MEN
602
MEN
Geometry. For tables of square measure, see Arithmetic,and Measure.
Prob. 1. To find the area of any parallelogram, whether it bea square, a rectangle, a rhombus, or a rhomboid. Fig. 17, 18,lf», 20.
Rule 1 . Multiply the length AB, by the perpendicular breadth,or height, and the product will be the area. If two sides and anincluded angle of a parallelogram are given to find the area, thenmake use of the following rule.
Rale 2. As radius to the sine of the angle of the parallelogram,so is the product of the sides to the area. AB x AD x nat. sineZ. A = area.
Prob. II. To find the area of a triangle. Fig. 21.
Rule 1. When the base and perpendicular height are given,multiply the base by the perpendicular height, and half the pro-
, . . „ ABx CD
duct is the area.-= area.
2
Rule 2. When two sides and their contained angle are given,multiply the two sides together, and take half their product: thensay, as radius to the sine of the given angle, so is that half prodnct
to the area.
AC X AB x nat. sine A-
area.
Rule 3. When the three sides are given, add together thethree sides, and take half their sum. Next subtract each side seve-rally from the said balf-sum, thus obtaining three remainders.Lastly, multiply the half-sum and those three remainders all toge-ther, and extract the square root of the last product for the area ofthe triangle.
Prob. III. To find the area of a trapezoid, fig. 22, AD, BC.Add together the two parallel sides; then multiply their half-sumby the perpendicular breadth or distance between them, and half
the product will be the area.
(BC -f AD) x BE2
= area.
Prob. IV. To find the area of any trapezium, fig. 23. Dividethe trapezium into two triangles by a diagonal; then find the areasof these triangles, and add them together. Or else let two perpen-diculars be drawn to the diagonal, from the opposite angles, thesum of these being multiplied by the diagonal; half the productshall be the area required.
Prob. V. To find the area of any irregular polygon. Fig. 24.Draw diagonals dividing the proposed polygon into trapeziumsand triangles. Then find the areas of all these separately, and addthem together for the content of the whole polygon.
Prob. VI. To find the area of any regular polygon. Fig. 25.
Rule 1 . Multiply the perimeter of the polygon, or sum of itssides, by the perpendicular drawn from its centre on one of itssides, and take half the product for the area.
Rule 2. Square the side of the polygon ; then multiply thatsquare by the area, or multiplier, set against its name in the follow-ing table, and the product will be the area.
N° ofSides.
Names,
Areas, orMultipliers.
3
Trigon or Triangle,
0.433013
4
1.000000
5
Pentagon,
1.720477
6
Hexagon,
2.598076
7
Heptagon,
3.633912
8
Octagon,
4.828427
9
Nonagon,
6.181824
10
Decagon,
7.694209
11
Undecagon,
9.365640
12
Dodecagon,
11.196152
jYtifc. The numbers in the above table express the areas of theregular polygons, when the linear sidu is unity.
Prob. Vll. To find the diameter and circumference of a cir-cle, the one from the other. Fig. 26.
As 7 to 22, so is the diameter to the circumference. Or, as 1*0 3.1416, so is the diameter to the circumference. ABx 3.1416=
circumference. The ratio of the diameter to the circumferencefound by the indefatigable Mr. Abraham Sharp to be as 1 to
3.1415926535897932384626433832795028841071693993751058*'
09749445923078164052 -f- which is true to the last figure,.thevalue of which amounts to only the ten hundred thousand billi° n '"part of the following very minute fraction
parts of unity, a degree of exactness millions of times more tin 111sufficient for any mathematical question.
Prob. VIII. To find the length of any arc of a circle.
Rule. Multiply the degrees in the given arc by the radius of tn*circle, and the product again by the decimal .01745 for the lengl"of the arc.
Prob. IX. To find the area of a circle.
Rule 1 . Multiply half the circumference by half the diameterOr multiply the whole circumference by the whole diameter an<*take i of the product for the area.
Rule 2. Square the diameter, and multiply that square by fi ,edecimal '7854, for the area.
Prob. X. To find the area of a circular ring.
Rule Take the difference between the areas of the circles,found by the last problem. Or, which is the same thing, subtractthe square of the least diameter from the square of the greater, an 1 *multiply their difference by ’7854.
Prob. XI. To find the area of the sector of a circle. Fig. 27-
Rule 1. Multiply the radius, or half-diameter, by half the arc °*the sector, for the area. Or, multiply the whole diameter bywhole arc of the sector, and take of the prodnct.
Rule 2. As 360 is to the degrees in the arc of the sector, so }*the area of the whole circle to the area of the sector. This is cv*'dent, because the sector is proportional to the length of the arc, ° rto the degrees contained in it.
Prob. XII. To find the area of a segment of a circle. Fig-
Rule. Find the area of the sector, having the same arc with d |ffsegment, by the last problem. Find also the area of the triangl®'formed by the chord of the segment and the two radii of the sect® 1 'Then take the sum of these two for the answer, when the segin el1is greater than a semicircle: or take their difference for the a 11 'swer, when it is less than a semicircle. This is evident byspection.
Prob. XIII. To find the area of an ellipse.
Rule. Multiply Ihe product of the transverse and conjugalaxis by the decimal ’7*54, the result will be the area. _ ,
Prob. XIV. To find the area of a parabola, its base and heig' !tbeing given. ,
Rule. Multipy the base by the height, and ■§ of the prod 11 cwill be the area.
Of Land-Surveying.
The most useful instruments for surveying are the CfUd*'Pi. ane Table, and Theodolite , see those articles. A statu 11 'acre of land being 1GO square poles, the chain is made 4 poles,
66 feet in length, that 10 square (bains, or 100,000 square lil>k; rmay make a square acre. This chain is commonly called Gunt cr j’chain. The plane table is used for drawing a plan of theand taking such angles as are necessary to calculate its area, h .of a rectangular form, and surrounded with a moveable frame, u,means of which, a sheet of paper may be fixed to its surface,is furnished with an index, by which a line may be drawn ll p°'the paper in the direction of any object in the field; and Vscales of equal parts ; by w hich such lines may be made prop 0 ,tional to the distances of the objects from the plane table, vmeasured by the chain ; and its frame is divided into degrees tobserving, angles. But the theodolite is the best instrument for lamg angles.
Piion. 1. To measure a field with the chain. v
Let A m BCD q, fig. 29, represent the field to be mcasur 1 - ■Let it be resolved into the triangles A»«B, ABD, BCD, -A?' 'Let all the sides of the large triangles ABD, BCD, and the | Jt ^ spendiculurs of the small ones, AwiB, AryD, from their vclt ! c .j/m, q, be measured by the chain, and the areas calculated: l “'amount is the area of the whole. But if, on account of the e,vature of its sides, the field cannot be wholly resolved intoangles, then either a straight line may be drawn over theside, so that the parts cut olf from the field, and those added