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Let ABCD, AEFD, be two rectan- D gles having the common altitude AD: they are to each other as their bases. AB, AE.
Suppose, first, that the bases are commensurable, and are to each other, for example, as the numbers 7 and 4. If AB be divided into 7 equal parts, AE will contain 4 of those parts: at each point of division erect a perpendicular to the base; seven partial rectangles will thus be formed, all equal to each other, because all have the same base and altitude. The rectangle ABCD will contain seven partial rectangles, while AEFD will contain four: hence the rectangle ABCD is to AEFD as 7 is to 4, or as AB is to AE. The same reasoning may be applied to any other ratio equally with that of 7 to 4: hence, whatever be that ratio, if its terms be commensurable, we shall have
ABCD AEFD :: AB : AE.
Suppose, in the second place, that the bases D AB, AE, are incommensurable: it is to be shown that we shall still have
ABCD AEFD :: AB : AE.
For if not, the first three terms continuing the same, the fourth must be greater or less A than AE. Suppose it to be greater, and that we have ABCD AEFD :: AB : AO.
ABCD: AIKD : : AB : AI.
But by hypothesis we have
Divide the line AB into equal parts, each less than EO. There will be at least one point I of division between E and 0: from this point draw IK perpendicular to AI: the bases AB, AI, will be commensurable, and thus, from what is proved above, we shall have
ABCD AEFD :: AB: AO.
In these two proportions the antecedents are equal; hence the consequents are proportional (Book II. Prop. IV.); and we find
AIKD AEFD :: AI
But AO is greater than AI; hence, if this proportion is correct, the rectangle AEFD must be greater than AIKD: on the contrary, however, it is less; hence the proportion is impossible; therefore ABCD cannot be to AEFD, as AB is to a line greater than AE.
Exactly in the same manner, it may be shown that the fourth term of the proportion cannot be less than AE; therefore it is equal to AE.
Hence, whatever be the ratio of the bases, two rectangles ABCD, AEFD, of the same altitude, are to each other as their bases AB, AE.
PROPOSITION IV. THEOREM.
Any two rectangles are to each other as the products of their bases multiplied by their altitudes.
Let ABCD, AEGF, be two rectangles; then will the rectangle,
ABCD: AEGF :: AB.AD: AF.AE.
Having placed the two rectangles, H, so that the angles at A are vertical, produce the sides GE, CD, till they meet in H. The two rectangles ABCD, AEHD, having the same al ́titude AD, are to each other as their bases AB, AE: in like manner the two rectangles AEHD, AEGF, having the same altitude AE, are to each other as their bases AD, AF: thus we have the two proportions,
ABCD AEHD :
Multiplying the corresponding terms of these proportions together, and observing that the term AEHD may be omitted, since it is a multiplier of both the antecedent and the consequent, we shall have
ABCD AEGF :: AB× AD: AE× AF.
Scholium. Hence the product of the base by the altitude may be assumed as the measure of a rectangle, provided we understand by this product, the product of two numbers, one of which is the number of linear units contained in the base, the other the number of linear units contained in the altitude. This product will give the number of superficial units in the surface; because, for one unit in height, there are as many superficial units as there are linear units in the base; for two units in height twice as many; for three units in height, three times as many, &c.
Still this measure is not absolute, but relative: it supposes
that the area of any other rectangle is computed in a similar manner, by measuring its sides with the same linear unit; a second product is thus obtained, and the ratio of the two products is the same as that of the rectangles, agreeably to the proposition just demonstrated.
For example, if the base of the rectangle A contains three units, and its altitude ten, that rectangle will be represented by the number 3 × 10, or 30, a number which signifies nothing while thus isolated; but if there is a second rectangle B, the base of which contains twelve units, and the altitude seven, this second rectangle will be represented by the number 12 × 7= 84; and we shall hence be entitled to conclude that the two rectangles are to each other as 30 is to 84; and therefore, if the rectangle A were to be assumed as the unit of measurement in surfaces, the rectangle B would then have 3 for its absolute measure, in other words, it would be equal to 34 of a superficial unit.
