COMPARISON AND MEASUREMENT OF THE SURFACES OF RECTILINEAR FIGURES. 1. DEFINITION. The area of a surface is its numerical measure, referred to some other surface as the unit; in other words, it is the ratio of the surface to the unit of surface (II. 43). The unit of surface is called the superficial unit. The most convenient superficial unit is the square whose side is the linear unit. 2. Definition. Equivalent figures are those whose areas are equal. PROPOSITION I.-THEOREM. 3. Two rectangles having equal altitudes are to each other as their bases. Let ABCD, AEFD, be two rectangles having equal altitudes, AB and AE their bases; then, ABCD AB F D C = AEFD AE D F А E Suppose the bases to have a common measure which is contained, for example, 7 times in AB, and 4 times in AE; so that if AB is divided into 7 equal parts, AE will contain 4 of these parts; then, we have If, now, at the several points of division of the bases, we erect perpendiculars to them, the rectangle ABCD will be divided into 7 equal rectangles (I. 120), of which AEFD will contain 4; consequently, we have The demonstration is extended to the case in which the bases are incommensurable, by the process already exemplified in (II. 51) and (III. 15). 4. Corollary. Since AD may be called the base, and AB and AE the altitudes, it follows that two rectangles having equal bases are to each other as their altitudes. Note. In these propositions, by "rectangle" is to be understood "surface of the rectangle." PROPOSITION II.-THEOREM. 5. Any two rectangles are to each other as the products of their bases by their altitudes. tangle R, and the same altitude h' as the rectangle R'; then we have, by (4) and (3), and multiplying these ratios, we find (III. 14), 6. Scholium. It must be remembered that by the product of two lines, is to be understood the product of the numbers which represent them when they are measured by the linear unit (III. 8). PROPOSITION III.-THEOREM. 7. The area of a rectangle is equal to the product of its base and altitude. Let R be any rectangle, k its base and h its altitude numerically expressed in terms of the linear unit; and let be the square whose side is the linear unit; then, by the preceding theorem, R k area, of the rectangle R (1); therefore, Area of RkX h. 8. Scholium I. When the base and altitude are exactly divisible by the linear unit, this proposition is rendered evident by dividing the rectangle into squares each equal to the superficial unit. Thus, if the base contains 7 linear units and the altitude 5, the rectangle can obviously be divided into 35 squares each equal to the superficial unit; that is, its area 5 X 7. The proposition, as above demonstrated, is, however, more general, and includes also the cases in which either the base, or the altitude, or both, are incommensurable with the unit of length. 9. Scholium II. The area of a square being the product of two equal sides, is the second power of a side. Hence it is, that in arithmetic and algebra, the expression "square of a number" has been adopted to signify "second power of a number." We may also here observe that many writers employ the expression " rectangle of two lines" in the sense of "product of two lines," because the rectangle constructed upon two lines is measured by the product of the numerical measures of the lines. 10. The area of a parallelogram is equal to the product of its base und altitude. Let ABCD be a parallelogram, k the numerical measure of its base AB, h that of its altitude AF; and denote its area by S; then, S=kXh. For, let the rectangle ABEF be constructed having the same base and alti h F D E C A k B tude as the parallelogram; the upper bases of the two figures will be in the same straight line FC (I. 58). The right triangles AFD and BEC are equal, having AF = BE, and AD = BC (I. 83). If from the whole figure ABCF we take away the triangle AFD, there remains the parallelogram ABCD; and if from the whole figure we take away the triangle BEC, there remains the rectangle ABEF; therefore the surface of the parallelogram is equal to that of the rectangle. But the area of the rectangle is k X h (7); therefore that of the parallelogram is also kXh; that is Sk× h. 11. Corollary I. Parallelograms having equal bases and equal altitudes are equivalent. 12. Corollary II. Parallelograms having equal altitudes are to each other as their bases; parallelograms having equal bases are to each other as their altitudes; and any two parallelograms are to each other as the products of their bases by their altitudes. PROPOSITION V.-THEOREM. 13. The area of a triangle is equal to half the product of its base and altitude. Let ABC be a triangle, k the numerical measure of its base BC, h that of its altitude AD; and S its area; then, S=kXh. E A B C D For, through A draw AE parallel to CB, and through B draw BE parallel to CA. The triangle ABC is one-half the parallelogram AEBC (I. 105); but the area of the parallelogram=kXh; therefore, for the triangle, we have S = 1 k × h. 14. Corollary I. A triangle is equivalent to one-half of any par allelogram having the same base and the same altitude. 15. Corollary II. Triangles having equal bases and equal altitudes are equivalent. 16. Corollary III. Triangles having equal altitudes are to each other as their bases; triangles having equal bases are to each other as their altitudes; and any two triangles are to each other as the products of their bases by their altitudes. PROPOSITION VI.-THEOREM. 17. The area of a trapezoid is equal to the product of its altitude by half the sum of its parallel bases. Let ABCD be a trapezoid; MN = h, its altitude; AD a, BC b, its parallel bases; and let S denote its area; then, E A M D F S = 1 (a + b) × h. B N For, draw the diagonal A C. The altitude of each of the triangles ADC and ABC is equal to h, and their bases are respectively a and b; the area of the first is a Xh, that of the second is b× h; and the trapezoid being the sum of the two triangles, we have Sah + 1 b × h = 1 (a + b) xh. 18. Corollary. The straight line EF, joining the middle points of AB and DC, being equal to half the sum of AD and BC (I. 124), the area of the trapezoid is equal to the product MN × EF. 19. Scholium. The area of any polygon may be found by finding the areas of the several triangles into which it may be decomposed by drawing diagonals from any vertex. The following method, however, is usually preferred, especially in surveying. Draw the longest diagonal AD of the proposed polygon |