the ascending ground to the distance of 264 feet, where it was evident that I was above the level of the top of the object; there the angles of depression were found to be, viz. of the mark left at the river's side 42°, of the bottom of the object 27°, and of its top 19°. Required then the height of the object, and the distance of the mark from its bottom? 57.26 Ans. Height, EXAM. XVII.-If the height of the mountain called the Peak of Teneriffe be 4 miles, and the angle taken at the top of it, as formed between a plumb line and a line conceived to touch the earth in the horizon, or farthest visible point, be 37° 25′ 55′′; it is required from hence to determine the magnitude of the whole earth, and the utmost distance that can be seen on its surface from the top of the mountain, supposing the form of the earth to be perfectly round? Ans. {Dist. 178-458 miles. Diam. 7957.818 EXAM. XVIII. Two ships of war, intending to cannonade a fort, are, by the shallowness of the water, kept so far from it, that they suspect their guns cannot reach it with effect. In order, therefore, to measure the distance, they separate from each other a quarter of a mile, or 440 yards; then each ship observes and measures the angles which the other ship and the fort subtends, which angles were 83° 45′ and 85° 15′. What then was the distance between each ship and the fort? Ans. $2292-26 yards. 2298.05 EXAM. XIX.-Being on the side of a river, and wanting to know the distance to a house which was seen at a distance on the other side; I measured out for a base 400 yards in a right line by the side of the river, and found that the two angles, one at each end of this line, subtended by the other end and the house. were 68° 2′ and 73° 15'. What then was the distance between each station and the house? Ans. 593'08 yards. 61238 { 98 I measured a base of EXAM. XX.-Wanting to know the breadth of a river, 500 yards in a straight line close by one side of it; and at each end of this line I found the angles subtended by the other end and a tree close on the bank on the other side of the river, to be 53° and 79° 12'. What then was the perpendicular breadth of the river? Ans. 529 48 yards. EXAM. XXI.-Wanting to know the extent of a piece of water, or distance between two headlands; I measured from each of them to a certain point inland, and found the two distances to be 735 yards and 840 yards; also, the horizontal angle subtended between these two lines was 55° 40'. What then was the distance required? Ans. 741-2 yards. EXAM. XXII.-A point of land was observed, by a ship at sea, to bear east-bysouth; and after sailing north-east 12 miles, it was found to bear south-east-byeast. It is required to determine the place of that headland, and the ship's distance from it at the last observation? Ans. 26.0728 miles. EXAM, XXIII.-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? Ans. 959 5866 yards. EXAM. XXIV.—Wanting to know my distance from an inaccessible object O, 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 direct line from the object O 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 O from each station A and B? Ans. SAO 53625 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, a foot, 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 the rectangle ABCD, and the little square E, whose side is one inch, be the measuring unit proposed: then, as often as the said little square is contained in the rectangle, so many square inches the rectangle is said to contain, which in the present case is 12. 3 D 4 A B E PROBLEM L 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. * The truth of this rule is proved in the Geometry, 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 equal parts, each equal to the lineal 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 farther, that the number of these little squares, or the area of the figure, is equal to the number of lineal measuring units in the length, repeated as often as there are lineal measuring units in the breadth, or height; that is, equal to the length drawn into the height; which here is 4 X 3 or 12. And it is proved (Geometry, Theor. 25, Cor. 2), that a rectangle is equal to any oblique parallelogram, of equal length and perpendicular breadth. Therefore, the rule is general for all parallelograms whatever. EXAMPLES. Ex. 1.-To find the area of a parallelogram, whose length is 12-25, and height 8.5. 12.25 length 8.5 breadth 6125 9800 104 125 area 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. 