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Example. Required the area of an octagon whose side is 20 feet

Here 202=400, and the tabular area 4-8284271 ;

Therefore 4-8284271 x 400=1931-37084 feet, area required. 1920. PROBLEM VII. To find the diameter and circumference of any circle, either from the ker. Rule 1. As 7 is to 22, or as 1 is to 3•1416, so is the diameter to the circumference. Or

as 22 is to 7, so is the circumference to the diameter. Example. Required the circumference of a circle whose diameter is 9.

Here 7 : 22::9: 283; or, *-*°=289, the circumference required. Required the diameter of a circle whose circumference is 36.

Here 22 : 7::36 : 114%; or, o=11the diameter required. 1991. PROBLEM VIII. To find the length of any arc of a circle. Rule 1. Multiply the decimal 01745 by the number of degrees in the given arc, and

that by the radius of the circle ; then the last product will be the length of the arc. This rule is founded on the circumference of a circle being 6.2831854 when the diameter is 2, or 3:1415927 when the diameter is 1. The length of the whole circumference then being divided into 360 degrees, we have 360° : 6.2831854

::1° : 01745. Example. Required the length of an arc of 30 degrees, the radius being 9 feet.

Here 01745 x 30 x 9=4-7115, the length of the arc. Rule 2. From 8 times the chord of half the arc subtract the chord of the whole arc,

and one third of the remainder will be the length of the arc nearly. Example. Required the length of an arc DCE (fig. 516.) whose chord DE is 48,

and versed sine 18.
Here, to find DC, we have 242 + 182=576 + 324 = 900,

and 1900=30= DC.
Whence

=64, the length of the arc required. 1299. PROBLEM IX. To find the area of a circle. Rule 1. Multiply half the circumference by half the diame

Or multiply the whole circumference by the whole diameter, and take of the product. Rule 2. Square the diameter, and multiply such square by ·7854. Rule 3. Square the circumference, and multiply that square by the decimal •07958. Example. Required the area of a circle whose diameter is io, and its circumference 31-416.

31.416x 10
By rule 1.,

=78-54.
By rule 2., 10° * *7854=100 •7854=78.54.

By rule 3., 31-416 * 31416 * •07958 = 78.54.
So that by the three rules the area is 78.54.

1223. PROBLEM X. To find the area of a circular ring, or of the space included between he circumferences of two circles, the one being contained within the other. Rule. The difference between the areas of the two circles will be the area of the ring.

Or, multiply the sum of the diameters by their difference, and this product again

by -7854, and it will give the arca required. Example. The diameters of two concentric circles being 10 and 6, required the area

of the ring contained between their circumferences.

Here 10+6=16, the sum, and 10-6=4, the difference.

Therefore •7854 x 16 x 4= -7854 x 64 = 50-2656, the area required. 1224. PROBLEM XI. To find the area of the sector of a circle. Rule 1. Multiply the radius, or half the diameter, by half the arc of the sector for the

Or multiply the whole diameter by the whole arc of the sector, and take of the product. This rule is founded on the same basis as that to Problem IX. Rule 2. As 360 is to the degrees in the arc of the sector, so is the area of the whole

circle to the area of the sector. This is manifest, because it is proportional to the

length of the arc. Example. Required the area of a circular sector whose arc contains 18 degrees, the

diameter being 3 feet.
By the first rule, 3.1416 * 3=9.4248, the circumference.

360 : 18::9.4248 : 47124, the length of the arc.
.47124 * 3+4=1.41372 +4= .35343, the area of the sector.

ter.

Fig. 516.

area.

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By the second rule, -7854 x 39=7-0686, area of the whole circle.

S60: 18::7-0686 : 35343, the area of the sector. 1225. Problem XII. To find the area of a segment of a circle

. Rule 1. Find the area of the sector having the same are with the segment by th

problem. Then find the area of the triangle formed by the chord of the seg
and the two radii of the sector. Take the sum of these two for the answer
the segment is greater than a semicircle, and their difference when less th

semicircle.
Example. Required the area of the segment ACBDA

(fig. 517.), its chord AB being 12, and the radius AE

or CE 10.
As AE : sin. 2 D 90°:: AD: sin. 36° : 524=36.87 degrees

in the arc AC.
Their double 73-74 degrees in are ACB.
Now, 7854 x 400=314.16, the area of the whole circle.
Therefore, 360 : 73 74:: 314.16 : 64.3504, area of the sector

ACBE.
Again, AE- AD = V100-36= 764=8-=DE.
Therefore, AD DE=6 x 8=48, the area of the triangle

AEB.
Hence the sector ACBE (64.350), less triangle AEB (48) Fig. 517.

