BOOK IV AREAS OF POLYGONS 332. DEF. The unit of surface is a square whose side is unit of length, e.g. a square inch, a square metre, etc. 333. The area of a surface is the number of units of su it contains. 334. Two figures are equivalent if their areas are equal. PROPOSITION I. THEOREM 335. Rectangles having equal altitudes are to other as their bases. Hyp. In rectangles ABCD and A'B'C'D', altitude A altitude A'D'. Proof. CASE I. AB and A'B' are commensurable. Let m be a common measure contained in AB 3 times, an A'B' 4 times. 164 [To be completed by the student. Compare (280).] CASE II. AB and A'B' are incommensurable. Divide AB into any number of equal parts, and lay off one of these parts on A'B' as often as possible. As AB and A'B' are incommensurable, there must be a remainder, EB', less than one of the equal parts. Draw EFLA'B'. As AB and A'E are commensurable, ABCD AB = A'EFD' A'E By increasing the number of parts into which AB is divided, we can diminish the length of these parts, and, therefore, the length of EB' indefinitely. Hence, A'E approaches A'B' as a limit, and A'EFD' approaches A'B'C'D' as a limit. 336. COR. Rectangles having equal bases are to each other as their altitudes. 337. NOTE. As a polygon can be measured by the unit of surface 99 66 only, the words "triangle,' quadrilateral," etc., are frequently used for the area of a triangle, area of a quadrilateral, etc. PROPOSITION II. THEOREM 338. The areas of two rectangles are to each other as the products of their bases and altitudes. a R Q R' Hyp. Rectangles R and R' have the bases b and b' and the altitudes a and a' respectively. Proof. Construct the rectangle Q, having the same base as R', and the same altitude as R. Ex. 752. Find the ratio of a rectangle 4 by 5 feet, and a square having a side of 10 feet. Ex. 753. The diagonal of a rectangle is 26 inches long, and one of its sides 24 inches. The diagonal of another rectangle is 25 inches long, and one of its sides 20 inches. Find the ratio of the areas of the two rectangles. PROPOSITION III. THEOREM 339. The area of a rectangle is equal to the product of its base and altitude. R a 1 2 ს Hyp. R is a rectangle with base b and altitude a. 340. SCHOLIUM. If the base and altitude of a rectangle are expressed by integral numbers, Prop. III may be proved by dividing the figure into squares. Thus, if the base contains 5 and the altitude 3 linear units, the figure can be divided into fifteen squares, each being the unit of surface. PROPOSITION IV. THEOREM 341. The area of a parallelogram is equal to the product of its base by its altitude. Hyp. ABCD is a parallelogram, having the base AB = b, and the altitude BE = h. To prove area of ABCD = bxh. Proof. Draw AFL AB, meeting CD produced in F. Then ABEF is a rectangle, having its base b and its alti 342. COR. 1. Parallelograms having equal bases and equal altitudes are equivalent. 343. COR. 2. Any two parallelograms are to each other as the products of their bases and altitudes. 344. COR. 3. Parallelograms having equal bases are to each other as their altitudes. 345. COR. 4. Parallelograms having equal altitudes are to each other as their bases. |