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1. Measure the Line AB, 12 foot long for the Base.
2. Measure the Perpendicular BC, of 8 foot.
3. Multiply the half of the Base upon the whole perpendicular, or the half of the per∣pendicular upon the whole Base, and the Sum,
(48 foot) will be the Area of the Triangle which was desired.
1. Let fall a perpendicular from F to its opposite side DE, which may divide the Tri∣angle into 2 right-angled Triangles, DGF, and FGE.
2. Measure these two Triangles by the foregoing Problem; i. e. multiply the whole Base DG and GE, by the half of the Perpen∣dicular FG, or the whole Perpendicular by the half of the Base DG and GE.
Thus for instance: Let the Base DG be 12 foot, and the Perpendicular FG 10: Take the half of the Perpendicular, i.e. 5 feet, and multiply it by the Base of 12 feet, and the sum will be 60 feet, which gives the A∣rea of the Triangle DFG.
3. Let the Base GE of the Triangle FGE be six feet, and the Perpendicular FG stands as it did, i. e. to feet. Multiply 5 by 6, or or 3 by 10; the sum will be 30 feet, which will give the Area of the Triangle. Then add 30 to 60, the sum 90 will be the number of feet in the whole Area of the given Triangle DEF.
Sect. 1. If one of the Lines to be multiplied has only fathoms, and the other fathoms and feet; when you multiply, instead of a fa∣thom take 6 foot. For instance: if you are to multiply 7 fathoms by 8 fathoms and 4 feet, instead of the first sum (seven fathoms) you must put 6 fathoms and 6 feet, because these 6 feet make but one fathom: For since there are no feet expressed, you cannot any other∣wise multiply the feet of the second sum: So the number will be 48 fathoms, and 24 feet: for 6 into 8 gives 48: and 6 into 4 gives 24.
Sect. 2. If one of these lines has only fathoms, and the other has fathoms, inches, and feet; for instance, if you are to multiply 6 fathoms by 8 fathoms, 4 feet, and 6 inches, this is the Method you are to take:
Multiply one side by the other: AB for instance of 12 feet, by BC of 12 feet: be∣cause the sides are equal: the product 144 feet will give you the Area which is sought for.
Multiply the shortest side BC of six foot, by the longest side AB of 12 foot: the product 72 foot will give you the Area of the Parallelogram which was required.
1. Let fall a Perpendicular from A to the Line DC, and AE.
2. Let fall another Perpendicular from B, equal to AE, as BF; and joyn CF together. So that instead of a Rhomb you have a Pa∣rallelogram ABEF: then work by the fore∣going Problem. See Prob. 28. Book. 1. Chap. 2.
1. Let fall a Perpendicular from A to the opposite Line DC, as AF.
2. Let fall another from B, equal to AF, as BE.
3. Joyn CE together, and instead of the Rhomboid ABCD you have a Parallelo∣gram ABEF: You will find its Area by the ••ourth Problem of this Chapter.
Let the given Quadrangle be ABCD.
1. Divide this Quadrangle into two Tri∣angles ABD, and CDB.
2. Divide each of these Triangles into two right-angled ones, as ABF, BCE, ADF, DEC.
3. When you have these 4 right-angled Triangles, search their Area's by the two first Problems of this Chapter: then joyn
the Sum of all these Triangles together, and it will give the Area which is required.
Let the regular Pentagon demanded be ABCDE.
1. Measure one side of the Polygon as AB, for instance, 6 fathoms: multiply this number by the number of the sides of the Polygon, as here by 5, because this Polygon has but 5 sides: the product of this multi∣plication will give you the length of all the sides.
2. Let fall a Perpendicular from any side to F the Center of the Polygon, as FG: then measure this Line, 5 fathoms for in∣stance.
3. Multiply the sum of all the sides, by the half of the Perpendicular: the product will give the Area which is required.
1. Divide the whole Figure into right an∣gled Triangles; as this irregular Pentagon ABCDE into ABG, BCG, ADH, CDH, AEF, and DEF.
2. When you have only right angled Tri∣angles, work by the first Problem of this Chapter; the sum of all the Triangles re∣duced to one, gives the Area which is fought for.
