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in the right triangle B D E, we have B D=(B E240 E2)
I 를 =(0.25 +0.0179) =0.5176. Now since B D is contained 12 times in the circumference, we have 12X0.5176 =6.2112 for the circumference when the radius is 1. This is the second approximate ratio, and is much nearer the truth than the first, because the chord B D differs much less from the arc B D, than the chord B C did from its
In order to make the third approximation, we draw A F perpendicular to the middle of the chord B D, and it will bisect the arc B D in F. Then the chord B F is the chord of of the circumference, and its value is found in the same manner as that of B D. Thus A B=1 and B G=Ž of B D=0.2588. Then A G=(A B2B G2) (1–0.0669) * =0.966.
Knowing A G we have F G=AF-A G=1-0.966=0.034. Then in the right triangle B F G, we have B F=(B'G2+F G2)#=(0.0669 +0.00115)#=0.2609. Now since B F is contained 24 times in the circumference, we have 24 x0.2609= 6.2616 for the value of the circumference when the radius is 1. This is the third approximate ratio of the circumference to the radius, but it is still too small because the chord B F is still too large to be confounded with its arc, or to be considered as an exact common measure between the circumference and radius. It is obvious, however, that the process now commenced may be carried on to any limit we please, and each approximate ratio will be nearer the truth than the preceding, because, each are being half the preceding, its chord, calculated exactly in the same manner as above, will constantly approach bearer and nearer to a coincidence with its arc. We shall not give the details of the calculation any further, because the preceding steps are sufficient to make the whole process intelligible. We shall only add that at the thirteenth division, where the arc is
of the circumference, and where the chord may be considered as almost exactly equal to its arc, the fourteenth approximate ratio is that of 6.2831852 to 1. If we call the diameter 1 the expression for the circumference is half the above, namely 3.1416926. Some persons, have had the patience to extend the calculation to
one hundred and forty decimals, but the above value is sufficiently accurate for all the purposes to which it ever becomes necessary to apply it. Universally, when the diameter of a circle is known—we obtain the circumference with sufficient accuracy by multiplying the diameter by 3.1415926, and we obtain the diameter with sufficient accuracy by dividing the circumference by the same number—. Moreover—the area of a circle may be found by multiplying the square of the radius by 3.1415926— We have already seen (105) that the area of a circle is equal to the circumference multiplied by half the radius, or Cx1 of R. But C=2 RX3.1415926. Hence the area=2 RX3.1415926 xì of R=R2X3.1415926. It is the practice of geometers to represent the above ratio 3.1415926 by the Greek character n. Then the expression for the circumference is a XD, and for the area N XR2. 114. – To make a square equivalent to any given fig
This general problem is sometimes enunciated thus--to find the quadrature of any given figure in explaining its solution it will not be necessary to have recourse to a diagram. All the figures whose properties we have considered, except irregular polygons, are measured by a product consisting of two factors; and we have seen (107) that irregular polygons may be converted into equivalent triangles, and then the same will be true of them. Accordingly-to make a square equivalent to any given figure, it is only necessary to find a mean proportionał between the two factors by which that figure is measured This can always be done by the process before explained (80), and the mean proportional thus found will be the side of the square required; for the two factors will then be the extremes of a proportion, and the square of a mean proportional is always equal to the product of the extremes. Thus if the given figure be a parallelogram, find a mean proportional between the base and altitude (101). If a triangle, between the base and half the altitude (102). If a trapezoid, between the altitude and half the sum of the parallel sides (103). If a regular polygon, between the perimeter and half the radius of the inscribed circle (104). If a circle, between the circumference and half the radius (105). If a sector, between the arc and half the radius (106). If an irregular polygon, between the base and half the altitude of the equivalent triangle (107).
Comparison of Surfaces.
115. Surfaces are compared by means of the products which measure them. Universally—any two surfaces are to each other as their areas- This is self evident. But —when the two products have one factor the same, it may be left out (66), and then the two surfaces will be to each other as the factors which are unlike, Thus if the radius of a circle is equal to the altitude of a triangle, then the circle will be to the triangle, as the circumference of the circle is to the base of the triangle, and so of all other similar cases. The comparison most frequently made is that where the figures are of the same kind. The following propositions need no demonstration, being only particular cases of the general proposition just enunciated and selfevident from the nature of measures. -If two parallelograms have the same altitude, they are to each other as their bases ; if they have the same base, they are to each other as their altitudes—If two triangles have the same base, they are to each other as their altitudes; if they have the same altitude, they are to each other as their bases—If two trapezoids have the same altitude, they are to each other as the sums of their parallel sides ; if the sums of their parallel sides are the same, they are to each other as their altitudes,
116. If the two surfaces to be compared are similar figures, we have other means of comparing them, than those just explained. We shall demonstrate the following general proposition. - Any two similar figures are to each other as the squares of their homologous sides—. We begin with two similar triangles. Let these be A B C and D EF (fig. 77), and let À G be the altitude of the first and D H F77 that of the second. Then by the preceding proposition, ABC:D EF::B CXA G:E FXD H. But from the similar triangles A B C and D E F we have (78) B C:E F::A B:D E. Moreover the triangles A B G and D E H are similar, since the angle B=E and they have each a right angle. Hence A B D E::AG:D H. The two last proportions have a common ratio A B:D E. Therefore B C:EF::AG:D H. Multiplying this, term by term, (67) by the identical proportion BC: EF::BC: É F,we have B C2: E F2::B CXAG:E FXD H. But we bad above A B C :D E F::B CXA G:EFXD H.
