A B feet, therefore, just as the lineal foot, for a similar purpose, is divided into inches, so the square foot is divided into square inches. For instance, suppose the square ABCD to represent a square foot. Divide AB, and AD, each into 12 equal parts (168, Part 11.); these equal parts will be inches, since AB-AD-1 foot-12 inches. From the several points of division in AB draw lines through the square parallel to AD; and from the several points of division in AD draw lines parallel to AB. These lines will obviously divide the square ABCD into 12 smaller squares repeated 12 times, that is, into 144 smaller squares; and each of these smaller squares is a square inch, that is, a square whose side is an inch in measure. Hence, if a proposed surface do not contain the square foot an exact integral number of times, it will contain a certain number of square feet, and a fraction of a square foot over, which fraction may be a certain number of square inches. And then the measure of the surface, or area, will be so many square feet together with so many square inches. For example, suppose ABCD to be a rectangle, of which the side AB is 4 feet, or 4 ft. 6 in., and AD one foot. Let AE=EF=FG=GH=1 foot; and through E, F, G, H draw Ee, Ff, Gg, Hh parallel to AD or BC. Then HB-half a foot, or 6 inches; and HBCh = half a square foot, or 72 square inches; therefore the measure of the rectangle ABCD is 44 square feet, or 4 square feet and 72 square inches. If we still find, that the proposed surface is not made up of an exact integral number of square feet and inches, but that a fraction of a square inch remains over, then the square inch must be subdivided into smaller squares, as the square foot was; and so on, until we either obtain the precise measure of the proposed surface, or approach so near to it, that the remainder is of no account for practical purposes. 222. To measure a given* square. 1 st. Let ABCD be the given square, of which each A B 1 2 3 4 5 3 4 side is a certain integral number of units, 5 suppose. Divide AB, and AD, each into 5 equal parts (168, Part II.), and through the several points of division draw lines parallel to the sides of the square, dividing the given square, as in the diagram, into a certain number of equal smaller squares, each of which is the unit of superficial measure. The only question then is, what is the number of such squares contained in the proposed square? That number is the measure required; and is in this case obviously 5 repeated 5 times, that is, 25. D 5 C Thus, if AB be 5 feel in length, the area of the square ABCD will be 5×5, or 25, square feet. Or, if AB be 5 yards in length, then the square ABCD will be 25 square yards, that is, equal to 25 squares, each of which has 1 yard for its side. The same rule will evidently hold whatever be the number of units in AB, that is, we must multiply that number by itself to find the Arithmetical measure of the square ABCD. 2nd. If the number of units in AB be not an integer, but fractional, as 51, each unit being reckoned in quarters, it is plain that each side of the square will contain 21 such quarters. Divide, then, AB and AD, each into Given, that is, as to the length of each of its sides. If the side be not given, but the surface, in the form of a square, be simply presented before us, then a side must be measured by Art. 218. 21 equal parts (168, Part 11.), and through the several points of division draw lines parallel to AD and AB, as before, dividing ABCD into equal smaller squares, each of which has a quarter of a unit for its side. The number of these smaller squares will obviously be 21 taken 21 times, or 21×21. But, as the side of the smaller square is of the lineal unit, the square unit contains 16 such squares. Therefore ABCD contains square units; that is, ABCD is measured by 21 21 21×21 16 -x or 51×51, the product of the side multiplied 4 4. by itself, as before. The same rule may be shewn to hold whatever fraction represents the side of the square, by dividing the lineal unit into as many equal parts as is expressed by the denominator of the fraction. Thus if AB be 3ğ, each unit must be divided into eighths, that is, AB must be divided in 29 equal parts, and then, as before, it will 29 29 readily be shewn that ABCD is measured by X 3 x 35. 8 8 or N.B. The square, of which any line, as AB, is a side, is commonly called the square of AB; and, for shortness, is written AB, and read AB square. Thus AB2+ OD3, which is read AB square plus OD square,' means that the square upon OD is to be added to the square upon AB. Hence 52 does not stand for 5 × 5, that is, 5 times 5, by definition merely, but is proved to be equal to that product: in other words, it is proved, that the square, whose side is 5 linear units, contains 25 square units. And so also, whatever the number may be, which measures the side of a square, the square itself is measured by that number multiplied by itself. Again, the half of the line AB is often written thus, AB; and its square, as will readily appear by drawing the diagram, is one-fourth of the square of AB, which is written AB. Similarly the square of AB is AB2; and so on-the fraction being multiplied by itself in all cases to obtain the square. 223. To measure a given* rectangle. 1st. Let ABCD be the given rectangle; and sup pose AB to contain the lineal unit 5 times, and AD to contain it 3 times. Divide AB into 5 equal parts, and AD into 3 equal parts; then each of these parts is the lineal unit; and, if through the several points of division A D B 1 2 3 4 5 2 3 C parallels to the sides of the rectangle be drawn, ABCD will obviously be cut up into a set of small squares, each of which is the superficial unit of measure (since its side is the lineal unit), and the number of these squares is plainly equal to 5 taken 3 times, or 3 × 5, the product of AD and AB. The same will hold whatever other whole numbers of units are contained in AB, and AD, that is, the measure of the rectangle ABCD is the product of AB and AD. Thus, if AB be 5 feet, and AD 3 feet, then ABCD will be 15 square feet. Or, if the lineal unit be a yard, then the measure of the same rectangle will be 15 square yards the unit of superficial measure being always the square which has the lineal unit for its side. 2nd. If the lineal unit is not contained an exact integral number of times in either or both of the sides, the process of measurement is the same as that employed in the 2nd case of Art. 222. Thus, suppose the lineal unit to be contained 51 times in AB, and 3 times in AD; then AB being divided into 21 equal parts, and AD into 14 equal parts (each lineal unit being 4 such parts), and the parallels being drawn through the several points of division, the whole rectangle ABCD will be cut up into a set of small squares, each of which is the 16th part of the square unit, and the number of the small squares will plainly be 21 taken 14 times, or 14×21. Therefore the measure of the rectangle will be the 16th part of this Given, that is, as to the length of each of two adjacent sides. If the sides be not given, but the rectangle, as a surface, simply stands before us, then each of the two sides must be measured by Art. 218. duct of the two adjacent sides, as before. The same rule may be shewn, in a similar manner, to hold, whatever be the fractions which measure the sides of the rectangle. Hence we conclude that in all cases the measure of a rectangle is the product of two adjacent sides. N.B. The adjacent sides, of which the product is taken, must be measured according to the same unit, before the multiplication takes place. And where some error of measurement is unavoidable, it is to be observed that the greatest care will be required in measuring the lesser of the two sides, because the error will be multiplied oftener when it has to be multiplied by the number of units in the longer side, than it would be, if the same amount of error is found in the longer side and is to be multiplied by the shorter. It is to be observed also, that, whereas in Parts I. and II. a rectangle, as ABCD, is always spoken of as the "rectangle contained by AB and BC, or more shortly 'the rectangle AB, BC,' it is now proved that it is equal to the product of AB and BC. 224. To measure a given parallelogram. F From B D Let ABCD be the given parallelogram. and A draw BE, and AF, perpendiculars to CD, and CD produced. Then it is easily shewn that the triangle AFD is equal in all respects to the triangle BEC (24, Part 1.); and thus it follows that Α B the parallelogram ABCD is equal to the rectangle ABEF. But the measure of the rectangle ABEF=AB × BE (223), therefore the measure of the parallelogram ABCD=AB× BE =the base xthe height, as it is usually stated. |