93 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 height 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 breadth 5 feet 3 inches. PROBLEM II. Ans. 21 square yards. To find the area of a triangle. RULE 1.-Multiply the base by the perpendicular height, and half the product will be the area,* Or, multiply the one of these dimensions by half the other. EXAMPLES. Ex. 1.—To find the area of a triangle, whose base is 625, and perpendicular height 520 links? 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? Ans. 663 square yards. Ex. 3.—To find the number of square yards in a triangle, whose base is 49 feet, and height 25 feet. Ans. 68, or 68 7361. Ex. 4.-To find the area of a triangle, whose base is 18 feet 4 inches, and height 11 feet 10 inches. 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. † * The truth of this rule is evident, because any triangle is the half of a parallelogram of equal base and altitude, by Geometry, Theor. 26. + For, let AB, AC, be the two given sides, including the given angle A. Now AB X CP is the area, by the first rule, CP being perpendicular. But, by Trigonometry, as sine angle P, or radius, is to sine angle A :: AC: CP = sine angle A X AC, taking radius = 1. Therefore, the area AB X CP is X sin, angle A, to radius 1; or, AB X AC A P B as radius: sin. angle A :: AB X AC: the area. Ex. 1.—What is the area of a triangle, whose two sides are 30 and 4), and their contained angle 28° 57′ 18′′ ? HereX 40 X 30 = 600, Therefore, 1:4841226 nat. sin. 28° 57' 18" 600 290 47356, the answer. Ex. 2.-How many square yards contains the triangle, of which one angle is 45o, and its containing sides 25 and 21 feet? 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. Lastly, 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. * Ex. 1.—To find the area of the triangle whose three sides are 20, 30, 40. = 84375. Then 45 X 25 X 15 X 5 Ex. 2.-How many square yards of plastering are in a triangle, whose sides are 30, 40, 50? Ans. 663. Ex. 3-How many acres, &c. contains the triangle, whose sides are 2569, 4900, 5025 links? Ans. 61 acres, 1 rood, 39 perches. PROBLEM III. To find the area of a trapezoid. Add together the two parallel sides; then multiply their sum by the perpendicular breadth or distance between them; and half the product will be the area; by Geometry, theorem 29. Ex. 1.-In a trapezoid, the parallel sides are 750 and 1225, and the perpendicular distance between them 1540 links: to find the area. 1225 1975 × 770 = 152075 square links = 15 acres, 33 perches, * For, let b denote the base AB of the triangle ABC (see the last fig.), also a the side AC, and c the side BC. Then, by Theor. 3, Trigonometry, as ba+c:: a-c: difference of the segments; аа - сс - AP-PC the the segment AP; hence VAC2- AP = perp. CP, 2a2b2 ̧2a2 b2 — aa + 2b2 c2 — b2 + 2a2 c2 —c1 v (a + b + c x 2 · œa + 2b2 c2 — ba + 2œ2 c2 — c2 4bb aa-bb. = 16 v - a+b+c 2 2 = v(s X 8-a × s − b × 3 — c), 4 = CP. x a+b-c) which is the rule, where s denotes half the sum of the three sides. Ex. 2.-How many square feet are contained in the plank, whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches? Ans. 13 feet. Ex. 3.—In measuring along one side AB of a quadrangular field, that side and the two perpendiculars let fall on it from the two opposite corners, measured as below: required the content. Divide the trapezium into two triangles by a diagonal: then find the areas of these triangles, and add them together. Note. If two perpendiculars be let fall on the diagonal, from the other two opposite angles, the sum of these perpendiculars being multiplied by the diagonal, half the product will be the area of the trapezium. Ex. 1.-To find the area of the trapezium, whose diagonal is 42, and the two perpendiculars on it 16 and 18. Ex. 2.-How many square yards of paving are in the trapezium, whose diagonal is 65 feet, and the two perpendiculars let fall on it 28 and 331 feet? Ans. 222 yards. Ex. 3.-In the quadrangular field ABCD, on account of obstructions there could only be taken the following measures, viz. the two sides BC 265, and AD 220 yards, the diagonal AC 378, and the two distances of the perpendiculars from the ends of the diagonal, namely, AE 100, and CF 70 yards. Required the area in acres, when 4840 square yards make an acre? Ans. 17 acres, 2 roods, 21 perches. PROBLEM V. To find the area of an irregular polygon. Draw diagonals dividing the proposed polygon into trapeziums and triangles. Then find the areas of all these separately, and add them together for the content of the whole polygon. EXAM. To find the content of the irregular figure ABCDEFGA, in which are given the following diagonals and perpendiculars; namely, |