Leaving out the common ratio in the two last proportions, we have A B C:D E F::B C2: E F2, which was first to be demonstrated. It will now be easy to generalize the demonstration for similar figures of any number of sides,
Take the two similar polygons A B C D E and F G H I F58 K (fig. 58). These are composed of the same number of
similar triangles (87). Hence ABC:FGH:: A C2: F H2. Also A CD:FHI:: A C2: F H2. Therefore A B C:FGH::ACD:FHI. In like manner A CD:F HI::A DE:FI K, and so on for any number of triangles. Thus we have the continued proportion A B C:F GH::AC D:F HI:: A DE:FIK. Here the sum of the antecedents is the polygon A B C D E, and the sum of the consequents is the polygon F G H I K. Therefore (69) A B C D E:F G HIK:: A B C:FGH. But A B C:F G H :: A B2: F G2. Therefore A B C D E:F GHIK:: A B2: F G2. In other words, similar figures are to each other as the squares of their homologous sides.
. 117. — Circles are to each other as the squares of their radii— No diagram is necessary for this demonstration. Let us call one cirele A, its circumference C, and its radius R; and let us call another circle a, its circumference cg and its radius r. Then, since surfaces are as their areas, we have A:a::Cxă R:cxă r. Doubling the second ratio, A:a::CXR:cXr. But (96) Cic::R:T. Multiplying this last, term by term, by the identical proportion R:r::R:r, we have CXR:cXr::R2 : r2. Now we had before A:a::CXR:cXr. Hence, leaving out the common ratio, A :a:: R2 :r2, that is, circles are as the squares of their radii which was to be demonstrated. 118. - qual perimeters do not always enclose equal
may be demonstrated by numbers without a diagram. Take, for instance, a square and an oblong of equal perimeters. Let the side of the square be 12 feet; then its pefieter is 12+12+12+12=48 feet. Let the base of the oblong be 16 feet and its altitude 8 feet; then its perimeter is 16+16+8+8=48 feet. Thus the perimeters are equal. But the area of the square is 12X12=144 square feet; and the area of the oblong is 16X8=128 square
feet. Therefore the areas are unequal. 119., We shall now consider some of the circumstances in which a given perimeter contains the greatest area. This investigation is highly useful in a practical point of
view, but the plan of this work requires us to confine ourselves to some of the simplest cases. The first proposition we shall demonstrate is the following. —If two triangles have the same base and equal perimeters, that is the greatest in which the two undetermined sides are equal. Let the two triangles be A C B and A M B (fig. 78) having the F 78 same base A B and equal perimeters. We say that if A C=C B and if A M is not equal to M B, then A C B is greater than A. M B. Produce A C till C D=C B. Join Ď and B and produce the line indefinitely. The angle A B D is a right angle, because if a circle were described with the centre © and radius C B, the angle A B D would be inscribed in a semicircumference. Now take M as a centre, and with a radius M B make an arc cutting D B produced in N, so that M N=M B. Then A M+MN=A M+M B=A C+C B=A C+CD=AD. But A MUM N is greater than A N, since A N is the shortest distance between A and N. Therefore A D is greater than A N. Now of two unequal oblique lines, that which is greater must be more distant from the perpendicular (31). Hence D B is greater than B N, and G B, half of D B, is greater than B P, half of B N. But G B is the altitude of AC B, and B'P is the altitude of A M B. Accordingly, since triangles of the same base are to each other as their altitudes, and since G B is greater than B P, the triangle A C B is greater than A M B, which was to be demonstrated.
120. -Among polygons of the same perimeter und the same number of sides, that is the greatest in which the sides are equal — Let there be the polygon A B C D E F (fig. 79). F 79 If the sides are not equal, the polygon may be enlarged without enlarging the perimeter. Draw BD. Then, by the preceding proposition, if B C is not equal to CD, the isosceles triangle B O D of the same base and perimeter is greater than B C D, and by substituting it, the polygon would be enlarged without enlarging its perimeter. The same might be proved with respect to all the other sides. Therefore the greatest polygon, of a given perimeter and a given number of sides, must be that in which all the sides are equal, which was to be demonstrated.
121. ---Among polygons of the same perimeter and the same number of sides, the regular polygon is the greatest. We have just proved that the sides must be equal, and we shall now prove that the angles must be equal. As the