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PROBLEM IX.

To find the Solidity of the two Parts, called Ungulas or Hoofs, into which the Fruftum of a Rectangular or a Square Pyramid, or a Rectangular Prifmoid, is cut by a Plane Inclined to its Bafe.

CASE I.

If the plane, paffing through A and B, cut the end in EF, between GH and DC; it will cut off the wedge AEHGFB, whofe base is EFGH, edge AB, and height the fame with that of the fruftum, or prifinoid; and the remaining part AEDCFBQP will be a prifmoid.

Then, by problem 7, find the wedge ABFEHG, and that of ABQPDCFE by problem 8.

EXAMPLE.

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If the fruftum of a fquare pyramid be cut by a plane, paffing through one fide of the lefs end, and through the middle of the greater end; what are the contents of the two parts, fuppofing each fide of the greater end to be 15 inches, each fide of the lefs 6, and the flant height 24 feet?

Since, by example i problem 6, the perpendicular height is 287.9649 inches, therefore,

I

× = Firft, (2 x 15 + 6) × 72 × 287.9649

45 × 287.964912958.4205 inches

feet, the content of the wedge.

7499086

`And [(30+6)×71⁄2+(12+15) × 6] × 287⋅9649 = (18×15+27 × 6) X 287.9649=72×287.9649 =20733 4728 inches 11.998536 feet, the prifmoid.

CASE II.

If the plane pafs through DC, as well as AB, the folid is thereby divided into two wedges or hoofs ABCGHD, ABCOPD, whofe two bafes are the ends or bafes of the folid.

And then the contents of the two parts will be found by problem 7.

EXAMPLE.

Let there be taken here the fame figure as in the laft example, to find the content of the two wedges into which it is cut.

Firft, (30+6)× 15 × 287.9649 = 90 × 287.9649 =25916-841 inches 14 99817 feet, the greater wedge.

And (12+15)×6×287.9649 = 27X2879649 77750523 inches 449945 feet, the lefs wedge.

CASE III.

If the plane cut the fide of the folid in ef; the part cut off ABQ fep will be a wedge, whofe bafe is AQ; and which being taken from the whole figure, will leave the content of the part A вfe DCG H.

EXAMPLE.

Let there be taken here the fame figure, fuppofing the plane to cut the fide pc in ef at the diftance of 10 feet from PQ, which let be now fuppofed one fide of the lefs end AQ; to find the folidity of the two parts.

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Wherefore (12 + 9°75) × 6 × 119.98536 =

21 X 119.98536 = inches

which taken from

87 × 29′99634 = 2609.68158 151023 feet, the folidity of the wedge AQe.

19:49762 the whole fruftum found by prob. 8,

leaves 17.98739 for the content of the

PROBLEM

part A Bfe DC HG,

X.

To find the Curve Surface of a Sphere, or of any Segment or Zone of it.

Multiply the circumference of the sphere by the height of the part required, and the product will be the curve furface, whether it be fegment, zone, hemifphere, or the whole fphere.*

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Note.

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D

F

E

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DEMONSTRATION,

Put d the diameter A B or 2 CG,

CE the height of the zone ADFB, y=DE, ≈ = AD, p = 3*14159, and the furface required.

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Then s=2py; buty: dx, A or 2yz dx; therefore s = pdx, and confequently spdx; viz. the product of the circumference of the fphere and height of the zone; for pd is the circumference of the circle whofe diameter is d.

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Note. The height of the whole fphere is its di

ameter.

EXAMPLE I.

If the diameter or axe of the earth be 7957 miles, what is the whole furface, fuppofing it a perfect sphere?

First, 7957 × 3.141592 = 25000 miles, very near, the circumference.

Then 79574 X 25000 = 198943750 square miles the whole furface required. And the half is 99471875 the furface of the hemifphere.

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=d, s will be pdd the furface of the hemifphere. And confequently that of the whole fphere is pdd, the product of the circumference and height or diameter.

Corol. 2. To or from pdd, the furface of the hemifphere, add or fubtract pdx, that of the zone ADFE, and the remainder pdx (d±x) = pdx GE will be the furface of the

A

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G

D

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fegment DG F, viz. the product of the circumference and height. So that the rule is general. 2. E.D.

Corol. 3. The furfaces of fpheres, and alfo of their fimilar parts, are to each other as the fquares of their diameters. For, by exterminating the common given quantity p, they are as d, the diameter, into the height of the part; but the heights of fimilar parts are as the diameters; therefore &c.

Corol. 4. The furfaces of any fegments or zones of a sphere are to each other, or to that of the whole fphere, as their heights. For pd is common to them all.

Corol. 5. Or the furface of any fegment or zone of a fphere, is as its height.

Corol.

EXAMPLE II.

To find the furface of the two frigid zones of the earth.

Note. The frigid zones are the two oppofite fegments DGF, dнf, in which each of the arcs DG, dн, or half the breadth of the zone, is 23 degrees.

Draw the radius CF; then the angle FCE = 231/ degrees, and its complement EFC 66 degrees; alfo the angle E = 90 degrees.

Then, as s. LE : S. LF :: FC

Which taken from CG

leaves the height GE

And the circumference is 25000.

= 3978: CE = 3648-8675054.

3978.8750000,

330'0074946.

Hence 25000 X 330.0074946 = 8250187.365 the furface of the fegment DGF or frigid zone.

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Corol. 6. The furface of any fegment or zone, is equal to 4 times a circle whofe diameter is a mean proportional between its height and the diameter of the fphere. For this circle is = 4p × (√/dx)2 = pdx.

Corol. 7. Hence the furface of a fegment DGF is equal to 4 times the circle whofe diameter is the chord DG drawn from the vertex to the extremity of the bafe; or equal to a circle whose radius is that chord. For DG = √GH X GE= = √/dx.

Corol. 8. And hence the furface of a fphere is equal to 4 times the area of a great circle of it; that is, 4 times the area of a circle of the fame diameter with the fphere. And confequently the furface of an hemisphere is double the area of its base.

Corol. 9. Or the furface of a fphere is equal to a circle whose diameter is double to that of the sphere.

Corol. 10. The furface of any fegment or zone of a fphere, is equal to the curve furface of a cylinder of the fame height with it, and whofe diameter is equal to that of the fphere.

Corol. 11. Hence the furface of a whole fphere, or of a hemifphere, is equal to the curve furface of the circumfcribed cylinder.

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