will appear very evident, if you cut a piece of paper in the form of a sector of a circle, as ABC, and bend the sides AB and AC together, till they meet, and you will find it to form a right cone.

I have omitted the demonstrations of the superfi cies of all the foregoing solids, because I thought it needless, they being all cotaposed of squares, parallelograms, triangles, &c. which figures are all demonstrated before. And if the area of all such figures as compose the surface of the solid, be found separately, and added together, the sum will be the superficial content of the solid.

2. The diameter of the base of a cone is 10, and its perpendicular height 65.1; required the solidity and superficies. Ans. S Solidity 1782.858 Superficies 1151.134076. 3. What is the solidity of an elliptical cone, the greater diameter of its base being 15 2, the less 10, and the perpendicular height 22 ? Ans. 875.4592.

4. What is the solidity and superficies of a cone, whose perpendicular height is 10.3 feet, and the circumference of its base 9 feet?

Ans. S Solidity 22.56093 feet. Superficies 54.1336 feet. 5. What is the solidity and superficies of a cone, whose slant height is 32 feet, and the circumference of

its base 24 feet?

6. What will the painting of a conical church spire come to at 80. per yard; supposing the circumference of the base to be 64 feet, and the perpendicular height 118 feet?

§ VII. Of the frustum of a PrRAMID.

A frustum of a pyramid is the remaining part, when the top is cut off by a plane parallel to the base.

To find the Solidity.

GENERAL RULE.

Multiply the areas of the two bases together, and to the square root of the product add the two areas; that sum, multiplied by one third of the height, gives the solidity of any frustrum.

RULE II.

If the Bases be Squares.

To the rectangle (or product) of the sides of the two bases add the sum of their squares; that sum, being multiplied into one-third part the frustum's height, will give its solidity.

RULE III.

If the Bases be Circles.

To the product of the diameters of the two bases. add the sum of their squares; this sum, multiplied by the height, and then by .2618, or one-third of .7854, the last product will be the solidity.

RULE IV.

If the Bases be regular Polygons.

Add the square of a side of each end of the frustum, and the product of those sides into one sum; multiply this sum by one-third of the tabular area be

Jonging to the polygon (sect. VII. p. 99,) and this pro duct by the height, for the solidity.

Sum

216

684 which multiplied by 6, one

third of the height, gives 4104; this divided by 144 gives 28.5 feet the solidity.

BY RULE 2.

Product of the sides of the bases 18 and 12 is

216

To find the Superficial Content.

RULE.

Multiply half the sum of the perimeters of the two bases by the slant height, and to that product add the areas of the two bases for the whole superficies.

NOTE. The slant height of any frustum, whose ends are regular polygons, is a line drawn from the middle of one side of the less end, to the middle of its parallel side at the greater end. And, the perpendicular height, is a line drawn from the centre of the less end, to the centre of the greater; or it is a perpendicular let fall from the middle of one of the sides of the less end, upon the surface of the greater end. Hence the slant height, and perpendicular height, will be two sides of a right angled triangle; the base of which will be equal to the difference between the radii of the inscribed circles of the two ends of the frustum. And this base may always be found, by multiplying the difference between a side at each end of the frustum, by the tabular perpendicular in section VIII. p. 100. The perpendicular height of the pyramid of which any frustum is a part, may readily be found, by saying, as the base found above, is to the perpendicular height of the frustum; so is the radius of the inscribed circle of the frustum's base, to the perpendicular height of the pyramid. The radius of the inscribed circle is found by multiplying a side of the base, by the tabular perpendicular, section VIII.

EXAMPLES. Required the superficies of the foregoing frustum.

The perimeter of the greater base 72 inches.
The perimeter of the less base is 48 inches.

From 18, a side of the greater end, take 12, a side of the less; the difference 6, multiplied by the tabular perpendicular .5, gives 3. The perpendicular height of the frustum in inches is 216: to the square of 216, which is 46656, add the square of 3, which is 9, the square root of the sum 46656 is 216.02083 inches, the slant height, which multiplied by 60 gives 12961.2498 inches. To this product add the areas of the two ends 324, and 144, and the whole surface will be 13429.2498 inches, or 93 2586 feet.

EXAMPLE 2. Let ABCD be the frustum of a triangular pyramid, each side of the greater base 25 inches, each side of the less base 9 inches, and the length 15 feet; the solid content of it is required.

The tabular area is .433013,

one-third of which is .1443377,

this multiplied by 931, and then

by 15, produces 2015.6759805, which divided by 144, the quotient is 13.9977 feet, the solidity.

The square of 25, multiplied by the tabular area, (section VIII.) gives the area of the greater base 270.633125: in a similar manner the area of the less base will be found to be 35.07405: the square root of the product of these two areas 9492.200569805625 is 97.427925,