Reduce this Circle to a right angled Tri∣angle, by Prob. 21. Book 1. Chap. 2. and then seek the Area by the Rules of the first Problem of this Chapter.
1. Divide the proposed Figure into four equal parts, as ABCD, so as that the line AB may make one Diameter, and CD ano∣ther.
2. Seek for a mean proportional between these two Diameters, which you may thus find:
Set the Diameter AB upon a long line, and upon the same line from the point B, set the other Diameter CD, as BC; then take the half of the distance between A and C, and carry it on to B, as EB; this shall be a mean proportional; then divide EB into two equal parts, as EF, FB; then draw
a Circle from the point F ••hro the point EB, which shall be equal to the given Oval Figure.
3. Having this Circle, find its Area, by the foregoing Problem; i. e•• raise a Perpendicu∣lar upon B, one extream point of the Dia∣meter, which shall have thrice the length of the Diameter EB, and a seventh part over, as BG.
4. Joyn F the Center of the Circle to the point G by a right line, which will give a right angled Triangle equal to the given Circle; its Area may be found by the Rules of the first Problem of this Chapter.
[This Triangle which he says is equal to the Area of the Circle is not expressed in the Scheme.]
1. Let ABCD be an equilateral Square, which shall be the Base of the Pyramid gi∣ven.
1. Find the Area of this Base by Prob. 3. of this Chapter, of 48 foot.
2. Multiply one of its sides, as AB, by the height of the Pyramid EF, that is 12 by 28, whose product will be 336 feet.
3. Multiply 336 by 4, the number of the sides of this Pyramid, and the product will be 1344 feet.
4. To this sum add the Area of the Base, 48 feet, and you will have the total sum of the Area of the Pyramid proposed, viz. 1392 feet.
1. Set the height of the Cylinder BD upon a long line.
2. Set upon the same line, beginning at the point D, the Diameter of the Base of the Cylinder marked CD, as DE.
3. Divide the line EB into two equal parts, BF, EF.
4. With the distance FD draw a Circle, which shall contain the Convex Surface of the given Cylinder: this you may find by Prob. 21. Chap. 2. of the first Book, which teaches you to make a right angled Tri∣angle equal to a given Circle; when you have this Triangle, find its Area by Prob. 1. of this Chapter.
5. When you have the Area of this right angled Triangle, which represents the Con∣vex Surface of the Cylinder, then find the Area of the two Bases of the Cylinder by
Prob. 1 of this Chapter•• Add the sum of the two Bases to the Area of that Triangle which gives you the Convex Surface of the Cylinder, without counting the Bases, and you will have the general sum of the whole Surface of the Cylinder, with both its Bases. See Archimedes de Sphaerâ & Cylindro, lib. 1. Prop. XIII.
Let the given Cone be ABD, and its dia∣meter AD.
1. Set the height of the Cone DB upon a ••ight line.
2. Set its semi-diameter CD upon the same line, as DE.
3. Divide EB into two equal parts, EF, FB.
4. Take the distance FD, and make a circle, as large as the Cone without the Base.
5. Reduce this circle to a right angled Triangle by Prob. 21. Chap. 2. of the first Book, and find its Area by Prob. 1. of this Chapter.
6. By the two forementioned Problems seek the area of the base of your Cone, and add this sum to the former; the sum total
will give the whole surface of the given Cone. See Archimed. Lib. cit. Prop. XIV.
1. Me••sure the diameter AB, of nine feet.
2. Find the circumference of the circle, whereof that is the diameter, which is always thrice as long, and a third part over.
3. Multiply the diameter by the circum∣ference; the product will give you the re∣quired surface of the given Globe.
You must work by the Rules of Problem 7. of this Chapter.
1. Measure the line AB, of fourteen fa∣thoms, and from its middle point draw a per∣pendicular to the arch, as DC, of four fa∣thoms.
2. Multiply the distance AD by the re∣mainder DB; i. e. 7 by 7; then divide by
4 the number of the perpendicular DC.
3. Add to this the length of the perpendi∣cular, i.e. four fathoms, and it will give you the length of the diameter, to compleat the base.
4. When you have found the base, seek for a right angled Triangle, equal to it, by Prob. 21. Chap. 2. Book 1; afterwards you may find the area of that Triangle by Prob. 1. of this Chapter.