Here h, the height of the frustum, is multiplied into a2 and b2, the areas of the two ends, and into a2b2 the square root of the products of these areas.

In this demonstration, the pyramid is supposed to be square. But the rule is equally applicable to a pyramid of any other form. For the solid contents of pyramids are equal, when they have equal heights and bases, whatever be the figure of their bases. (Sup. Euc. 14. 3.) And the sections parallel to the bases, and at equal distances, are equal to one another. (Sup. Euc. 12. 3. Cor. 2.)*

Ex. 1. If one end of the frustum of a pyramid be 9 feet square, the other end 6 feet square, and the height 36 feet, what is the solidity?

The areas of the two ends are 81 and 36.
The square root of their product is 54.

And the solidity of the frustum=(81+36+54) X12=2052.

2. If the height of a frustum of a pyramid be 24, and the areas of the two ends 441 and 121; what is the solidity? Ans. 6344.

3. If the height of a frustum of a hexagonal pyramid be 48, each side of one end 26, and each side of the other end 16; what is the solidity? Ans. 56034.

PROBLEM VI.

To find the LATERAL SURFACE of a FRUSTUM of a regular

pyramid.

51. MULTIPLY HALF THE SLANT-HEIGHT BY THE SUM OF THE PERIMETERS OF THE TWO ENDS.

Each side of a frustum of a regular pyramid is a trapezoid, as ABCD. (Fig. 19.) The slant-height HP, (Def. 11.) though it is oblique to the base of the solid, is perpendicular to the line AB. The area of the trapezoid is equal to the product of half this perpendicular into the sum of the parallel sides AB and DC. (Art. 12.) Therefore the area of all the equal trapezoids which form the lateral surface of

See note F.

the frustum, is equal to the product of half the slant-height into the sum of the perimeters of the ends.

Ex. If the slant-height of a frustum of a regular octagonal pyramid be 42 feet, the sides of one end 5 feet each, and the sides of the other end 3 feet each; what is the lateral surface?, Ans. 1344 square feet.

52. If the slant-height be not given, it may be obtained from the perpendicular height, and the dimensions of the two ends. Let GD (Fig. 17.) be the slant-height of the frustum CDGL, RN or GP the perpendicular height, ND and RG the radii of the circles inscribed in the perimeters of the two ends. Then PD is the difference of the two radii:

And the slant-height GD =√(GP2+PD3).

Ex. If the perpendicular height of a frustum of a regular hexagonal pyramid be 24, the sides of one end 13 each, and the sides of the other end 8 each; what is the whole surface? √(BC2 – BP3)=CP, (Fig. 7.) that is, √(13a — 6.52)=11.258 And 83-42

2

The difference of the two radii is, therefore,

The slant-height =√(24+4.332)=24.3875

The lateral surface is

And the whole surface,

= 6.928

4.33

1536.4

2141.75.

53. The height of the whole pyramid may be calculated from the dimensions of the frustum. Let VN (Fig. 17.) be the height of the pyramid, RN or GP the height of the frustum, ND and RG the radii of the circles inscribed in the perimeters of the ends of the frustum.

Then, in the similar triangles GPD and VND,

DP GP: DN: VN.

The height of the frustum subtracted from VN, gives VR the height of the small pyramid VLG. The solidity and lateral surface of the frustum may then be found, by subtracting from the whole pyramid, the part which is above the cut

ting plane. This method may serve to verify the calculations which are made by the rules in arts. 50 and 51.

Ex. If one end of the frustum CDGL (Fig. 17.) be_90 feet square, the other end 60 feet square, and the height RN 36 feet; what is the height of the whole pyramid VCD: and what are the solidity and lateral surface of the frustum ?

DN-GR=45-30=15.

And GP=RN=36.

DP Then 15: 36:45: 108=VN, the height of the whole pyramid. And 108-36=72=VR, the height of the part VLG. The solidity of the large pyramid is 291600 (Art. 48.) of the small pyramid

86400

of the frustum CDGL 205200

The lateral surface of the large pyramid is 21060 (Art. 49.) of the small pyramid

of the frustum

9360

11700

PROBLEM VII.

