D CASE II. When AB and AE are incommensurable. Divide AB into any number of equal parts, and apply one of them to AE as often as it will be contained in AE. Since AB and AE are incommensurable, a certain number of these parts will extend from A to a point K, leaving a remainder KE less than one of the parts. Draw KH to EF. Since AB and AK are commensurable, rect. AH AK rect. AC AB Case I. These ratios continue equal, as the unit of measure is indefinitely diminished, and approach indefinitely the limiting ratios rect. AF AE and rect. AC respectively. AB (if two variables are constantly equal, and each approaches a limit, the limits are equal). Q. E. D. 361. COR. The areas of two rectangles having equal bases are to each other as their altitudes. For AB and AE may be considered as the altitudes, AD and AD as the bases. NOTE. In propositions relating to areas, the words " ' rectangle," "triangle," etc., are often used for “area of rectangle,” “area of triangle," etc. PROPOSITION II. THEOREM. 362. The areas of two rectangles are to each other as the products of their bases by their altitudes. Let R and R' be two rectangles, having for their bases b and b', and for their altitudes a and a'. Proof. Construct the rectangle S, with its base the same as that of R, and its altitude the same as that of R'. Then R α S a' $361 (rectangles having the same base are to each other as their altitudes); (rectangles having the same altitude are to each other as their bases). By multiplying these two equalities, Ex. 288. Find the ratio of a rectangular lawn 72 yards by 49 yards to a grass turf 18 inches by 14 inches. Ex. 289. Find the ratio of a rectangular courtyard 18 yards by 15 yards to a flagstone 31 inches by 18 inches. Ex. 290. A square and a rectangle have the same perimeter, 100 yards. The length of the rectangle is 4 times its breadth. Compare their areas. Ex. 291. On a certain map the linear scale is 1 inch to 5 miles. How many acres are represented on this map by a square the perimeter of which is 1 inch? PROPOSITION III. THEOREM. 363. The area of a rectangle is equal to the product of its base and altitude. Let R be the rectangle, b the base, and a the altitude; and let U be a square whose side is equal to the linear unit. (two rectangles are to each other as the product of their bases and altitudes). 364. SCHOLIUM. When the base and altitude each contain the linear unit an integral number of times, this proposition is rendered evident by dividing the figure into squares, each equal to the unit of measure. Thus, if the base contain seven linear units, and the altitude four, the figure may be divided into twenty-eight squares, each equal to the unit of measure; and the area of the figure equals 7 × 4 units of surface. PROPOSITION IV. THEOREM. 365. The area of a parallelogram is equal to the product of its base and altitude. Let AEFD be a parallelogram, AD its base, and CD its altitude. To prove the area of the AEFD=AD × CD. Proof. From A draw AB to DC to meet FE produced. Then the figure ABCD will be a rectangle, with the same base and altitude as the □ AEFD. In the rt. A ABE and DCF ABCD and AE= DF, (being opposite sides of a □). .. ΔΑΒΕ= Δ DCF, $179 § 161 (two rt. A are equal when the hypotenuse and a side of the one are equal respectively to the hypotenuse and a side of the other). Take away the ▲ DCF, and we have left the rect. ABCD. Take away the ▲ ABE, and we have left the .. rect. ABCD=□ AEFD. But the area of the rect. ABCD = a × b, AEFD. Ax. 3 § 363 Ax. 1 Q. E. D. 366. COR. 1. Parallelograms having equal bases and equal altitudes are equivalent. 367. COR. 2. Parallelograms having equal bases are to each other as their altitudes; parallelograms having equal altitudes are to each other as their bases; any two parallelograms are to each other as the products of their bases by their altitudes. PROPOSITION V. THEOREM. 368. The area of a triangle is equal to one-half of the product of its base by its altitude. Let ABC be a triangle, AB its base, and DC its altitude. To prove the area of the ▲ ABC= AB × DC. (the diagonal of a divides it into two equal ▲). The area of the □ ABCH is equal to the product of its base by its altitude. § 365 Therefore the area of one-half the O, that is, the area of the ▲ ABC, is equal to one-half the product of its base by its altitude. Hence, the area of the ▲ ABC AB× DC. = Q. E. D. 369. COR. 1. Triangles having equal bases and equal altitudes are equivalent. 370. COR. 2. Triangles having equal bases are to each other as their altitudes; triangles having equal altitudes are to each other as their bases; any two triangles are to each other as the product of their bases by their altitudes. |