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the remaining sub-divisions, which are thus proved to be respectively, 3-10ths, 4-10ths, &c., of the value assigned to C 9.
In order to take any distance off the scale-suppose 2.45–place one foot of the compasses at D, and extend the other foot to the division on DC, marked 4. Then move the compasses upward, still keeping one foot on the line D B, till the line marked 55 is reached. Extend the compasses to that point at which 55 is intersected by the diagonal 4 s, and the required length, 2.45, will be obtained. Distances too great to be measured directly from the scale, can, of course, be readily made up in aggregate parts.
To find the centre of a given circle.
Draw within the circle any straight line, as A B, and bisect it at the point D. Through this point draw the perpendicular E F, which will be a diameter of the circle, and bisect E F at C. C is the centre of the given circle. (Euclid, Book III, Prop. 1.)
2nd Method.-Draw any two chords as A B, and A C. From A, B, and C as centres, with any radius greater than v half of either line, describe four arcs, intersecting in mn and st. Draw m n and s t and produce them until they meet. Their point of intersection, 0, is the centre of the of circle.
This method is the same in principle as the preceding, but the bisection of two chords increases the accuracy of the construction.
An arc, or portion of the circumference of a circle being given, to find the centre of that circle.--Let B A C (last fig.) be the given arc. From any point in it, A, draw straight lines A B, A C, to its extremities. Bisect A B, A C, by the perpendiculars m n, s t, and produce these lines until they meet. The point of intersection, 0, is the centre of the circle of which B A C is an arc. (Euclid III., 25.)
The mode of describing an ellipsis and parabola will be explained a little further on, when treating of the areas of those figures.
(10.) COMPUTATION OF AREAS.—The term area has the same meaning as superficies or quantity of surface, but is applied in Mensuration exclusively to the space comprised within the boundaries of plane figures. Thus, it is customary to speak of the extent of « surface of a sphere or cylinder,” the « amount of area of a rectangle or trapezium." Although the word “ area" should properly be confined to expressions of superficial magnitude only, yet, in the practice of gauging, it is convenient, for reasons which will be explained in the next chapter, to call by this name, what is really the content or capacity of regular vessels, such as maltcisterns, at one inch of their depth.
Areas in general are estimated numerically, by ascertaining the linear dimensions of the figure in certain directions, and then inferring from the relation which geometry teaches us subsists between the parts so measured, how often the squaro described on the linear unit, whatever that may be, is contained in the given superficies. The knowledge of an area, therefore, is arrived at, not by the trouble. some and often impracticable process of directly applying a measuring unit of surface to each particular figure, and finding, by actual trial, how many times, and parts of a time, it is necessary to repeat that unit in order to make up the space before us ; but simply by determining the length of certain lines, and computing the superficial extension of the figure from the data so obtained.
All measurement, it should be borne in mind, consists in the comparison of one magnitude with another of the same kind,-a line with a line, a surface with a surface, a solid with a solid. No intelligible relation or proportion can be said to exist between magnitudes of different species. It would be irrational to consider, for instance, which is greater or less, a foot in length, or a square inch ; a cubic foot or a square yard. For surfaces, the unit of measure is invariably a square, constructed on the linear unit as a side—a square inch, a square foot, or, generally, the square of some known unit of length. For solids, the measuring unit must be itself a solid ; that which is universally adopted is the cube formed upon the linear unit as an edge, or on the superficial unit as a base. Thus, it appears there are three distinct units of measure which can never be substituted the one for the other,-a known length for the measurement of lines, a known square for tbe measurement of surfaces, and a known cube for the measurement of solids. The extension of any magnitude, or the quantity of space comprised within its limits, is expressed numerically by stating how often it contains its peculiar unit of measure.
To determine the number of square units contained in the area of any plane rectilinear figure, we apply the principle laid down in the following theorem :
If the sides of a rectangle be measured in linear units, such as inches, feet, Sc., the number of superficial units in its area, corresponding to the linear unit so employed, will be found by multiplying the number of linear units in its height or length by the number of linear units in its base or breadth. Or, more shortly, the area of a rectangle in square units is the product obtained by multiplying together the linear units in its length and breadth respectively.
