PROBLEM XV.

To find the Area of an Elliptic Segment.

FIND the area of a corresponding circular segment, having the same height and the same vertical axis or diameter. Then say, as the said vertical axis is to the other axis, parallel to the segment's base, so is the area of the circular segment before found, to the area of the elliptic segment sought. This rule also comes from cor. 2, theor. 3, of the Ellipse.

Ex. 1. To find the area of the elliptic segment, whose height is 20, the vertical axis being 70, and the parallel axis 50. Ans. 648-13. Ex. 2. Required the area of an elliptic segment, cut off parallel to the shorter axis; the height being 10, and the two axes 25 and 35. Ans. 162.03. Ex. 3. To find the area of the elliptic segment, cut off parallel to the longer axis; the height being 5, and the axes 25 and 35. Ans. 97-8425.

PROBLEM XVI.

To find the Area of a Parabola, or its Segment.

MULTIPLY the base by the perpendicular height; then take two-thirds of the product for the area. As is proved in theorem 17 of the Parabola, in the Conic Sections.

Ex. 1. To find the area of a parabola; the height being 2,

and the base 12.

Ex. 2. Required the area of the parabola, whose height is 10, and its base 16. Ans. 1063.

MENSURATION OF SOLIDS.

By the Mensuration of Solids are determined the spaces included by contiguous surfaces; and the sum of the measures of these including surfaces is the whole surface or su perficies of the body.

The measure of a solid, is called its solidity, capacity, or

content.

Solids are measured by cubes, whose sides are inches, or feet, or yards, &c. And hence the solidity of a body is said to be so many cubic inches, feet, yards, &c. as will fill its capacity or space, or another of an equal magnitude.

The least solid measure is the cubic inch, other cubes being taken from it according to the proportion in the following table, which is formed by cubing the linear proportions.

Table of Cubic or Solid Measures.

1728 cubic inches make 1 cubic foot

To find the Superficies of a Prism or Cylinder.

MULTIPLY the perimeter of one end of the prism by the length or height of the solid, and the product will be the sur. face of all its sides. To which add also the area of the two ends of the prism, when required*.

Or, compute the areas of all the sides and ends separately, and add them all together.

Ex. 1. To find the surface of a cube, the length of each side be ng 20 feet. Ans. 2400 feet. Ex. 2. To find the whole surface of a triangular prism, whose length is 20 feet, and each side of its end or base 18 inches. Ans. 91.948 feet.

Ex. 3. To find the convex surface of a round prism, or cylinder, whose length is 20 feet, and the diameter of its base is 2 feet. Ans. 125-664.

Ex. 4. What must be paid for lining a rectangular cistern with lead, at 3d. a pound weight, the thickness of the lead being such as to weigh 71b. for each square foot of surface; the inside dimensions of the cistern being as follow, viz. the length 3 feet 2 inches, the breadth 2 feet 8 inches, and depth 2 feet 6 inches? Ans. 31. 5s. 93d.

* The truth of this will easily appear, by considering that the sides of any prism are parallelograms, whose common length is the same as the length of the solid, and their breadths taken all together make up the perimeter of the ends of the same.

And the rule is evidently the same for the surface of a cylinder.

PROBLEM II.

To find the Surface of a Pyramid or Cone.

MULTIPLY the perimeter of the base by the slant height, or length of the side, and half the product will evidently be the surface of the sides, or the sum of the areas of all the triangles which form it. To which add the area of the end or base, if requisite.

Ex. 1. What is the upright surface of a triangular pyramid, the slant height being 20 feet, and each side of the base 3 feet? Ans. 90 feet. Ex. 2. Required the convex surface of a cone, or circular pyramid, the slant height being 50 feet, and the diameter of its base 8 feet. Ans. 667.59.

PROBLEM III.

To find the Surface of the Frustum of a Pyramid or Cone, being the lower part, when the top is cut off by a plane parallel to the base.

ADD together the perimeters of the two ends, and multiply their sum by the slant height, taking half the product for the answer. As is evident, because the sides of the solid are trapezoids, having the opposite sides parallel.

Ex. 1. How many square feet are in the surface of the frustum of a square pyramid, whose slant height is 10 feet; also each side of the base or greater end being 3 feet 4 inches, and each side of the less end 2 feet 2 inches? Ans. 110 feet.

Ex. 2. To find the convex surface of the frustum of a cone, the slant height of the frustum being 12 feet, and the circumferences of the two ends 6 and 8.4 feet.

