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 for its absolute measure, in other words, it would be equal to 34 of a superficial unit. A 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. G PROPOSITION V. THEOREM. The area of any parallelogram is equal to the product of its base by its altitude. 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. E C B 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, BXC 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. PROPOSITION VI. THEOREM. The area of a triangle is equal to the product of its base by half its altitude. 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 BC AD (Prop. V.); hence that of the triangle must be BC × AD, or BC ×÷AD. B D E 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 alti- DE tude, AB and CD its parallel bases; then will its area be equal to EFX (AB+CD). T Through I, the middle point of the side BC, draw KL parallel to the opposite side AD; and produce DC till it meets KL. A F C I L 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=EFX 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 2 pressed by EFX III: it is therefore equal to the altitude of the trapezoid multiplied by the line which connects the middle points of its inclined sides. PROPOSITION VIII. THEOREM. If a line is divided into two parts, the square described on the whole line is equivalent to the sum of the squares described on the parts, together with twice the rectangle contained by the parts. H D Let AC be the line, and B the point of division; then, is = = F B C The square ACDE is made up of four parts; the first ABIF is the square described on AB, since we made AF-AB: the second IDGH is A the square described on IG, or BC; for since we have AC AE and AB AF, the difference, AC-AB must be equal to the difference AE-AF, which gives BC=EF; but IG is equal to BC, and DG to EF, since the lines are parallel; therefore IGDH is equal to a square described on BC. And those two squares being taken away from the whole square, there remains the two rectangles BCGI, EFIH, each of which is measured by AB BC: hence the large square is equivalent to the two small squares, together with the two rectangles. Cor. If the line AC were divided into two equal parts, the two rectangles EI, IC, would become squares, and the square described on the whole line would be equivalent to four times the square described on half the line. Scholium. This property is equivalent to the property demonstrated in algebra, in obtaining the square of a binominal; which is expressed thus: (a+b)2=a2+2ab+b2. PROPOSITION IX. THEORFM. The square described on the difference of twe lines, is equivalent to the sum of the squares described on the lines, minus twice the rectangle contained by the lines. Let AB and BC be two lines, AC their difference; then is AC2, or (AP-BC)2=AB2+BC2-2AB × BC. Describe the square ABIF; take AE L F =AC; draw CG parallel to to BI, HK parallel to AB, and complete the square K EFLK. The two rectangles CBIG, GLKD, are each measured by AB× BC; take them away from the whole figure ABILKEA, which is equivalent to G I H E D AB+BC, and there will evidently remain the square ACDE; hence the theorem is true. Scholium. This proposition is equivalent to the algebraical formula, (a-b)2=a2-2ab+b2. PROPOSITION X. THEOREM. The rectangle contained by the sum and the difference of two lines, is equivalent to the difference of the squares of those lines. Let AB, BC, be two lines; then, will (AB+BC) x (AB-BC)-AB-BC?. On AB and AC, describe the squares ABIF, ACDE; produce AB till the produced part BK is equal to BC; and complete the rectangle AKLE. The base AK of the rectangle EK, is the sum of the two lines AB, BC ; its altitude AE is the difference of the same lines; therefore the rectangle AKLE is equal to (AB+BC) × (AB E F G I H D C B K BC). But this rectangle is composed of the two parts ABHE +BHLK; and the part BHLK is equal to the rectangle EDGF, because BH is equal to DE, and BK to EF; hence AKLE is equal to ABHE+EDGF. These two parts make up the square ABIF minus the square DHIG, which latter is equal to a square described on BC: hence we have (AB+BC) × (AB-BC)=AB2-BC2. Scholium. This proposition is equivalent to the algebraical formula, (a+b) x (a—b)=a2—b2. G* |