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be easily verified upon the figures, are contained in the enunciation of a theorem by the celebrated Euler, and which is translated by the formula

V+FE+2:

V designating the number of vertices, F the number of faces, and E the number of edges. This formula has been previously given.

There are a great many theorems more or less important upon polyhedrons as well as upon polygons, similar to those in a previous appendix, for which, see a memoir of M. Poinsot, in the Journal de l'Ecole Polytechnique, 10a cahier, t. iv., p. 6, et seq. Also, a memoir of M. Cauchy, in the same journal, 16o cahier, t. ix., p. 77. Also, Annales de Mathematiques of M. Gergonne, particularly tome xv., page 157.

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 each 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. In other words, the area of a surface is the numerical ratio of this surface to its unit. 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 3 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.

PROBLEM I.

D

A

4

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B

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. 60, Schol.

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 linear measuring unit, being here four for the length and three 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

I

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

Ex. 2. To find

is 35.25 chains.

9800

104.125 area.,

the area of a square whose side Ans. 124 acres, 1 rood, 1 perch. the area of a rectangular board whose length is 12 feet, and breadth 9 inches.

Ex. 3. To find

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. Ans. 21 square yards.

PROBLEM II.

To find the area of a triangle.

Or,

RULE I. Multiply the base by the perpendicular height, and half the product will be the area.* multiply the one of these dimensions by half the other.

little squares, or the area of the figure, is equal to the number of linear measuring units in the length, which is the same as the number of square units in a horizontal row, repeated as often as there are linear measuring units in the breadth or height, which is the same as the number of horizontal rows, that is here 4 X 3 or 12.

And it is proved (Geometry, theor. 22) that a rectangle is equal to any oblique parallelogram 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 Geometry, Theor.

23.

EXAMPLES.

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

Here

625 × 260162500 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.†

Ex. 1. What is the area of a triangle whose two sides are 30 and 40, and their contained angle 28° 57' 18/?

Here

Therefore,

[ocr errors]

× 40 × 30 = 600,

4841226 nat. sin. 28° 57′ 18′′
600

290-47356, the answer.

Ex. 2. How many square yards contains the tri

* 100 links make a chain, 10,000 square links a square chain, and 10 square chains an acre.

C

+ The following demonstration requires an acquaintance with Trigonometry. 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, CP sine angle A X AC, taking radius=1. Therefore, the area AB AB × AC × sin. angle A, to radius 1; or, as radius: sin. angle A::AB X AC: the area.

× CP is

P

B

feet?

angle, of which one angle is 45°, and its containing sides 25 and 21 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.

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40

2)90

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25, first rem. 15, second rem. 5, third rem.

45, half sum.

Then

45 X 25 X 15 X 5=84375.

The root of which is 290.4737, the area.

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

* For, let a, b, c denote the sides opposite respectively to A, B, C, the angles of the triangle A B C (see last figure); then, by theor. 29, Geom., we have BC2AB+ AC2-2AB.AP, or a2 = b2 + b2 + c2: c2-2c. AP. AP=.· ; hence we have 2c

CP2-62

a2

(b2+c2-a2)2_4b2c2—(b2+c2—a2)2__(2bc+b2+c2—a2)•(2bc-b2—c2+a2)

[blocks in formation]

..4c2.CP2={(b+c)2—a2 } · { a2—(c—b)2 }=(a+b+c) (−a+b+c) (a−b+c) (a+b—c)

[blocks in formation]

where s=(a+b+c)= half the sum of the three sides.

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