30. The Altitude of a pyramid is the perpendicular distance from its vertex to its base; as A B.

D

A

E

B

31. A Right Pyramid is one whose base is a regular polygon and in which the perpendicular from the vertex passes through the centre of the base.

32. The Slant Height of a right pyramid is the perpendicular distance from the vertex to the base of any one of its lateral faces; as A C.

D

A

33. A Cone is a right pyramid whose base is a regular polygon of an infinite number of sides, that is, whose base is a circle. A cone can be described by the revolution of a right triangle about one of its sides which remains fixed. The other side describes the circular base, and the hypothenuse the convex surface. Thus the right triangle A B C revolving about A B would describe the cone, BC the base, and the hypothenuse AC the convex surface.

B

E

C

34. The Axis of a cone is the line from the vertex to the centre of the base; or it is the fixed side of the right triangle whose revolution describes the cone; as A B.

35. Corollary. The axis of a cone is perpendicular to the base, and is therefore the altitude of the cone.

36. A Frustum of a pyramid is a part of the pyramid included between the base and a plane cutting the pyramid parallel to the base; as D E.

Ꭰ .

GF

B

E

37. The Altitude of a frustum is the perpendicular distance between the two parallel planes or bases; as FB.

38. The Slant Height of a frustum of a right pyramid is the perpendicular distance between the parallel edges of the bases; as G C.

THEOREM VII.

39. If a pyramid is cut by a plane parallel to its base, 1st. The edges and altitude are divided proportionally; 2d. The section is a polygon similar to the base.

Let A-BCDEF be a pyramid whose altitude is AN, cut by a plane GI parallel to the base; then

1st. The edges and the altitude are divided proportionally.

For suppose a plane passed through the vertex A parallel to the base; then the edges and altitude, being cut by three parallel planes, are divided proportionally (IV. 12), and we have

B

G

A

N

F

E

AB: AGAC: AHAD: AIAN: AM

2d. The section GI is similar to the base BD.

D

For the sides of GI are respectively parallel to the sides of BD (IV. 9), and similarly situated; therefore the polygons GI, BD are mutually equiangular. Also, as GL is parallel to BF,

and LK to FE, the triangles A B F and A G L are similar, and the triangles A FE and ALK; therefore

GL: BFAL: A F, and LK: FE=AL: AF

Therefore

GL: BFLK: FE

In the same manner we should find

LK: FE KI:ED=IH: DC, &c.

Therefore the polygons GI and BD are similar (II. 19).

40. Corollary. A section of a cone made by a plane parallel to the base is a circle.

THEOREM VIII.

41. The convex surface of a right pyramid is equal to the perimeter of its base multiplied by half its slant height.

Let A-BCDEF be a right pyramid whose slant height is AH; its convex surface is equal to BCCD+DE+EF+FB multiplied by half of A H.

B

The edges AB, AC, AD, AE, AF, being equally distant from the perpendicular AN (II. 34), are equal (IV. 6); and the bases BC, CD, DE, &c. are equal; therefore the isosceles triangles ABC, ACD, ADE, &c. are all equal (I. 48); and their altitudes are equal. The area of ABC is BC ×

H

Α

[blocks in formation]

of ACD is CDXAH; and so on. Therefore the sum of the areas of these triangles, that is, the convex surface of the right pyramid, is (BC+CD+DE+ E F + FB) 1⁄2 A II.

42. Corollary. As a cone is a right pyramid (33), this demonstration includes the cone. If, therefore, R = the radius of the base, and S the slant height of a cone,

its convex surface 2 R S = TRS

If a plane parallel to the base and bisecting the altitude be

drawn, as the section will be a circle (40) with a radius and circumference one half the radius and circumference of the base, therefore, if r' = the radius of this section,

the convex surface 2πr' S

THEOREM IX.

43. The convex surface of a frustum of a right pyramid is equal to the sum of the perimeter of its two bases multiplied by half its slant height.

Let G D be the frustum of a right pyramid; its convex surface is equal to GH+ HI+IK+K L+LG+B C + C D +DE+EF+FB multiplied by half

MN.

B

Η

G

M

N

D

F

E

The lateral faces of a frustum of a right pyramid are equal trapezoids (39; II. 6); and their altitudes are all equal. The area of GC (II. 14) is (GH + BC) × MN; of HD is (HI + CD) × & M N ; and so on. Therefore the sum of the areas of these trapezoids, that is, the convex surface of the frustum of the right pyramid, is GHHI+IK+KL+LG+ B C + CD + DE+EF+ FB multiplied by half MN.

44. Cor. 1. If the frustum is cut by a plane parallel to its two bases, and at equal distances from each base, this plane will bisect the edges G B, HC, ID, &c. (39); and the area of each trapezoid is equal to its altitude multiplied by the line joining the middle points of the sides which are not parallel (II. 15). Therefore the convex surface of a frustum of a right pyramid is equal to the perimeter of a section midway between the bases multiplied by its slant height.

45. Cor. 2. As a cone is a right pyramid (33), this demonstration includes the frustum of a cone. If, therefore, R and

r the radii of the two bases of the frustum of a right cone, and Sits slant height,

its convex surface = (2 π R + 2 π r) 1⁄2 S = (« R + π r) S

П

If r' the radius of a section midway between and parallel to the bases,

the convex surface 2πr' S

THEOREM X.

46. If two pyramids having equal altitudes are cut by planes parallel to their bases and at equal distances from their vertices, the sections are to each other as their bases.

[blocks in formation]

LMNOP: QRS BCDEF: HIK

For as the polygons LMNOP and BCDEF are similar (39) LMNOP: BCDEF-LP: BFAL: ABAV2: AT2 In like manner

Q R S : HIK = GY2 : G W2

But as AVG Y and AT GW

therefore

or (Pn. 15)

LMNOP: B C D E F =QRS: HIK

LMNOP: QRS BCDEF: HIK

=

47. Corollary. If two pyramids have equal altitudes and equivalent bases, sections made by planes parallel to their bases and at equal distances from their vertices are equivalent.

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