II. TABLE OF SURFACE MEASURES. 144 Square Inches make 1 Square Foot. 4 Roods make 1 Acre. 640 Acres make 1 Square Mile. THEOREM I. Parallelograms having equal bases and equal altitudes are equal. Let the parallelograms, A B CD and EFG H, have equal bases and equal altitudes; then will they be equal. For, let E G be applied to A C so that their lower bases shall coincide. Since they have equal altitudes their upper bases will lie in D C, or D C produced; and E G will take the position A I. In the triangles t and t', Denote the whole quadrilateral, A BID, by Q. Cor. I. Rectangles having equal bases and equal altitudes are equal. For a rectangle is a species of parallelogram, and what is true of the whole is true of a part. Cor. II. A parallelogram is equal to a rectangle having an equal base and equal altitude. QUERIES. 1. What are the essential points in the demonstration proper?- First, to superpose the parallelograms; second, to prove t and t'equal; third, to take t and t' from the entire figure. 2. What principles are used in proving the equality of t and t'? 3. What axiom is involved in subtracting t and t' from the quadrilateral? THEOREM II. Rectangles having equal altitudes are proportional to their bases. There may be two cases: the bases may be commensurable, or they may be incommensurable. R b will R R' :: b : b'. R First: Let R and R' be two rectangles having equal altitudes, whose bases, b and b', are commensurable; then For, suppose the common unit of measure to be contained in b five times, and in b' three times. Then, evidently, bb' 5: 3. Now, conceive b to be divided into five equal parts, and b′ into three, each equal to the common unit. At the points of division, erect perpendiculars to the bases. R will then be divided into five rectangles, and R' into three, all equal, each to each, since they have equal bases and equal altitudes (Th. I, Cor. I); hence, But, R: R':5: 3. b: b' :: 5:3; R: R'b: b'. Second: Let R and R' be two rectangles whose altitudes are For, if the fourth term of the proportion is not b', it is either greater or less than b'. Suppose, then, that R: R'b: AB; in which AB is greater than b′. Conceive b to be divided into equal parts, each less than B C; and let one of these parts be applied, as a measure, to AB, commencing at A. Between B and C there will be at least one point of division, as D. At D, erect the perpendicular D E, meeting the upper base produced, in E. Since b and AD are commensurable, we have, by the preceding case, R: AE b: AD. But, by hypothesis, R: R':: b: A B; hence, But, DE, R' AE: AB: AD Ch. III, Th. VII. : R'<AE; .. AB < AD, which is false. Since the conclusion is false, the supposition which led to it is false; that is, the fourth term of the proportion cannot be greater than b′. By a similar process, it may be shown that the fourth term cannot be less than b'; and, since it can be neither greater nor less, it must be equal to b'. .. R R': b : b'. Q. E. D. Cor.-Rectangles having equal bases, are to each other as their altitudes. For, the bases may be considered as the altitudes and the altitudes as the bases. THEOREM III. Any two rectangles are proportional to the products of their bases by their altitudes. b' For, construct a third rectangle whose alti tude shall be that of R, and whose base shall be that of R'. Since R'' and R' have equal bases, we have Then, by Theorem I, (1) R R": b: b'. What Problem of Ch. V is involved in the construction of R"? THEOREM IV. The area of a rectangle is equal to the product of its base by its altitude. the unit of length, construct a square, as S. Since a square is a species of rectangle, we shall have, by Theorem III, RS: ab: a'b'. But a' and b' are each equal to the linear unit, and S is there fore the superficial unit, or unit of area; hence, That is, the number of superficial units in the rectangle equals the number of linear units in the base multiplied by the number of linear units in the altitude. Q. E. D. Cor. The area of a square is equal to the second power of one of its sides. For, in this case, the base and the altitude are equal. THEOREM V. The area of any parallelogram is equal to the product of its base by its altitude. Let A B be any parallelogram, b its base, and a its altitude; then will AB = a b. For, construct the rectangle A C, A b having the same base and altitude as the parallelogram. C B Cor.-Parallelograms are to each other as the products of their bases by their altitudes. For, let P and P' denote any two parallelograms, b and b' their bases, a and a'their altitudes. Then, by the Theorem just proved, |