Of Ratios and Proportions. Sch. The proposition may be verified by the following proportions: viz. If four quantities are in proportion, their squares or cubes will also be proportional. Then, if we square both members, we have B2 D2 and if we cube both members, we have B3 D3 and then, changing these equalities into a proportion, we have for the first, Sch. We may verify the proposition by the proportion, 2 : 4 :: 6 : 12, and by squaring each term we have, 4 : 16 :. 36 : 144 Of Ratios and Proportions. numbers which are still proportional, and in which the ratio If we have two sets of proportional quantities, the products o the corresponding terms will be proportional. and this by Th. II, gives AXE B×F :: D×H. Sch. The proposition may be verified by the following GEOMETRY. BOOK IV OF THE MEASUREMENT OF AREAS, AND THE PROPORTIONS OF FIGURES. DEFINITIONS. 1. Similar figures, are those which have the angles of the one equal to the angles of the other, each to each, and the sides about the equal angles proportional. 2. Any two sides, or any two angles, which are like placed in the two similar figures, are called homologous sides or angles. 3. A polygon which has all its angles equal, each to each, and all its sides equal, each to each, is called a regular polygon. A regular polygon is both equiangular and equilateral. 4. If the length of a line be computed in feet, one foot is the unit of the line, and is called the linear unit. If the length of a line be computed in yards, one yard is the linear unit 5. If we describe a square on the unit of length, such square is called the unit of surface. Thus, if the linear unit is one foot, one square foot will be the unit of surface, or superficial unit. 1 foot. unit 1 Of Para lelograms. 6. If the linear unit is one yard, one square yard will be the unit of surface; and this square yard contains nine square feet. 1 yd.=3 feet. 7. The area of a figure is the measure of its surface. Th unit of the number which expresses the area, is a square, the side of which is the unit of length. 8. Figures have equal areas, when they contain the same measuring unit an equal number of times. 9. Figures which have equal areas are called equivalent. The term equal, when applied to figures, implies an equality in all respects. The term equivalent, implies an equality in one respect only: viz. an equality in their areas. The sign. , denotes equivalency, and is read, is equivalent to. THEOREM I. Parallelograms which have equal bases and equal altitudes, are equivalent. Place the base of one parallel E D B ogram on that of the other, so that AB shall be the common base of the two parallelograms ABCD and ABEF. Now, since the parallelograms have the same altitude, their upper bases, DC and FE, will fall on the same line FEDC, parallel to AB. Since the opposite sides of a parallelogram are equal to each other (Bk. I. Th. xxiii), AD is equal to BC. Also, DC and FE are each equal to AB: and consequently, they are equal to each Of Triangles and Parallelograms. other (Ax. 1). To each, add ED: then will CE be equal to DF. But since the line FC cuts the wo parallels CB and DA, the angle BCE will be equal to the D angle ADF (Bk. I. Th. xiv): hence, the two triangles ADF and BCE have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each; consequently, they are equal (Bk. I. Th. iv). If then, from the whole space ABCF we take away the triangle ADF, there will remain the parallellogram ABCD; but if we take away the equal triangle BEC, there will remain the parallelogram ABEF: hence, the parallelogram ABEF is equivalent to the parallelogram ABCD (Ax. 3). Cor. A parallelogram and a rectangle, having equal bases and equal altitudes, are equivalent. I THEOREM II. Triangles which have equal bases and equal altitudes, are equivalent. Place the base of one triangle F on that of the other, so that ABC and ABD shall be two trian gles, having a common base AB, and for their altitude, the distance E between the two parallels AB, FC: then will the triangle ABC be equivalent to the triangle ADB. For, through A draw AE parallel to BC, and AF parallel to BD, forming the two parallelograms BE and BF Then |