For, the polygons abcd and ABCD, being similar, are

to each other as the squares of their homologous sides ab

and AB (B. IV., P. XXVII); but,

Cor. 2. If the bases are equal, any sections at equal distances from the bases will be equal.

Cor. 3. The area of any section parallel to the base, is proportional to the square of its distance from the vertex.

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

Let S be the vertex, ABCDE the

base, and SF, perpendicular to EA, the slant height of a right pyramid: then will the convex surface be equal to,

(AB + BC + CD + DE + EA) × †SF.

Draw SO perpendicular to the plane of the base.

E

A

From the definition of a right pyramid, the point O is the centre of the base (D. 11): hence, the lateral edges, SA, SB, &c., are all equal (B. VI., P. V.); but the sides of the base are all equal, being sides of a regular polygon : hence, the lateral faces are all equal, and consequently their altitudes are all equal, each being equal to the slant height of the pyramid.

Now, the area of any lateral face, as SEA, is equal to its base EA, multiplied by half its altitude SF: hence, the sum of the areas of the lateral faces, or the convex surface of the pyramid, is equal to,

(AB + BC + CD + DE + EA) × †SF;

which was to be proved.

Scholium. The convex surface of a frustum of a right pyramid is equal to half the sum of the perimeters of its upper and lower bases, multiplied by the slant height.

Let ABCDE-e be a frustum of a right pyramid, whose vertex is S: then will the section abcde be similar to the base ABCDE, and their homologous sides will be parallel, (P. III.). Any lateral face of the frustum, as AEea, is a trapezoid, whose altitude is equal to Ff, the slant height of the frustum;

E

(EA + ea) × Ff

But the area of the con

A

hence, its area is equal to (B. IV., P. VII.). vex surface of the frustum is equal to the sum of the areas of its lateral faces; it is, therefore, equal to the half sum of the perimeters of its upper and lower bases, multiplied by half the slant height.

PROPOSITION V. THEOREM.

If the three faces which include a triedral angle of a prism are equal to the three faces which include a triedral angle of a second prism, each to each, and are like placed, the two prisms are equal in all their parts.

Let B and b be the vertices of two triedral angles, included by faces respectively equal to each other, and similarly placed then will the prism ABCDE-K be equal to

the prism abcde-k, in all of its parts.

sides fg and gh, of one upper base, will coincide with the homologous sides of the other upper base; and because the upper bases are equal, they must coincide throughout; consequently, each of the lateral faces of one prism will coincide with the corresponding lateral face of the other prism: the prisms, therefore, coincide throughout, and are therefore equal in all their parts; which was to be proved.

Cor. If two right prisms have their bases equal in all their parts, and have also equal altitudes, the prisms themselves will be equal in all their parts. For, the faces which include any triedral angle of the one, will be equal to the faces which include the corresponding triedral angle of the other, each to each, and they will be similarly placed.

PROPOSITION VI. THEOREM.

In any parallelopipedon, the opposite faces are equal, each to each, and their planes are parallel.

Let ABCD-H be a parallelopipedon : then will its opposite faces be equal and their planes will be parallel.

A

E

H

G

F

D

B

For, the bases, ABCD and EFGH are equal, and their planes parallel by definition (D. 7). The opposite faces AEHD and BFG C, have the sides AE and BF parallel, because they are opposite sides of the parallelogram BE; and the sides EH and FG parallel, because they are opposite sides of the parallelogram EG ; ; and consequently, the angles AEH and BFG are equal (B. VI., P. XIII.). But the side AE is equal to BF, and the side EH to FG; hence, the faces AEHD and BFGC are equal; and because AE is parallel to BF, and EH to FG, the planes of the faces are parallel (B. VI., P. XIII.). In like manner, it may be shown that the parallelograms ABFE and DCGH, equal and their planes parallel hence, the opposite faces are equal, each to each, and their planes are parallel; which was to be proved.

Cor. 1. Any two opposite faces of a parallelopipedon may be taken as bases.

Cor. 2. In a rectangular parallelopipedon, the square of either of the diagonals is equal to the sum of the squares of the three edges which meet at the same vertex.

For, let FD be either of the diagonals, and draw FH.

Then, in the right-angled triangle FHD, we have,

FD2 = DH2 + FH3.

But DH is equal to FB, and FH is equal to FA plus AH or FC2:

-2

hence,

FD2

FB2 + FA2 + FC2

B

Cor. 3. A parallelopipedon may be constructed on three lines AB, AD, and AE, intersecting in a common point A, and not lying in the same plane. For, pass through the extremity of each line, a plane parallel to the plane of the other two lines; then will these planes, together with the planes of the given lines, determine a parallelopipedon.

PROPOSITION VII. THEOREM.

If a plane be passed through the diagonally opposite edges of a parallelopipedon, it will divide the parallelopipedon into two equal triangular prisms.

Let ABCD-N be a parallelopipedon, and let a plane be passed through the edges BF and DH: then will the prisms ABD-H and BCD-H be equal

in volume.

For, through the vertices F and B let planes be passed perpendicular to FB, the former cutting the other lateral edges in the points e, h, g, and the latter cutting those edges produced, in the points a, d, and C. The sections Fehg and Bade will be parallelograms,