It being understood, that by the words “ length” and « breadth” are always signified the number of linear units in each of those dimensions, it is possible to present the foregoing theorem in a still more compendious and easily recollected form, viz. :
Area of Rectangle = Length x Breadth. Whichever of the dimensions is taken as the multiplier must in strictness bo regarded as an abstract number,-not as representing so many inches or feet, &c., but as standing for a certain number of times. The length of a rectangle, when repeated as many times as there are inches, &c., in the breadth, or vice versa, furnishes the number of square inches, &c., in the surface. It is customary to say, for example, that, “ 12 inches long by 9 inches broad, gives 108 area.” Such an expression as this, rationally interpreted, means that a rectangle 12 inches long and 1 inch broad, contains 12 square inches; and if the breadth be increased to 9 inches, the rectangle will have a total area of 9 x 12 or 108 square inches.
It is of great importance to possess a clear conception of the truth of this leading axiom in mensuration, as it constitutes the basis of all the processes used to determine the magnitude of plane rectilineal figures of whatever shape or variety of outline.
To illustrate the case of dimensions both in whole numbers and fractions :
(1.) Suppose a rectangle as A B D C in the margin, having the side A C, 5 inches, and CD, 4 inches long. Divide AC and B D, which by the property of the rectangle are of equal length, into 5 inches each, and A B and CD, which are also equal, into 4 inches each, and connect the opposite points of division by parallel lines as shown in the drawing. The entire figure is thus divided into 20 equal squares, as A E, each of which has obviously the area of one square inch.
(2.) Next, let the dimensions of the opposite figure ABCD, be supposed to be-A C 6%, and A B 5.1 inches. Produce A C to F, so that A F shall be three times the length of A C, and produce A B to E, so that A E shall be twice the length of A B. Then complete the figure A F G E as in H the diagram. Now, since A F is three times as long as A CC it is equal to 3 x 69 or 20 inches ; and since A E is double the length of A B it is equal to 2 x 54 or 11 inches. The entire figure or rectangle A F G E, contains therefore, as above proved, 20 X 11 or 220 square inches, and consists of six rectangles, each having the same dimensions and area as A B C D. Accordingly, A B C D must have one-sixth part of the area of AFGE, that is, it must contain 220 - 6 or 365 square inches. But 364 is the product of 6, the Fi. length of A C, multiplied by 51, the length of A B. It is thus seen, that whether the sides of a rectangle be expressed in whole or fractional values, its area or the number of square inches in its surface will be found by multiplying together the dimensions of two of its adjacent sides.
That which is here shown to be true of rectangles or right-angled parallelograms specially, viz. :-That the area = the product of the beight x the base, may now be proved to hold good as respects oblique parallelograms also.
By Euclid, Book I, Prop. 35, it is a established that “parallelograms upon the same base and between the same parallels are equal to one another." Now, the perpendicular distance, D C, between the sides, DF and B C, of the oblique parallelogram B DFC, represents the height of that figure, while B C may be taken as its base. Complete the rectangle C D A B, the area of which is found by multiplying together the lengths of B C and D O. But as the rectangle and the oblique parallelogram stand on the same base, BC, and between the same parallels, B C and A F, their areas, by the prop. above quoted, must be equal, and consequently the area of BDFC will be obtained in the same way as that of its equivalent rectangle-by multiplying the base B C by the height D C.
The area of a square is, of course, the product of any side multiplied into itself.
Areas of Parallelograms. GENERAL ROLE.—Multiply the length by the breadth, or the base by the height, and the product will be the area in square units.
Example (1.) Find the area of a rectangle 37 inches long and 32 inches broad.
Answer. 1,184 square inches. Example (2.) The side of a square is 427 inches. What is its area in square feet ?
Answer. 12 77 square feet. Example (3.) Required the area of a rhombus, the side of which is 7 feet 6 inches, and the perpendicular height 3 feet 4 inches ?
Answer. 25 square feet. Example (4.) If the area of a rectangle 49 feet long be 236 square yards, 7 feet, 72 inches, what must be its breadth ?
Answer. 43.5 feet. NOTE. As the area of a rectangle is the product of its length x its breadth, its breadth must be the quotient of the area divided by the length.
