14 MENSURATION long are commensurable ; for, if ^ inch be taken as unit of length, the former contains the unit nine times and the latter seven times. If no common measure can be found, the two magnitudes are said to be "incommensurable." For instance, 1 and V2 have no common measure ; for /2 = l 4142 ... an interminable decimal, and hence no unit, however small, can be found which will be contained in each an exact number of times. If, however, we take /2 = l 4, the error will be less than y 1 ^; if */2 = 1 4 14, the error will be less than y^ni ^ rc - Hence, by taking a sufficient number of figures, we can find a fraction which will differ from /2 by less than any assignable quantity, and therefore we can always find two commensurable magnitudes that will represent two incommensurable ones to any degree of accuracy we please. In what follows we need therefore only consider commensurable lines. 3. Area of a Rectangle. Let the side AB (fig. 1) con tain a units and the side BC b units of length. If we divide AB into a equal parts, A ^ each equal to the unit of length, and similarly BC into b equal parts, and if through the points of division we draw lines parallel to the sides of the rectangle, these lines will di vide the rectangle into a series of rectangles, each of which is the unit of area, since each is ~ p ic , j a square whose sides are of unit length. As we have a rows of these rectangles, and b in each row, the whole number of rectangles will be ab. Therefore area of ABCD = ab units of area =ai. PART I. PLANE FIGURES. SECTION I. PLANE FIGURES CONTAINED BY STRAIGHT LINES. A. The Rectangle. 4. Let ABCD (fig. 2) be a rectangl BC = DA = &, AC = c, and the angle A BAC = a ; it is required to find its area. Since a rectangle is completely de termined when two independent data, one of which at least is a length, are given connecting its parts, we can de termine its area in the following cases. (a) When its length a and its breadth and let b are given. It has already been proved ( 3) that Fig. 2. area of ABCD = ab; GV the area of a rectangle is equal to its length multiplied by its breadth. Example. Let a=12 feet 6 inches and 6 = 9 inches, then area of ABCD = 12 5 x 75 = 9 375 square feet. If we make use of logarithms in the above calculation we have log area = log a + log b . loga = log!2-5 =1-0969100 log b = log 75 = 1-8750613 therefore logarea= 9719713; hence area.= 9 375. (j8) When a side a and the diagonal c arc given. By Euclid i. 47 we have t 2 = c 2 - 2 , or b = Vc 2 - a 2 , therefore area of ABCD = rt6 = Vc 2 -a 2 ; or logarea = logrt + ilog(c + a) + ^log(c-) . Example. Let = 238 and c = 456, then log= Iog238 = 2 3765770 therefore hence J log(c + a) -ilog694- 1-4206797
log(c - a) = |log21 8 = 1 -1692282
log area = 4 9664849 ; area 92573 1. (7) When a side a and a. its inclination to the diagonal are given. Since = tana, 6 = atana, a and therefore area of ABCD = a& = a 2 tan a ; or logarea = 21oga + Ltana-10. Example. Let a = 36 and a = 32 25 15", then 21oga = 21og36 = 31126050 Ltana = Ltan32 25 15"= 9-8028622 therefore logarea = 12 9154672 - 10 = 2 9154672 ; hence area = 823 127. (5) When the diagonal c and a its inclination to either of the sides are given. We have = c cos a , and 5 = c sin a , = & = c 2 sinaooso = ^c 2 sin2a; therefore area of or 2area = c-siu2a, and hence Iog2area = 21ogc + Lsin2a - 10 . 5. A square being a rectangle whose sides are equal, we can at once determine its area. When one datum, which must be a length, is given the square is completely determined, and hence we have only two cases to consider. (o) When the side is given. From 4, a, we have at once area of square = ab = a x a = a?. (18) When the diagonal c is given. From 4, , we have a 2 + a 2 = c 2 , or a 2 = Jc 2 ; hence area of square = a" = c^, or 2 area = c 2 ; and therefore log 2 area = 2 log c. B. Right-angled Triangles. 6. The diagonal of every rectangle divides it into two equiva lent right-angled triangles (Eucl. i. 34), and hence the area of the right-angled triangle ABC (fig. 2) is equal to half the area of the corresponding rectangle ABCD. C. Triangles Generally. 7. In every triangle there are six elements to be considered, namely, the three sides and the three angles. If any three of these six be given, provided one is a length, the triangle is com pletely determined, and hence its area can be found. 8. Length of Perpendiculars of a Triangle. In the triangle ABC (fig. 3) let BC = a, CA = 6, AB = c, AD the perpendicular from A on BC = A, BD = x, and CD = 7/. Since BDA and CDA are right angles, we have c 2 = x" + 7t 2 , and 6 2 = y 1 + h- , and therefore whence But y + X-- and, by solving these equations, we obtain y ~ 2a Again 2a 4a 2 hence h = V(+ & + c )(& + c ~ )( c + a ~ &X a + & - , then 2(s- a) , Now let and a + Therefore, on substituting and reducing, we obtain 2 , _ h = Vs(s - a)(s - b)(s - c) . Similarly the psrpendiculars from B and C on the opposite sides are respectively _ _ -T- Vs(s - a)(s - b)(s - c) , and Vs(s - a)(s - b)(s - c). c 9. We now proceed to investigate formulae for the area of a triangle in the following important cases.
(a) When the base a and the altitude h are given. Since a