=16:9504, area of segment ACBDA.
Rule 2. Divide the height of the segment by the diameter, and find the quotient in

column of heights in the following table. Take out the corresponding area in next column on the right hand, and multiply it by the square of the circle's diam for the area of the segment. This rule is founded on the principle of similar p figures being to one another as the squares of their like lineal dimensions, segments in the table are those of a circle whose diameter is 1. In the first colo is contained the versed sines divided by the diameter. Hence the area of similar segment taken from the table and multiplied by the square of the diam gives the area of the segment to such diameter. When the quotient is not fo

exactly in the table, a proportion is used between the next less and greater area. Example. As before, let the chord AB be 12, and the radius 10 or diameter 20.

Having found as above DE=8: then CE-DE=CD=10-8=2. He

by the rule CD+CF=2+20='1, the tabular height; this being found the first column of the table, the corresponding tabular area is 0408 then *040875 x 20= *040875 × 400=16.340, the area nearly the same

before. Areas of the SEGMENTS OF A Circle WHOSE DIAMETER, UNITY, IS LUPPOSED TO E

DIVIDED INTO 1000 EQUAL Parts.

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1226. PROBLEM XIII. To find the area of an ellipsis.
Rule. Multiply the longest and shortest diameter together, and their product by 78

which will give the area required. This rule is founded on Theorem 3. Cor. 2.

Conic Sections. (1098, 1100.)
Example. Required the area of an ellipse whose two axes are 70 and 50.

Here 70 x 50 x 7854=2748-9.
1227. PROBLEM XIV. To find the area of any elliptic segment.
Rrile. Find the area of a circular segment having the same height and the sa

vertical axis or diameter ; then, as the said vertical axis is to the other axis (para to the base of the segment), so is the area of the circular segment first found to area of the elliptic segment sought. This rule is founded on the theorem allud to in the previous problem. Or, divide the height of the segment by the vert: axis of the ellipse; and find in the table of circular segments appended to Prob. (1224.) the circular segment which has the above quotient for its versed sine ; th

multiply together this segment and the two axes of the ellipse for the area. Example. Required the area of an elliptic segment whose height is 20, the verti axis being 70, and the parallel axis 50.

Here 20=70= -2857142, the quotient or versed sine to which in

table answers the segment .285714.

Then .285714 x 70 x 50 = 648-342, the area required. 1228. PROBLEM XV. To find the area of a parabola or its segment. Rule. Multiply the base by the perpendicular height, and take two thirds of the pi

duct for the area. This rule is founded on the properties of the curve alrea described in conic sections, by which it is known that every parabola is of

circumscribing parallelogram. (See 1097.) Example. Required the area of a parabola whose height is 2 and its base 12.

Here 2 x 12 = 24, and 3 of 24=16 is the area required.

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MENSURATION OF SOLIDS. 1229. The measure of every solid body is the capacity or content of that body, co sidered under the threefold dimensions of length, breadth, and thickness, and the measu of a solid is called its solidity, capacity, or content. Solids are measured by units whi are cubes, whose sides are inches, feet, yards, &c. Whence the solidity of a body said to be of so many cubic inches, feet, yards, &c. as will occupy its capacity or spa or another of equal magnitude.

1230. The smallest solid measure in use with the architect is the cubic inch, from whi other cubes are taken by cubing the linear proportions, thus,

1728 cubic inches=1 cubic foot ;

27 cubic feet =1 cubic yard. 1231. Problem I. To find the superficies of a prism. Multiply the perimeter of one end of the prism by its height, and the product will be t

surface of its sides. To this, if wanted, add the area of the two entds of the pris

Or, compute the areas of the sides and ends separately, and add them together. Example 1. Required the surface of a cube whose sides are 20 feet.

Here we have six sides ; therefore 20 x 20 x 6=2400 feet, the area required.
Example 2. Required the surface of a triangular prism whose length is 20 feet a

each side of its end or base 18 inches.
Here we have, for the area of the base,
1:52 – 752 = (2.25 – "5625 =) 1.68759 for the perpendicular of triangle

base;
and 71:6875=1.299, which inultiplied by 1.5=1.948 gives the area of t

two bases;

then, 3 x 20 x 1.5 +1.948 = 91 ·948 is the area required. Example 3. Required the convex surface of a round prism or cylinder whose leng is 20 feet and the diameter of whose base is 2 feet.

