443. The ALTITUDE of a pyramid is a perpendicular drawn from the vertex to the plane of the base.

444. A pyramid is triangular, quadrangular, &c., according as its base is a triangle, a quadrilateral, &c.

445. A RIGHT PYRAMID is one whose base is a regular polygon, and the perpendicular drawn from the vertex to the base passes through the centre of the base. In this case the perpendicular is called the axis of the pyramid.

446. The SLANT HEIGHT of a right pyramid is a line drawn from the vertex to the middle of one of the sides of the base.

447. A FRUSTUM of a pyramid is the part of the pyramid included between the base and a plane cutting the pyramid parallel to the base.

448. The ALTITUDE of the frustum of a pyramid is the perpendicular distance between its parallel bases.

449. The SLANT HEIGHT of a frustum of a right pyramid is that part of the slant height of the pyramid which is intercepted between the bases of the frustum.

450. The AXIS of the frustum of a pyramid is that part of the axis of the pyramid which is intercepted between the bases of the frustum.

451. The DIAGONAL of a polyedron is a line joining the vertices of any two of its angles which are not in the same face.

452. SIMILAR POLYEDRONS are those which are bounded by the same number of similar faces, and have their polyedral angles respectively equal.

453. A REGULAR POLYEDRON is one whose faces are all equal and regular polygons, and whose polyedral angles are all equal to each other.

PROPOSITION I.-THEOREM.

454. The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude. Let ABCDE-K be a right prism; then will its convex surface be equal to the perimeter of its base,

AB+BC+CD+DE+E A,

multiplied by its altitude A F.

A

K

F

G

H

G

B C

For, the convex surface of the prism is equal to the sum of the parallelograms AG, BH, CI, DK, EF (Art. 436). Now, the area of each of those parallelograms is equal to its base, A B, BC, CD, &c., multiplied by its altitude, AF, BG, CH, &c. (Prop. V. Bk. IV.). But the altitudes AF, BG, CH, &c. are each equal to A F, the altitude of the prism. Hence, the area of these parallelograms, or the convex surface of the prism, is equal to (AB+ B C + CD+DE+E A) × AF; or the product of the perimeter of the prism by its altitude.

455. Cor. If two right prisms have the same altitude, their convex surfaces are to each other as the perimeters of their bases.

456. In every prism, the sections formed by parallel planes are equal polygons.

Let the prism ABCDE-K be intersected by the parallel planes NP, SV; then are the sections NOPQR, STVXY equal polygons.

For the sides ST, NO are parallel, being the intersections of two parallel planes with a third plane ABG F

b

K

I

F

H

Y G

S

X

RT

N

Q

E

A

B

(Prop. XIII. Bk. VII.); these same sides ST, NO, are included between the parallels NS, OT, which are sides of the prism; hence NO is equal to ST. For like reasons, the sides OP, PQ, QR, &c. of the section NOPQR, are respectively equal to the sides TV, VX, XY, &c. of the section STVXY; and since the equal sides are at the same time parallel, it follows that the angles NOP, OPQ, &c. of the first section are respectively equal to the angles STV, TVX of the second (Prop. XVI. Bk. VII.). Hence, the two sections NOPQR, STVXY, are equal polygons.

457. Cor. Every section made in a prism parallel to its base, is equal to that base.

PROPOSITION III.-THEOREM.

458. Two prisms are equal, when the three faces which form a triedral angle in the one are equal to those which form a triedral angle in the other, each to each, and are similarly situated.

equal to the base L M N O P Q, the parallelogram ABG F equal to the parallelogram LMSR, and the parallelogram HG equal to MOTS; then the two prisms are equal.

equal each to each, and similarly situated (Prop. XXI. Sch. 2, Bk. VII.); hence the edge BG will fall on its equal M S, and the face BH will coincide with its equal MT, and the face BF with its equal MR. But the upper bases are equal to their corresponding lower bases (Art. 436); therefore the bases FGHIK, RSTVY are equal; hence they coincide with each other. Therefore HI coincides with T V, IK with V Y, and KF with YR; and consequently the lateral faces coincide. Hence the two prisms coincide throughout, and are equal.

459. Cor. Two right prisms, which have equal bases and equal altitudes, are equal.

For, since the side AB is equal to LM, and the altitude BG to MS, the rectangle A B GF is equal to the rectangle LMSR; so, also, the rectangle BG HC is equal to MSTO; and thus the three faces which form the triedral angle B, are equal to the three faces which form the triedral angle M. Hence the two prisms are equal.

PROPOSITION IV.-THEOREM.

460. In every parallelopipedon the opposite faces are equal and parallel.

Let ABCD-H be a parallelopipedon; then its opposite faces are equal and parallel.

The bases ABCD, EFGH are equal and parallel (Art. 436), and it remains only to be shown that the same is

E

三

I

B. VIL), and the planes DAE, CEP are paralel; hence, also, the panelemn BCGF is epal to the parallelogram. ADHE. In the same way. it may be shown that the opposite faces ABFE. DOGH are equal and parallel.

481. Cor. Any two opposite faces of a parallelopipedon may be assumed as its bases, since any face and the one opposite to it are equal and parallel

PROPOSITION V.-THEOREM.

462. The diagonals of every parallelopipedon bisect each other.

Let ABCD-H be a parallelopipedon; then its diagonals, as BH, DF, will bisect each other.

For, since BF is equal and parallel to DH, the figure BFHD is a parallelogram; hence the diago nals BH, DF bisect each other at

A

E

H

F

G

B

C

the point O (Prop. XXXIV. Bk. I.). In the same manner it may be shown that the two diagonals A G and CE bisect each other at the point 0; hence the several diagonals bisect each other.

463. Scholium. The point at which the diagonals mutually bisect each other may be regarded as the centre of the parallelopipedon.