Areas of Triangles.-Every triangle is exactly one half of a parallelogram on the same base and between the same parallels (Euclid, Book I., Prop. 41). It follows, therefore, from what has been above stated respecting parallelograms, that the area of any triangle, the base and perpendicular height of which can be determined, may be found by multiplying together those dimensions and taking half the product.
Example. Let it be required to compute the area of the triangle A B C, having given the base A B = 12 inches, and the altitude or perpendicular height C D = 10 inches. Complete the parallelogram A m as shown by the dotted lines. This figure is the double of the triangle A B C, as it stands on the same base A B, and between the same parallels, AB and C m, (Euclid, I., 41). But the area of the ASD parallelogram is A B X CD. Therefore the area of the triangle is A B x CD.
By pursuing a similar construction, triangles of overy form can, it is evident, be readily extended into parallelograms of double the size, and hence the generality of the method just laid down for determining the areas of the former. Accordingly, we have :
RULE I. (Applicable to all ordinary cases.) Multiply (the units of length in) any side taken as base, by (the units of length in) the perpendicular let fall upon that side from the opposite angle. Half the product will be the area required. In order to draw the perpendicular with exactness, when the given triangle is not right-angled, we may adopt the following process :
Bisect either of the two sides containing the angle from which the perpendicular is to be let fall, as B C, in the point m. Then with the distance in B or m o as radius, and from m as a centre, describe an arc cutting A C (or A C produced, as in the second figure, where the perpendicular would not fall within the triangle) in the point D. The line joining B and D will be perpendicular to A C, or A C produced. *
• The proof of this method depends on Euclid, Book I., Prop. 5, in which it is shown that the angles at the base of an isosceles trianglo are equal to one another (in the prosent instance, the angle m C D, and the angle m D C that would be formed by drawing a line from m to D; and also the angles m D B and m B D); and, secondly, on the consideration, that if one angle of a triangle be equal to the sum of the other two angles, (B DO = DBC + BCD) that angle must be a right angle, as all the three angles are equal to two right angles. (Euclid I., 32.)
It thus appears, that the area of any triangle, such as A B C, may be found by taking any of the three sides as base, and drawing a perpendicular to it from the opposite angle; multiplying together the dimensions so obtained, and dividing the product by 2; that is, the area of the triangle A BC= A B x perpen, from C or, BC x perpen. from A, or, AC x perpen. from B.
Instead of first forming the product of the base x the perpendicular, and then dividing by 2, it will amount to the same thing, and may sometimes be more convenient, to multiply half the base by the perpendicular, or balf the perpendicular by the base. Where one of the sides of a triangle is horizontal, as A C, it will generally prove the least troublesome plan, to make that side the base.
Having given the three sides of a triangle to calculate its area, and also the length of the perpendicular let fall upon any side from the opposite angle.
RULE II.-Add together the three sides and take half the sum. From this half-sum subtract each side separately, and multiply the half-sum and the three remainders into one another. Extract the square-root of the final product, and the result will be the area of the triangle.
The area being thus determined, in order to find the perpendicular to any side, divide twice the area by that side.
It is not possible, without some knowledge of elementary algebra and geometry to understand the reason of RULE II. A demonstration is inserted in most modern works on Mensuration.
Examples by the two methods :
Example (1.) In the first of the preceding figures, let the side or base A C be given = 18.4, and the perpendicular B D = 13.2 units; required the area of the triangle A B C. By Rule I.: AC = 18.4 AO= 18.4 .
BD= 132 BD= 13.2 À BD= 6:6
AO= 9.2 368 1104
264 2392 1104
1188 2)242.88 121.44 121.44
Answer. 121.44 square units. Next, let the sides of the same triangle be given respectively, -AC= 18.4, AB= 16.5, and BC= 15.7 units. Required the area, and also the perpendicu. lar B D. By Rule II.
Sum 25.3 25.3 25-3 16.5
Sides 18.4 16.5 15.7 15.7
Diffs. 6.9 8.8 9.6 2)50-6 Sum of sides = 25.3 And multiplying together the half-sum and these three differences, we have as the product, 14747.6736, the square-root of which is 121.44, the area as before ; that is, 25-3 X 6.9 x 8.8 x 9.6 = 14747.6736, and A/14747-6736 = 121.44 area.