Here, 2 x 3.1416 = 6•2832,

and 3:1416 * 20 = 125-664, the convex surface required. 1232. Proslem II. To find the surface of a pyramid or cone. Rule. Multiply the perimeter of the base by the length of the slant side, and half t1

product will be the surface of the sides or the sum of the areas of all the sides, of the areas of the triangles whereof it consists. To this sum add the area of ti

end or base. Example 1. Required the surface of the slant sides of a triangular pyramid who

slant height is 20 feet and each side of the base 3 feet.

Here. 20 x 3 (the perimeter) * 3+2 =90 feet, the surface required.

A

B

Example 2. Required the convex surface of a cone or circular pyramid whose slant height is 50 feet and the diameter of its base 84 feet.

Here, 8.5 x 3:1416 x 50-2=667-5, the convex surface required. 1233. PROBLEM III. To find the surface of the frustum of a pyramid or cme, being the lower pert where the top is cut off by a plane parallel to the base. Rale. Add together the perimeters of the two ends and multiply their sum by the slant

height. One half of the product is the surface sought. This is manifest, because

the sides of the solid are trapezoids, having the opposite sides parallel. Example 1. Required the surface of the frustum of a square pyramid whose slant

height is 10 feet, each side of the base being 3 feet 4 inches and each side of the

top 2 feet 2 inches. Here, 3 feet 4 inches *4=13 feet 4 inches, and 2 feet 2 inches x 4 = 8 feet 8 inches; and 13 feet 4 inches + 8 feet 8 inches =22. Then 22; 2x10=110 feet, the surface

required. Example 2. Required the convex surface of the frustum of a cone whose slant height

is 124 feet and the circumference of the two ends 6 and 8·4 feet. Here, 6 +8.4=14:4; and 14.4 x 12.5+2 = 180-2 =90, the convex surface required. 1234. PROBLEM IV. To find the solid content of any prism or cylinder. llule. Find the area of the base according to its figure, and multiply it by the length of

the prism or cylinder for the solid content. This rule is founded on Prop. 99. (Geometry, 980.). Let the rectangular parallelopipedon be the solid to be measured, the small cube P(fig. 518.) being the measuring unit, its side being 1 inch, 1 foot, &c. Let also the length and breadth of the base ABCD, and also let the height AH, be divided into spaces equal to the side of the base of the cube P; for instance, H here, in the length 3 and in the breadth 2, making 3 times 2 or 6 squares in the base AC each equal to the base of the cube P. It is manifest that the parallelopipedon will contain the cube P as many times as the hase AC contains the base of the cube, repeated as often as the height AH contains the height of the cube. Or, in other words, the contents of a parallelopipedon is found by multiplying the area of the base by the altitude of the solid. And because all prisms and cylinders are equal to parallelopipedons of equal bases and

altitudes, the rule is general for all such solids whatever the figure of their base. Example 1. Required the solid content of a cube whose side is 24 inches.

Here, 24 x 24 x 24=13824 cubic inches. Example 2. Required the solidity of a triangular prism whose length is 10 feet and

the three sides of its triangular end are 3, 4, and 5 feet. Here, because (Prop. 32. Geometry, 907.) 32 + 42=52, it follows that the angle con.

tained by the sides 3 and 4 is a right angle. Therefore x 10=60 cubic feet,

the content required. Example 3. Required the content of a cylinder whose length is 20 feet and its diameter 5 feet 6 inches.

Here, 5.5 x 5.5 x 7854 =- 23.75835, area of base;

and 23.75835 x 20=47:5167, content of cylinder required. 1335. PROELEM V. To find the content of any pyramid or cone. Kule. Find the area of the base and multiply that area by the perpendicular height.

One third of the product is the content. This rule is founded on Prop. 110

(Geometry, 991.) Example 1. Required the solidity of the square pyramid, the sides of wiose base are 30, and its perpendicular height 25.

Here, 30 * 25=7500, content required. Example 2. Required the content of a triangular pyramid whose perpendicular height is 50 and each side of the base 3.

3+3+3 Here, '=f=4.5, half sum of the sides : and 4:5—3=1.5, one of the three equal remainders. (See Trigonometry, 1052.) but v4.5 x 1.5 x 1.5 x 1.5 30:3 = 3.897117 ~ 10, or 38.97117, the solidity

required. Esample 3. Required the content of a pentagonal pyramid whose height is 12 fect and each side of its baze 2 feet. Here, 1.7204774 (tabular area, Prob. 6. 1218.) 4 (square of side) = 6.8819096

area of base; and 6.8819096 x 12 = 82.5829152. Then =275276384, content required.

z

Fig. 518.

3x4

2

30 x 30

2

82.5829152

3

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