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

the area of the triangle ABC =

=of (8 × 3) =

of AB × CD

of 24 = 12;

that is to say, if the base and 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 name; and it is of no consequence whether the dimensions be integral, fractional or irrational, as appears from Article (204).

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

find the numerical value of the diagonal BD and of each 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 by measuring we have found that BD = 5, AG = 4 and CH = 11⁄2 lineal units; we shall have

the area of ABCD = the area of ABD + the area of BCD = 1⁄2 BD × AG + ¦ BD × CH

= (5 × 4)+(5 × 1.5)

20 7.5

= +

=

2 2

= 13.75 = 133 superficial units;

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 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.

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

Also, if the superficial units comprised in the area of a square whose side is AB, be 1521; then,

AB2 = 1521:

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, each of its sides will be 39 lineal units.

Again, an acre, being a parallelogram 40 poles in length and 4 poles in breadth containing 4840 square yards, will be equal to a square whose side

69.57 &c. 69 yards, nearly.

=

4840

THE THEORY OF SOLID OR CUBIC MEASURE.

=

208. DEF. An Unit of solid or cubic measure is a cube or rectangular parallelepiped whose length, breadth and thickness are each equal to the lineal unit; as the solid af represented hereafter, wherein ab ac = ad= the lineal unit, denotes the solid or cubic unit: and the solid content or volume of any body of three dimensions will 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.

209. The numerical representative of the solid Content or Volume of a rectangular parallelepiped is equal to the product of the magnitudes representing its length, breadth and thickness.

Let ABFM represent a rectangular parallelepiped whereof the length AB = 5, the breadth AC-4 and the thickness AD = 3 lineal units, the denominations of the dimensions being the same in each:

M

R

G

B

take AH = AK = AL = the lineal unit and complete the construction as in the diagram; then AG will be a cube equal to the solid unit af; and by EUCLID, xi. 25, we have

the parallelepiped AF: the parallelepiped AI

:: CD: - CL :: AD : AL :: 3 : 1; and the parallelepiped AF = 3 x the parallelepiped AI; also, the parallelepiped AI: the parallelepiped AP :: - BI : - BP :: AC: AK :: 4 : 1;

and the parallelepiped AI-4 x the parallelepiped AP; again, the parallelepiped AP: the parallelepiped AG :: 0 BL : □ HL :: AB : AH :: 5 : 1;

and the parallelepiped AP = 5 x the parallelepiped AG; whence, we have the parallelepiped AF

=

3 × the parallelepiped AI

= 3 × 4 × the parallelepiped AP

= 3 × 4 × 5 × the parallelepiped AG;

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

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

If the three edges AB, AC, AD be equal to one another and their magnitude be 3 lineal feet or 1 yard, the parallelepiped 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.

foot.

Similarly, 1728 cubic inches are equal to 1 cubic

Also, it follows, from EUCLID, XI. 31, that the content of any parallelepiped is equal to the product of the area of its base and its altitude.

210. Hence may be found the length of an edge of the cube which is of equal solid content with any proposed parallelepiped.

Thus, if a parallelepiped be 7 inches in length, 3 inches in breadth and 12 inches in depth, its solid content will be

7 × 3 × 12 = 42 inches :

whence, the edge=423-42.875-3.5 = 34 inches. In the same manner the altitude of a parallelepiped may be found by dividing the solid content by the area of the base; and vice versa.

211. It will not here be necessary to pursue these subjects further; and we shall now only give directions for ascertaining the measures of such magnitudes as most frequently present themselves to our notice, without attempting their investigations, which belong to higher departments 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 Hypotenuse: and the square root of the difference of the squares of the hypotenuse 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 hypotenuse = √1 + 12 = √2.

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

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

If the base be 2 and the perpendicular be 1, the hypotenuse = √√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 (202).

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.

(1)

Examples for Practice.

Required the hypotenuse of a right-angled

triangle whose sides are 24 and 32 feet.

Answer: 40 feet.

(2) Find the diameter of a rectangle whose sides are 2 of an inch and .99 of an inch.

Answer: 1.01 inches.