350 



THE POPULAR EDUCATOR. 



and height. Thus the area of the triangle A B G (Fig. 7) is 

 equal to half A B into G H, or its equal B c ; that is, 

 O = |ABXBC, or^AB x G H. 



For the area of the parallelogram ABCD is ABX BC; and 

 by Euclid I. 41, i? a parallelogram and a triangle are upon the 

 same base, and between the same parallels, the triangle is half 

 the parallelogram. 



Hence, an algebraical expression may be obtained for the 

 area of any figure whatever which is bounded by right lines. 

 For every such figure may be divided into triangles. 



D 



Fig. 7. 



Thus the right-lined figure ABODE (Fig. 8) is composed of 

 the triangles A B c, A c E, and BCD. 



The area of the triangle ABC^-ACXBL; 

 That of the triangle A c E = -- A c X EH; 



That of the triangle ECD = 4 EC XDG. 



The area of the whole figure is therefore equal to 



( A C X B L) + (^ A C X E H) -f (^- E C X D G). 



The expression for the superficies has here been derived from 

 that of a line or lines. It is frequently necessary to reverse 

 this order ; to find a side of a figure, from knowing its area. 



If the number of square inches in the parallelogram A B c D 

 (Fig. 3), whose breadth, B c, is 3 inches, be divided by 3, the 

 quotient will be a parallelogram, A B E F, one inch wide, and of 

 the same length with the larger one. But the length of the 

 small parallelogram is the length of its side, A B. The number 

 of sauare inches in one is the same as the mimber of linear 

 inches in the other. If, therefore, the area of the large parallelo- 

 gram be represented by a, the s-ide A B = ; that is, the 

 length of a parallelogram is found by dividing the area by the 

 breadth; and B c = . 



AB 



If a be put for the area of a square whose side is A B, 

 Then a = (A s) 2 ; 



And extracting both sides, Va = A B. 



That is, the side of the square is found by extracting the square 

 root of the number of measuring units in its area. 



If A B be the base of a triangle, and B c its perpendicular 

 height, 



Then a = ^BcXAB, or^AB X BC; 



And dividing by A B c, , = AB, and BC . 



2 BC ^ AB 



That is, the base of a triangle is found by dividing the area by 

 half the height, and the height by dividing tlie area by half the 

 base. 



As a surface is expressed by the product of its length and 

 breadth, the contents of a soacJ may be expressed by the pro- 

 ! duct of its length, breadth, and depth. It is necessary to bear 

 in mind, that the measuring unit of solids is a cube ; and that 

 the side of a cubic inch is a square inch ; the side of a cubic 

 foot, a square foot, etc. 



Let A s c D (Fig. 3) represent the base of a parallelepiped, 

 five inches long, three inches broad, and one inch deep. It is 

 evident there must be as many cubic inches in the solid, as 

 there are square inches in its base. And as the product of the 

 lines A B and B c gives the area of this base, it gives, of course, 

 the contents of the solid. But suppose that the depth of the 

 parallelepiped, instead of being one inch, is four inches, its 

 contents must be four times as great. If. then, the length be 

 A B, the breadth B c, and the depth C O, the expression for the 

 Bolid contents will be A B X B c X CO. 



By means of algebraical notation, a geometrical demon- 



stration may often be rendered much more simple and concise 

 than in ordinary language. The proposition (Euclid II. 4), that 

 when a straight line is divided into two parts, the square of the 

 whole line is equal to the squares of the two parts, together 

 with twice the product of the parts, is demonstrated by 

 squaring a binomial. 



Let the side of a square be represented by s ; 

 And let it be' divided into two parts, a and b. 

 By the supposition, s = a + b ; 



And squaring both sides, s 2 = a 2 -f- 2ab + b 2 . 



That is, s 2 , the square of the whole line, is equal to o 2 and b 2 , 

 the squares of the two parts, together with 2ab, twice the pro- 

 duct of the parts. 



Algebraical notation may also be applied with great ad- 

 vantage to the solution of geo- 

 metrical problems. In doing this 

 it will be necessary, in the first 

 place, to form an algebraical equa- 

 tion from the geometrical rela- 

 tions of the quantities given and 

 required ; and then by the usual 

 reductions, to find the value of the 

 unknown quantity in this equa- 

 tion. 



EXAMPLES. 



1. Given b the base, and a the 



sum of the hypothenuse and perpendicular of the right-angled 

 triangle ABC (Fig. 9), to find the perpendicular B c. 



Let x = the perpendicular B c. The sum of hypothenuse and 

 perpendicular, x + A c = a. Then transposing x, A c = a x. 



(1.) By Euclid I. 47, (B c) 2 + (A s) 2 = (A c) 2 . 



(2.) That is, by the notation, # 2 -f- b i = (a - x)- = a"- 2ax+ a; 2 . 



Fig. 9. 



And x = 



= B c, the side required. Hence, 



In a right-angled triangle, the perpendicular is equal to the 

 square of the sum of the hypothenuse and perpendicular, dimin- 

 ished by the square of the base, and divided by twice the sum of 

 the hypothenuse and perpendicular. 



It is applied to particular cases by substituting numbers 

 for the letters a and b. Thus, if the base .is 8 feet, and the 

 sum of the hypothenuso and perpendicular 16, the expression 



2 72 1 fi2 QC 



~ becomes = 6, the perpendicular ; and this sub- 



2a 2 X 16 



tracted from 16, the sum of the hypothenuse and perpendicular, 

 leaves 10, the length of the hypothenuse. 



2. Given the base of a right-angled triangle ABC (Fig. 10) 

 = b, and the difference between the hypothenuse and perpen- 

 dicular = d, to find the perpendicular B c. Apply this where 

 b = 20 and d = 10. 



D me 



Fig. 10. 



Fig. 1L 



Let B c, the perpendicular, = x ; then A c, the hypothenuse, 

 = x + d. Now, by Euclid I. 47, (A c) 2 = (A s) 2 + (B c) 2 ; and 

 by substitution, (x + &)* = 6 2 + as 2 , or as 2 + 2dx + d 2 = 6 2 + a; 2 . 



52 _ d i 400 - 100 

 .-. x = - _ _ = = ID = BC. 



2d 



20 



LESSONS IN ENGLISH LITERATURE. XXIV. 



SWIFT. 



IT would scarcely be possible to point to a more strongly-marked 

 contrast than that between the character and career of the great 

 humorist of whom we spoke in our last lesson, and the still 

 greater, whose stormy, unhappy life, and strange, powerful 

 genius we have to consider in the present. 



Jonathan Swift was of English descent ; but his father having 

 held an office in Dublin, the son was born in that city, a 

 \ osthumous child, in 1667. His childhood was passed amid 



