is equal to half of the rectangular parallelogram whose base is AB and whose perpendicular altitude is CD: whence,

the area of the triangle ABC = 1⁄2 of AB × CD

= 1⁄2 of (8 × 3) = 1 of 24 = 12;

that is to say, if the base and perpendicular altitude of a triangle be equivalent to 8 and 3 lineal units respectively, then will its area be represented by 12 superficial units of the same denomination; and it is of no consequence whether the dimensions be integral, fractional or irrational, as appears from Article (191).

193. If we take the four-sided figure ABCD, called a trapezium, and

D

G

II

find the lineal magnitudes of the diagonal BD, and of the perpendiculars AG and CH let fall upon it from the angles A and C, the area of the figure, being the sum of the areas of the two triangles ABD and BCD, may be ascertained.

=

Thus, if it be found that BD = 5, AG = 4, and CH = 1 lineal units, respectively; we shall have

the area of ABCD = the area of ABD + the area of BCD

and the same result must evidently have been obtained, if perpendiculars had been let fall upon the other diago

nal AC, from the angles B and D, because the area of the same figure cannot have two different magnitudes.

Similarly, the area of any rectilineal figure may be found by adding together the areas of the triangles which compose it.

194. Conversely, if the area of a rectangular parallelogram, or of a triangle, and either its base or perpendicular altitude, be given, the other of these magnitudes will manifestly be obtained by division.

Also, if the superficial units comprised in the area of a square, whose base is AB, be 1521; it is evident that

from which, by the extraction of the square root, we have AB = 39:

that is, if the area of a square surface be 1521 superficial units, every one of its sides will be 39 lineal units, which may be inches, feet, yards, &c.

Again, an acre, being a rectangular parallelogram 40 poles in length and 4 poles in breadth, contains 4840 square yards, and will therefore be equal to a square whose side√4840 69.57 &c. = 69 yards, nearly.

=

THE THEORY OF SOLID OR CUBIC MEASURE.

195. DEF. An Unit of solid or cubic measure, is a cube or rectangular parallelopiped whose length, breadth and thickness are each equal in magnitude to the lineal unit; as the solid af represented below, wherein ab= acad the lineal unit, which may be an inch, a foot, a yard, &c., as before, denotes the solid or cubic unit: and the solid content or volume of any other body of three dimensions will evidently be ascertained by finding what multiple, part or parts, it is of this unit, the lineal dimensions, or the length, breadth and thickness being supposed first to be numerically exhibited.

196. The numerical representative of the Solid Content or Volume of a rectangular parallelopiped, is equal to the continued product of the magnitudes representing its length, breadth and thickness.

Let ABFM represent a rectangular parallelopiped, whereof the length AB = 5, the breadth AC = 4, and the thickness AĎ = 3 lineal units, the denominations of the dimensions being the same in each, whether inches, feet, yards, &c.:

M

G

B

take AH = AK =

AL = the lineal unit, and complete the construction as in the diagram; then it is manifest that AG will be a cube, whose magnitude is equal to that of the solid unit af; and, by EUCLID, XI. 25, we have the parallelopiped AF: the parallelopiped AI :: CD: CL :: AD: AL :: 3 : 1; and the parallelopiped AF = 3 × the parallelopiped AI; also,

the parallelopiped AI: the parallelopiped AP :: - BI : - BP :: AC : AK :: 4 : 1; and the parallelopiped AI = 4 x the parallelopiped AP; again,

the parallelopiped AP : the parallelopiped AG :: □ BL : – HL :: AB : AH :: 5 : 1; and the parallelopiped AP = 5 x the parallelopiped AG; whence, we have now, the parallelopiped AF

= 3 × the parallelopiped AI

= 3 × 4 × the parallelopiped AP

=

3 × 4 × 5 × the parallelopiped AG;

but the parallelopiped AG being equal to the solid unit, is represented by 1; consequently, the numerical magni

tude of the rectangular parallelopiped, whose three contiguous edges are equivalent to 3, 4 and 5 lineal units, will be represented by

that is, the content of the parallelopiped ABFM AB × AC × AD = 5 x 4 x 3 = 60.

=

If the three edges AB, AC, AD be all equal to one another, and their magnitude be 3 lineal feet or 1 yard, the parallelopiped becomes a cube, whose magnitude 3 × 3 × 3 = 27 solid feet: that is, 27 solid or cubic feet are equal to 1 solid or cubic yard.

=

Similarly, 1728 cubic inches are equal to 1 cubic foot and so on for other denominations.

197. Hence, also may be found the length of an edge of the cube which is of equal solid content with any proposed parallelopiped or solid, whose dimensions or volume are given.

Thus, if a parallelopiped be 7 inches in length, 3 inches in breadth, and 18 inches in depth, its solid content will be

7 × 3 × 13 = 427 cubic inches,

which is therefore equal to the solid content of a cube whose edge

3

=√42=√42.875 = 3.5 = 3 lineal inches.

In the same manner, the edge of a rectangular parallelopiped may be found by dividing the solid content by the area of the surface to which it is at right angles; and vice versa.

198. It will not be necessary to pursue these subjects further in this place; and we shall here only insert directions for ascertaining the measures of such magnitudes as most frequently present themselves to our notice, without attempting their investigations, which more properly belong to other parts of Mathematics.

THE PRACTICE OF LINEAL MEASURE.

(1) Right-angled Triangle. The square root of the sum of the squares of the sides forming the right angle is equal to the Hypothenuse: and the square root of the difference of the squares of the hypothenuse and either side is equal to the other side.

(2) Circle. The circumference is equal to the product of twice the radius by 3.14159, nearly and the radius is equal to the quotient of the circumference by 6.28318, nearly.

(3) Hence, the homologous lines in similar triangles, and in all circles, are proportional.

Ex. 1.

If the base of a triangle be 1, and the perpendicular be 1, the hypothenuse = √12 + 12 = √2. If the base be√2, and the perpendicular be 1, the hypothenuse = √2+1 = √3.

If the base be√3, and the perpendicular be 1, hypothenuse = √3+1 = √4 = 2.

the

If the base be 2, and the perpendicular be 1, the hypothenuse = √4+1 = √√5, and so on: and in all these, only approximate arithmetical values of the surds can be found by evolution; also, it is worth noticing how all the primitive surds successively originate from these geometrical considerations, as has been hinted before at the end of Article (189).

Ex. 2. The wheels of a carriage are 2 yards asunder, and the inner wheel describes the circumference of a circle whose radius is 20 yards: find the difference of the paths of the two wheels.

The circumference of the inner circle = 3.14159 × 40: the circumference of the outer circle = 3.14159 × 45: whence, their difference will evidently = 3.14159 × 5 = 15.70795 yards = 15 yards, nearly.

Examples for Practice.

(1) Required the hypothenuse of a right-angled triangle whose sides are 24 and 32 feet.

Answer: 40 feet.

(2) Find the base of the right-angled triangle whose other sides are 4 and 48.

Answer: 4√2.

(3) If a ladder 103.44 feet long, be placed so as to reach a window 40 feet high on one side of a street,