To find the SOLIDITY of a WEDge.

54. Add the LENGTH OF THE EDGE TO TWICE THE LENGTH OF THE BASE, AND MULTIPLY THE SUM BY OF THE PRODUCT OF THE HEIGHT OF THE WEDGE AND THE BREADTH OF THE BASE.

Let L-AB the length of the base. (Fig. 20.)
1=GH the length of the edge.
b=BC the breadth of the base.
h=PG the height of the wedge.

Then L-1=AB-GH=AM.

If the length of the base and the edge be equal, as BM and GH, (Fig 20.) the wedge MBHG is half a parallelopiped of the same base and height. And the solidity (Art. 43.) is equal to half the product of the height, into the length and breadth of the base; that is to bhl.

If the length of the base be greater than that of the edge, as ABGH; let a section be made by the plane GMN, par

allel to HBC. This will divide the whole wedge into two parts MBHG and AMG. The latter is a pyramid, whose solidity (Art. 48.) is bh × (L-1)

The solidity of the parts together, is, therefore,

bhl+bh × (L-1)=1bh3l+¦bh2L—}bh2l=}bh× (2L+?) If the length of the base be less than that of the edge, it is evident that the pyramid is to be subtracted from half the parallelopiped, which is equal in height and breadth to the wedge, and equal in length to the edge.

The solidity of the wedge is, therefore,

bhl - bh×

(l—L)=1bh3l-1bh2l+}bh2L=÷bh × (2L+1) Ex. 1. If the base of a wedge be 35 by 15, the edge 55, and the perpendicular height 12.4; what is the solidity?

2. If the base of a wedge be 27 by 8, the edge 36, and the perpendicular height 42; what is the solidity?

Ans. 5040.

PROBLEM VIII.

To find the SOLIDITY of a rectangular PRISMOID.

55. TO THE AREAS OF THE TWO ENDS, ADD FOUR TIMES THE AREA OF A PARALLEL SECTION EQUALLY DISTANT FROM THE ENDS, AND MULTIPLY THE SUM BY OF THE HEIGHT.

Let L and B (Fig. 21.) be the length and breadth of one end, the length and breadth of the other end, the length and breadth of the section in the middle.

l and b M and m

The solid may be divided into two wedges, whose bases are the ends of the prismoid, and whose edges are L and 1. The solidity of the whole, by the preceding article, is ¿Bh×(2L+1)+¦bh×(21+L)=¿h(2BL+Bl+2b1+bL) As M is equally distant from Land 1, 2M=L+1,2m=B+b, and 4Mm=(L+1)(B+b)=BL+BI+

[bL+lb.

Substituting 4Mm for its value, in the preceding expression for the solidity, we have

h(BL+bl+4Mm)

That is, the solidity of the prismoid is equal to of the height, multiplied into the areas of the two ends, and 4 times the area of the section in the middle.

This rule may be applied to prismoids of other forms. For, whatever be the figure of the two ends, there may be drawn in each, such a number of small rectangles, that the sum of them shall differ less, than by any given quantity, from the figure in which they are contained. And the solids between these rectangles will be rectangular prismoids.

Ex. 1. If one end of a rectangular prismoid be 44 feet by 23, the other end 36 by 21, and the perpendicular height 72; what is the solidity?

The area of the larger end

of the smaller end

=44X23=1012

=36 X21 756 of the middle section =40 x 22= 880 And the solidity =(1012+756+4×880)×12=63456 feet.

2. What is the solidity of a stick of hewn timber, whose ends are 30 inches by 27, and 24 by 18, and whose length is 48 feet? Ans. 204 feet.

Other solids not treated of in this section, if they be bounded by plane surfaces, may be measured by supposing them to be divided into prisms, pyramids, and wedges. And, indeed, every such solid may be considered as made up of triangular pyramids.