It is more common and more simple, to assume the square as the unit of surface; and to select that square, whose side is the unit of length. In this case the measurement which we have
regarded merely as relative, becomes absolute: the number 30, for instance, by which the rectangle A was measured, now represents 30 superficial units, or 30 of those squares, which have each of their sides equal to unity, as the diagram exhibits.
In geometry the product of two lines frequently means the same thing as their rectangle, and this expression has passed into arithmetic, where it serves to designate the product of two unequal numbers, the expression square being employed to designate the product of a number multiplied by itself.
The arithmetical squares of 1, 2, 3, &c. are 1, 4, 9, &c. So likewise, the geometrical square constructed on a double line is evidently four times greater than the square on a single one; on a triple line it is nine times greater, &c.
The area of any parallelogram is equal to the product of its base by its altitude.
PROPOSITION V. THEOREM.
For, the parallelogram ABCD is equivalent F D to the rectangle ABEF, which has the same base AB, and the same altitude BE (Prop. I. Cor.): but this rectangle is measured by AB × BE (Prop. IV. Sch.); therefore, AB× BE A is equal to the area of the parallelogram ABCD.
Cor. Parallelograms of the same base are to each other as their altitudes; and parallelograms of the same altitude are to each other as their bases: for, let B be the common base, and C and D the altitudes of two parallelograms:
then, B×C: B×D: : C: D, (Book II. Prop. VII.) And if A and B be the bases, and C the common altitude, we shall have
AXC BXC :: A : B.
And parallelograms, generally, are to each other as the products of their bases and altitudes.
For, the triangle ABC is half of the parallelogram ABCE, which has the same base BC, and the same altitude AD (Prop. II.); but the area of the parallelogram is equal to BCX AD (Prop. V.); hence that of the triangle must be BC × AD, or BC × †AD.
PROPOSITION VI. THEOREM.
The area of a triangle is equal to the product of its base by half its altitude.
Cor. Two triangles of the same altitude are to each other as their bases, and two triangles of the same base are to each other as their altitudes. And triangles generally, are to each other, as the products of their bases and altitudes.
PROPOSITION VII. THEOREM.
The area of a trapezoid is equal to its altitude multiplied by the half sum of its parallel bases.
Let ABCD be a trapezoid, EF its altitude, AB and CD its parallel bases; then will its area be equal to EF× (AB+CD). H Through I, the middle point of the side BC, draw KL parallel to the opposite side AD; and produce DC till it meets KL.
In the triangles IBL, ICK, we have the side IB=IC, by construction; the angle_LIB=CIK; and since CK and BL are parallel, the angle IBL=ICK (Book I. Prop. XX. Cor. 2.); hence the triangles are equal (Book I. Prop. VI.); therefore, the trapezoid ABCD is equivalent to the parallelogram ADKL, and is measured by EFX AL.
But we have AL=DK; and since the triangles IBL and KCI are equal, the side BL=CK: hence, AB+CD=AL+ DK=2AL; hence AL is the half sum of the bases AB, CD; hence the area of the trapezoid ABCD, is equal to the altitude EF multiplied by the half sum of the bases AB, CD, a result AB+CD which is expressed thus: ABCD=EF× 2
Scholium. If through I, the middle point of BC, the line IH be drawn parallel to the base AB, the point H will also be the middle of AD. For, since the figure AHIL is a parallelogram, as also DHIK, their opposite sides being parallel, we have AH=IL, and DH IK; but since the triangles BIL, CIK, are equal, we already have IL IK; therefore, AH=DH.
It may be observed, that the line HI-AL is equal to AB+CD ; hence the area of the trapezoid may also be ex
pressed by EF HI: it is therefore equal to the altitude of the trapezoid multiplied by the line which connects the middle >oints of its inclined sides.