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PROPOSITION XII. THEOREM

568. The volume of a triangular prism is equal to the product of its base by its altitude.

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Hyp. V denotes the volume, B the base, and a the altitude of the triangular prism ABC-H.

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Proof. Upon the edges AB, BC, BH, construct the parallelopiped ABCD-H.

ABC-
H1 ABCD-H.

( (557) The volume of ABCD-H= base ABCD X a. (Why?)

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Ex. 1099. In the diagram for Prop. XII, find the volume of ABC-H, if AB = 2, BC = 3, BH = 5, Z ABC = 30°, and the projection of BH upon base ABC equals 4.

Ex. 1100. In the same diagram find the area of A ABC, if the volume of ABC-H is 200 and the altitude is 20.

PROPOSITION XIII. THEOREM

569. The volume of any prism is equal to the product of its base by its altitude.

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To prove

Hyp. V denotes the volume, B the base, and a the altitude of any prism.

V=B x a. Proof. Pass planes through one lateral edge BG, and the diagonals BE, BD, etc., of the base.

The prism will be divided into triangular prisms, all of which will have the altitude a.

The volume of the given prism is equal to the sum of the volumes of the triangular prisms.

The volume of each triangular prism equals its base multiplied by a.

(Why ?) .. the volume of the given prism equals the sum of the bases of the triangular prisms multiplied by the common altitude a. But the sum of the bases of the triangular prism equals B.

...V= B x a.

570. Cor. 1. Prisms are to each other as the products of their bases by their altitudes.

571. COR. 2. Prisms that have equivalent bases are to each other as their altitudes.

572. Cor. 3. Prisms that have equal altitudes are to each other as their bases.

573. Cor. 4. Prisms that have equal altitudes and equivalent bases are equivalent.

Ex. 1101. In the diagram for Prop. XIII, find the volume V, if ABCDE = 40 sq. in., AF = 13 in., and the projection of AF upon the base equals 5 in.

Ex. 1102. In the diagram for Prop. XIII, find AF, if V = 200, ABCDE = 25, and the projection of AF upon the base is 6.

Ex. 1103. In the same diagram find the inclination of AF to the base, if AF = 6, ABCDE = 20, and V = 60.

PYRAMIDS

574. DEF. A pyramid is a polyedron, one of whose faces is a polygon, and whose other faces are triangles having a common vertex.

575. DEF. The base of the pyramid is the polygon, the lateral faces are the triangles, the vertex of the pyramid is the common vertex of the triangles, the lateral edges are the intersections of the lateral faces, and the lateral area is the sum of the areas of the lateral faces.

576. DEF. A pyramid is triangular, quadrangular, etc., according as its base is a triangle, a quadrilateral, etc.

NOTE. A tetraedron is a triangular pyramid.

577. DEF. The altitude of a pyramid is the length of the perpendicular from the vertex of the pyramid to the plane of the base.

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578. DEF. A regular pyramid is one whose base is a regular polygon, the center of which coincides with the foot of the altitude.

579. DEF. The axis of a regular pyramid is its altitude.

Note. — The lateral edges of a regular pyramid are equal since they cut off equal distances from the foot of the altitude.

(473) Hence the lateral faces of a regular pyramid are equal isosceles triangles.

580. DEF. The slant height of a regular pyramid is the altitude of any one of the lateral faces.

581. DEF. A truncated pyramid is the portion of a pyramid included between the base and a section formed by a plane cutting all the lateral edges.

582. DEF. A frustum of a pyramid is a truncated pyramid in which the plane of the section is parallel to the base. 583. DEF. The altitude of a frustum is the perpendicular distance between its bases.

584. DEF. The lateral area of the frustum is the sum of the areas of its lateral faces; the lateral faces of a frustum of a regular pyramid are equal isosceles trapezoids.

585. DEF. The slant height of a frustum of a regular pyramid is the altitude of one of these trapezoids which form the lateral faces.

PROPOSITION XIV. THEOREM 586. If a pyramid be cut by a plane parallel to its base :

(1) The lateral edges and the altitude are divided proportionally.

(2) The section is a polygon similar to the base.

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Hyp. Pyramid 0-ABCD is cut by plane | to the base, E, F, G, H are the points of intersection with edges, forming section EFGH, and N the point of intersection with altitude OM.

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OB

OM
1st.
ОЕ OF

ON This follows directly from 488.

To prove

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