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On the numerical relation between the area of

rectangles and the length of their sides. If A B C D be a rectangle, the sides of which contain

respectively 4 and 5 times a certain unit of length (A E), and the sides be divided into 4 and 5 of this unit, and lines be drawn through the points of division parallel to the sides of the rect

angle, it is obvious that the whole A E

rectangle will be divided into 4 x 5 equal squares, each of which is the square on A E.

It will be easily seen that the same relation holds good for any other numbers. That is (using algebraical notation), if the sides that contain a rectangle are divisible exactly into a and b linear units respectively (a and b being whole numbers), the rectangle will contain ab times the square on that unit. The same relation holds good if a and b are fractional

3 numbers. Suppose the sides of a rectangle contained 7

and 23 5 of the linear unit. If we divide the unit into 35 equal parts, the sides will contain respectively 3 x 5, or 15, and 23 x 7, or 161 of these sub-units, and the rectangle will contain (3 x 5) x (23 x 7), or 15 x 161 times the square on this subunit. But since it takes 35 x 35 times the square on the sub-unit of length to make up the square on the original

15 x 161 unit, it follows that the rectangle contains

35 x 35 (3 x 5) x (23 x i.e.

times the square on the original (3 x 5) x (23 x 7) 3 x 23 unit. But

Therefore, (5 x 7) x ( 5 x 7) 7 x 5° here again the whole or fractional number of times that the rectangle contains the square on the linear unit is obtained by multiplying the fractional number that denotes how much of the linear unit is contained in one side, hy

3)

m

n

the fractional number that denotes how much of the linear unit is contained in the other side.

As it is obvious that similar reasoning appiies to every case, it may be stated generally, that if the sides of a rectangle contain

and

P of a certain linear unit, the rect

9

тр angle contains of the square on the linear unit. Or,

ng if a and b (whole or fractional) denote the lengths of two sides of a rectangle in terms of some linear unit, ab denotes the area of the rectangle in terms of the square on that unit. If the sides are equal, and each contains a linear units, the area contains asquare units.

It is sometimes stated loosely that the area of a rectangle is the product of the sides that contain it. This, taken literally, has no signification whatever. It is absurd to talk of multiplying a line by a line. Before any sense can be attached to the words, they must be translated into the proposition previously stated. The sign x can only be placed between symbols that denote numbers. If ever the expression AB x CD be employed, A B and CD being two lines, it must be understood that, for the purpose in hand, A B and C D are numbers, denoting the number of linear units contained in A B and C D respectively.

The following propositions are obviously true :-“Those magnitudes are equal which are represented in terms of the same unit) by equal numbers, or numerical expressions;” and the quantity of space denoted by a number is independent of the form of the space; m square units make the same quantity of space, whatever the way in which they are arranged or grouped together. Hence (for example), if a + b + cd= m (that is, if the final result of the operation indicated on the first side be m), the quantity of space filled by m square units will be equal to what we get if to the space occupied by a square units we add successively the spaces occupied by b and c square units, and then take away the space occupied by d square units. Thus, if a = m + n, the area denoted by ab is equal to the area denoted by 6 (m + n), because the numbers ab and b (m + n) are equal. But b (m + n) = bm + bn. Consequently, the quantity of space represented by b (m + n) is equal to the quantity of space represented by bm + bn; and therefore the space represented by ab is equal to the quantity represented by bm + bn. But bm denotes the area of a rectangle whose sides contain respectively b and m units; and bn denotes the area of a rectangle whose sides contain respectively b and n units. Hence, the area of a rectangle, whose sides contain respectively a and b units, is the same as the sum of two rectangles whose sides contain respectively b and m and b and n linear units, when a = m + n.

The application of these principles enables us to exhibit the propositions of the Second Book of Euclid in a very

easy form.

PROPOSITION I. Let two lines, A and B, contain a and b linear units, and let b = m + n + p ; that is, let B be divided into parts containing respectively m, n, and p units. The rect. A, B contains ab square units.

But ab = a (m + n + p) and a (m + N + p) = am + ап + ар ..

ab = am + an + ap; and therefore the area denoted by ab is equal to that denoted by am + an + ap.

Now the area denoted by am + an + ap is the same as that of the sum of three rectangles whose sides contain respectively a and m, a and n, and a and p linear units.

Therefore, the rectangle contained by A and B is equal to the sum of the rectangles contained by B and the several parts of the divided line.

Propositions II. and III. may be proved in a similar manner. It was shown (p. 139), that they are only particular cases of Prop. I.

PROPOSITION IV. (See p. 142.) Let A B contain a linear units, and let a = b + c. The square on A B contains a? square units. But since a = b + c, a= (b + c)2 ...a = 62 .t cz + 2 bc.

But the areas represented by equal numbers are equal, therefore the area denoted by a’ is equal to the area denoted by b2 + c2 + 2 bc.

But this area is equal to the area denoted by 52 + the area denoted by c + the area denoted by 2 bc; that is, to the on a line containing b linear units + the on a line containing c linear units + twice the rectangle under these two lines. Therefore the square on A B is equal to the sum of the squares on the two parts of A B, together with twice the rectangle under the parts.

PROPOSITION V. (See p. 144.) Let A C or C B contain a linear units, and CD b linear units. A D contains a + b, and D B contains a b units. (a + b) (a - b) = a62. Consequently, the space represented by (a + b) (a --6) is equal to the space represented by a? 62.

But (a + b) (a b) represents the area of a rectangle whose sides contain respectively a + b and a

b linear units; a' represents the upon a line containing a linear units, b2 the upon a line containing 6 linear units.

Therefore the rectangle contained by the sum and the difference of two lines is equal to the difference between the squares on the lines.

PROPOSITION VI. (See p. 147.) The only difference between this proposition and the last is that b is larger than a, so that we get (b + a) (6 a) = 12 — a?. But in this case, as in the last, b + a is the sum of the lines, and b - a is their difference.

PROPOSITION VII. (See p. 149.) Let A B contain a linear units, and BC 6 of the same. AC contains a b units. (a 5)2 = a2 + b2 — 2 ab .. (a - b)2 + 2 ab = a + 62.

Since the numerical expression on the first side of the equation stands for the same number as the numerical expression on the second side, the number of square units (or the quantity of space) represented by the first side is equal to the number of square units (or the quantity of space) represented by the second side.

(a - b)2 represents the on AC; 2 ab represents twice the rect. A B, BC; a2 + b2 represents the sum of the Os on A B and BC. Therefore on A B + 0 on BC= 2ce rect. A B, BC + on A C.

PROPOSITION IX. (See p. 152.) Let A D contain a, and D B 6 linear units. A C contains a + 6

'a 6 and CD contains

units. 2

2 a + b

a? + 32 + 2 ab al + 62 2 ab +

+
2
4

4
a +

2 a2 + 2 62 a+ 62 + 2

2 ... 2

+ 2

=a? + 62.

The quantity of space denoted (numerically) by the first side is equal to the quantity denoted (numerically) by the second side. a? + 62 denotes the sum of the squares on lines containing respectively a and b units, that is, the sum of the squares on A D and D B. a +

denotes the on half the sum of the same lines

2

2

2

2

and

denotes the on half their difference. Therefore the sum of the Os on A D and D B is equal to twice the on A C (half their sum), together with twice the

on CD (half their difference). Proposition X, may be proved in a similar manner.

In what precedes it has been assumed that the lines dealt with are commensurable, that is, admit of being expressed exactly in terms of some common unit. Now it is

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