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Exam. XII. What is the perpendicular height of a hill; its angle of elevation, taken at the bottom of it, being 46°, and 200 yards farther off, on a level with the bottom, the angle was 31o?

Ans. 286.28 yards. EXAM. XIII. Wanting to know the height of an inaccessible tower; at the least distance from it, on the same horizontal plane, I took its angle of elevation equal to 58o; then going 300 feet directly from it, found the angle there to be only 32°: required its height, and my distance from it at the first station?

height 307:53

Ans. { distance 192:15

EXAM. XIV. Being on a horizontal plane, and wanting to know the height of a tower placed on the top of an inaccessible hill, I took the angle of elevation of the top of the hill 40°, and of the top of the tower 51°; then measuring in a line directly from it to the distance of 200 feet farther, I found the angle to the top of the tower to be 33° 45'. What then is the height of the tower?

Ans. 93.33148 feet. EXAM. XV. From a window near the bottom of a house, which seemed to be on a level with the bottom of a steeple, I took the angle of elevation of the top of the steeple equal 40°; then from another window, 18 feet directly above the former, the like angle was 37° 30': what then is the height and distance of the steeple ?

Ans. { distance 250-79 EXAM. XVI. Wanting to know the height of, and iny distance from, an object on the other side of a river, which appeared to be on a level with the place where I stood, close by the side of the river; and not having room to measure backward, in the same line, because of the immediate rise of the bank, I placed a mark where I stood, and measured in a direction from the object, up 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, z. of the mark left at the river's side 42, of the bottom of the object 27°, and of its top 19o. Required then the height of the object, and the distance of the mark from its bottom?



distance ‘150.50 EXAM. XVII. If the height of the mountain called the Peak of Teneriffe be 2 miles, as it is nearly, and the angle


Ans. { height

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 87° 58'; it is required from these measures 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 globular?

dist. 140.876

Ans. { diam.

7936 } miles.

ExAM. XVIII. Two ships of war, intending to cannonade à 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 angle which the other ship and the fort subtends, which angles are 830 45' and 85° 15'. What then is 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.

612:38 EXAM. xx. Wanting to know the breadth of a river, I measured a base of 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 to 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. 7412 yards. EXAM. XXII. A point of land was observed, by a ship at sea, to bear east-by-south; and after sailing north-east 12 miles, it was found to bear south-east-by-east. It is required

to determine the place of that headland, and the ship's di. stance 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 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 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 ?


AO 536.25

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 3275 yards. Now, having to erect a battery againsä 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 Asl 13° 30', and the angle Bsc 29° 50'; required the three distances, SA, SB, sc; the object B being situated nearest to me, and between the two others a and c?

SA 757.14 Ans. 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 330 45', and ssc 22° 30', also the three distances, AB 600, AC 800, BC 400 yards ?

SA 709 Ans. SB 1042





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 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

A rectangle is said to contain, which in the present case is 12.



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 *.


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* 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 1 ; 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


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. 2172 square yards.


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 X 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 perpendicular breadth. Therefore the rule is general for all parallelograms 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.


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