Ans. 90 feet.

PROBLEM IV.

To find the Solid Content of any Prism or Cylinder.

FIND the area of the base, or end, whatever the figure of it may be; and multiply it by the length of the prism or cylinder, for the solid content*.

*This rule appears from the Geom. theor. 110, cor. 2. The same is more particularly shown as follows: Let the annexed rectangular paral·

Note. For a cube, take the cube of its side by multiplying this twice by itself; and for a parallelopipedon, multiply the length, breadth, and depth all together, for the content.

Ex. 1. To find the solid content of a cube, whose side is 24 inches. Ans. 13824.

Ex. 2. How many cubic feet are in a block of marble, its length being 3 feet 2 inches, breadth 2 feet 8 inches, and thickness 2 feet 6 inches? Ans. 21.

Ex. 3. How many gallons of water will the cistern contain, whose dimensions are the same as in the last example, when 277 cubic inches are contained in one gallon?

Ans. 131.53.

Ex. 4. Required the solidity of a triangular prism, whose length is 10 feet, and the three sides of its triangular end or base are 3, 4, 5 feet.

Ans. 60.

Ex. 5. Required the content of a round pillar, or cylinder, whose length is 20 feet, and circumference 5 feet 6 inches. Ans. 48.1459 feet.

PROBLEM V.

To find the Content of any Pyramid or Cone.

FIND the area of the base, and multiply that area by the perpendicular height; then take of the product for the content*.

B

F

G

lelopipedon be the solid to be measured, and the cube P the solid measuring unit, its side being 1 inch, or 1 foot, &c.; also, let the length and breadth of the base ABCD, and also the height AH, be each divided H into spaces equal to the length of the base of the cube P, namely, here 3 in the length and 2 in the breadth, making 3 times 2 or 6 squares in the base Ac, each equal to the base of the cube P. Hence it is manifest that the parallelopipedon will contain the cube P, as many times as the base ac contains the base of the cube, repeated as often as the height AH contains the height of the cube. That is, the content of any parallelopipedon is found, by multiplying the area of the base by the altitude of that solid.

A

B

P

And because all prisms and cylinders are equal to parallelopipedons of equal bases and altitudes, by Gecm. theor. 108, it follows that the rule is general for all such solids, whatever the figure of the base may be. This rule follows from that of the prism, because any pyramid is of a prism of equal base and altitude; by Geom, theor. 115, cor. 1 and 2.

Ex. 1. Required the solidity of a square pyramid, each side of its base being 30, and its perpendicular height 25. Ans, 7500. Ex. 2. To find the content of a triangular pyramid, whose perpendicular height is 30, and each side of the base 3. Ans. 38.971143. Ex. 3. To find the content of a triangular pyramid, its height being 14 feet 6 inches, and the three sides of its base 5, 6, 7 feet. Ans. 71-0352. Ex. 4. What is the content of a pentagonal pyramid, its height being 12 feet, and each side of its base 2 feet?

Ans. 27.5276. Ex. 5. What is the content of the hexagonal pyramid, whose height is 6-4 feet, and each side of its base 6 inches ? Ans. 1.38564 feet.

Ex. 6. Required the content of a cone, its height being 10 feet, and the circumference of its base 9 feet,

Ans. 22-56093.

PROBLEM VI.

To find the Solidity of a Frustum of a Cone or Pyramid.

:

ADD into one sum, the areas of the two ends, and the mean proportional between them and take of that sum for a mean area; which being multiplied by the perpendicular height, or length of the frustum, will give its content*.

*Let ABCD be any pyramid, of which BCDGFE is a frustum. And put a2 for the area of the base BCD, b2 the area of the top, EFG, h the height 1 of the frustum, and c the height A1 of the top part above it. Then ch = AH is the height of the whole pyramid.

E

A

H

B

Ꮐ

Hence, by the last prob. Ja (c+h) is the content of the whole pyramid ABCD, and 36'c the content of the top part AEFG; therefore the difference Ja2(c + h) — }b2c is the content of the frustum BCDGFE. But the quantity c being no dimension of the frustum, it must be expelled from this formula, by substituting its value, found in the following manner. By Geom. theor. 112, a : 1a : : (c+ h)2; c2, or ab :: c+h: c, hence (Geom. th. 69) a − b : b : : h : 6, bh and a-ba::h: c-h; hence therefore c = " a-b,

and ch

ah

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