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to determine the place of that headland, and the ship's distance from it at the last observation ? Ans. 26-0728 miles.

ExAM. xxii.1. Wanting to know the distance between a house and a mill, which were seen at a distance on the other side of a river, I measured a base line along the side where I was, of 600 yards, and at each end of it took the angles subtended by the other end and the house and mill, which were as follow, viz. at one end the angles were 58° 20' and 95°20', and at the other end the like angles were 53° 30' and 98° 45'. What then was the distance between the house and mill 2 Ans. 959:5866 yards.

ExAM. xxiv. Wanting to know my distance from an inaccessible object 0, on the other side of a river; and having no instrument for taking angles, but only a chain or cord. for measuring distances; from each of two stations, A and B, which were taken at 500 yards asunder, I measured in a di- . rect line from the object 0 100 yards, viz. Ac and BD each equal to 100 yards; also the diagonal AD measured 550 yards, and the diagonal BC 560. What then was the distance of the object 0 from each station A and B 2

-- - Ao 536°25

- Ans. { BO 500'09

ExAM. xxv. In a garrison besieged are three remarkable objects, A, B, C, the distances of which from each other are discovered by means of a map of the place, and are as follow, viz. AB 266#,Ac 530, Bc 3273 yards. Now, having to erect a battery against it, at a certain spot without the place, and being desirous to know whether my distances from the three objects be such, as that they may from thence be battered with effect, I took, with an instrument, the horizontal angles subtended by these objects from my station s, and . found them to be as follow, viz. the angle Ass 13° 30', and the angle Bsc. 29° 50'; required the three distances, sa, se, sc; the object B being situated nearest to me, and between the two others A and c 2 SA 757'14

Ano

SB 537: 10

SC 655°30

ExAM. xxvi. Required the same as in the last example,

when the object B is the farthest from my station, but still

seen between the two others as to angular position, and those

angles being thus, the angle ASB 33° 45', and Esc 22° 30', also the three distances, AB 600, Ac 800, Bc 400 yards 2

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MENSURATION OF PLANES.

THE Area of any plane figure, is the measure of the space contained within its extremes or bounds; without any regard to thickness.

This area, or the content of the plane figure, is estimated by the number of little squares that may be contained in it; the side of those little measuring squares being an inch, or a foot, or a yard, or any other fixed quantity. And hence the area or content is said to be so many square inches, or square feet, or square yards, &c.

Thus, if the figure to be measured be I) 4- C the rectangle ArcD, and the little square E, whose side is one inch, be the mea

suring unit proposed: then as often as 3. the said little square is contained in the rectangle, so many square inches the A B rectangle is said to contain, which in the present case is 12.

PROBLEM I.

To find the Area of any Parallelogram ; whether it be a Square, a Rectangle, a Rhombus, or a Rhomboid.

MULTIPLY the length by the perpendicular breadth, or height, and the product will be the area”.

EXAMPLES.

* The truth of this rule is proved in the Geom. theor. 81,

cor. 2. The same is otherwise proved thus: Let the foregoing rectangle be the figure proposed; and let the length and breadth be divided into several parts, each equal to the linear measuring unit, being here 4 for the length, and 3 for the breadth; and let the opposite points of division be connected by right lines.— Then it is evident that these lines divide the rectangle into a number of little squares, each equal to the square measuring unit E ; and further, that the number of these little squares, or the area of the figure, is equal to the number of linear measuring units in the length, repeated as often as there are linear - measuring

ExAMPLEs.

Ex. 1. To find the area of a parallelogram, the length being 12:25, and breadth or height 8-5. 12:25 length 8'5 breadth

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Ex. 2. To find the area of a square, whose side is 35:25 chains. Ans. 124 acres, 1 rood, 1 perch.

Ex. 3. To find the area of a rectangular board, whose length is 12# feet, and breadth 9 inches. Ans. 9; feet.

Ex. 4. To find the content of a piece of land, in form of a rhombus, its length being 6'20 chains, and perpendicular breadth 5'45. Ans. 3 acres, 1 rood, 20 perches.

Ex. 5. To find the number of square yards of painting in a rhomboid, whose length is 37 feet, and height 5 feet 3 inches. Ans. 21 or square yards.

PROBLEM II.
To find the Area of a Triangle.

RULE 1. MULTIPLY the base by the perpendicular height, and take half the product for the area". Or, multiply the one of these dimensions by half the other.

measuring units in the breadth, or height; that is, equal to the length drawn into the height; which here is 4 × 3 or 12. And it is proved (Geom. theor. 25, cor. 2), that any oblique parallelogram is equal to a rectangle, of equal length and perndicular breadth. Therefore the rule is general for all paralo: whatever.

* The truth of this rule is evident, because any triangle is the half of a parallelogram of equal base and altitude, by Geom. theor. 26. \ EXAMPLE5.

EXAMPLES.

Ex. 1. To find the area of a triangle, whose base is 625, and perpendicular height 520 links 2

Here 625 x 260 = 162500 square links,
or equal 1 acre, 2 roods, 20 perches, the answer.

Ex. 2. How many square yards contains the triangle,

whose base is 40, and perpendicular 30 feet 2 Ans. 663 square yards.

Ex. 3. To find the number of square yards in a triangle,

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Ex. 4. To find the area of a triangle, whose base is 18 feet

4 inches, and height 11 feet 10 inches 2 Ans. 108 feet, 5; inches.

RULE II. When two sides and their contained angle are given : Multiply the two given sides together, and take half their product: Then say, as radius is to the sine of the given angle, so is that half product, to the area of the triangle.

Or, multiply that half product by the natural sine of the said angle, for the area”.

Ex. 1. What is the area of a triangle, whose two sides are 30 and 40, and their contained angle 28°57' By Natural Numbers. By Logarithms. First, # x 40 x 30 = 600, then, 1 : 600 :: '4S4046 sin. 28° 57' log. 9'684887

600 2-7781.51 Answer 290,4276 the area answ. to 2,463038 * For, let AB, AC, be the two given sides, (;

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Ex. 2. How many square yards contains the triangle, of which one angle is 45°, and its containing sides 25 and 21+. feet 2 Ans. 20°86947.

RULE III. When the three sides are given: Add all the three sides together, and take half that sum. Next, subtract each side severally from the said half sum, obtaining three remainders. Then multiply the said half sum and those three remainders all together, and extract the square root of the last product, for the area of the triangle *.

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triangle ABC. Q. E. D. Otherwise. Because the rectangle AG. CF = the A. ABC, and since

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