HF 7 Th. Equiangular parallelograms are to one another as the products of the numbers proportional to their sides. Dem. Because parallelograms upon G C the same base, or upon equal bases, and between the same parallels, are equal to each other (a); therefore the parallelograms AC and DF may be reduced to equivalent rectangles, as AG and DH. E Now AG is the product of ABXBG (6), and DH is the product of DEXEH. Moreover BG is to BC as EH is to EF (c); and alternately, BG : EH :: BC: EF (d). Therefore AG : DH :: ABXBG : DEXEH; that is, AC: DF : : ABXBC: DEXEF. Wherefore, equiangular parallelograms, &c. Q. E. D. (d) p. 16 of b, 5. B 8 Th. A perpendicular drawn from the centre of a circle to the chord of any arc, is a mean proportional between half the radius and the straight line composed of the radius and its excess above the subsine of the arc: and the chord of the arc is a mean proportional between the diameter and the subsine. B В Let C be the centre of the circle ABD; let DB be any arc, and DBE twice the arc; draw the chords DB, DE; also CG and CF at right angles to DB, DE; produce CF to meet the circumference in B and A ; bisect AC in H; join AD. BF is A the subsine of the arc DB. (P. T. def. 10) Then AH : CG : :CG: AF; also, AB: BD:: BD: BF. ADB is a right angle, being in a semicircle; and CĞB is also a right angle: the triangles ADB, CGB, are therefore equiangular; and AB : AD : :-CB: CG (p. 4 of b. 6); or, alternately, AB : CB : : AD: CG. And because AB is twice CB, AD is twice CG; and AD=4CG”. But because of the perpendicular DF, from the right angle, on AB; therefore AB : AD :: AD: AF (p. 8 of b. 6). Hence AD = ABXAF (p. 17 of b. 6); or, since AB=4AH, AD=4AH XAF= 4CG”. Therefore CG2-AHXAF, and CG is therefore a mean proportional between AH and AF; that is, half the radius and the excess of the radius above the subsine BF. Again, the chord BD is a mean proportional between the diame. ter and the subsine of the arc : for, (by p. 8 of b. 6,) AB : BD :: BD: BF. 9 Th. The circumference of a circle exceeds three times its diameter, by a line less than ten of the parts of which the diameter contains seventy ; but greater than ten of the parts of which the diameter contains seventy-one. Let C be the centre of the circle ABD, and AB its diameter; the circumference exceeds three times AB by a line less than 10-70ths, or 1-7th part of AB, but greater that 10-7 1st parts of AB. Apply, in the circle, the chord BD, A equal to the radius, or the side of an inscribed hexagon (p. 15 of b. 4); draw DF perpendicular on BC, and produce it to E; draw CG at right angles to DB; produce BC to A; bisect AČ in H, and join AD. The arcs BD, BE contain each 1-6th, and DBE 1-31 of the circumference; and (p. 8 of this) AH : CG:: CG: AF. Now, because the triangle BDC is equilateral, DF bisects BC in F. Therefore, if AC or BC=1000, AH=500, CF=500, and AF=1500 ; and as CG is a mean proportional between AH and AF, CG-=AH X AF=500 X 1500=7 50000. Wherefore, CG=V750000=866.0254 +. Hence AC+CG=1866.0254+. Now, as CG is the perpendicular from C on the chord of 1-6th of the circumference, if P be the perpendicular from C on the chord of 1-12th of the circumference, P will be a mean proportional between AH and AC+CG; and P=AHX(ACT-CG)=500 X 1866.0254+ 933012.7 +. Wherefore P=7933012.7 +: 965.9258+ Hence AC+P=1965.9258+. Again, if Q be the perpendicular from C on the chord of 1-24th of the circumference, Q will be a mean proportional between AH and AC +P; and Q2 AHX (AC +P)=500 X 1965.9258 + = 982962.9. Wherefore Q=V982962.9 - 991.4449 +. Hence AC +Q=1991.4449+: Likewise, if S be the perpendicular from C on the chord of 1-48th of the circumference, S will be a mean proportional between AH and AC+Q: and Sa=AHX(AC+Q)=500 X 1991.4449 + 995722.45+. Wherefore S=V995722.45+=997.8589+. Hence ACES=1997.8589+. Lastly, if T be the perpendicular from C on the chord of 1-96th of the circumference, T will be a mean proportional between AH and AC+S; and T=AHX(AC+8)=500X1997.8589+=998929.45+. Wherefore T=V 998929.45+=999.46458+. Thus, the perpendicular from the centre on the chord of 1-96th part of the circumference, exceeds 999.46-158 of those parts of which the radius contains 1000. But (p. 8 of this), the chord of 1-96th part of the circumference is a mean proportional between the diameter and the excess of the radius above the perpendicular from C on the chord of 1-48th part of the circumference. Therefore the square of the chord of 1-96th part=ABX(AC-5)= 2000X(1000—997.8589+)=2000X2.1411-=4282.2- and the chord itself=1 4282.2-=65.4366—; which is the side of a regular polygon of 96 sides, inscribed in the circle ; and the perimeter of the polygon is equal to 96 X 65.4385-=6282.10566, Let the perimeter of the circumscribed polygon of 96 sides be M, then T: AC :: 6282.1056—:M; that is (since T=999.46458+, as already shown), 999.46458+: 1000 :: 6282.1056—:M; if then N be such that 999.46458 : 1000 :: 6282.1056—: N; ex æquo perturbato 999.46458+: 999.46458 ::N:M; and since the first is greater than the second, the third is greater than the fourth, or N is greater than M. Now, if a fourth proportional be found to 999.46458, 1000 and 6282.1056, viz. 6285.461-, then because 999.46458 : 1000 :: 6282.1056 : 6285.461–, and as before, 6282.1056 : 6282.1056—:: 6285.461-:N; and as the first of these proportionals is greater than the second, the third is greater than the fourth, viz. 6285.461—is greater than N. But N was shown to be greater than M; therefore 6285.461 is still greater than M, the perimeter of a polygon of 96 sides described about the circle; that is, the perimeter of that polygon is less than 6285.461; but the circumference of the circle is less than the perimeter of the polygon, and still less than 6285.461. Wherefore the circumference of a circle is less than 6285.461 of those parts of which the radius contains 1000. The diameter, therefore, has to the circumference a less ratio (p. 8 of b. 5), than 2000 has to 6285.461, or than 1000 has to 3142.7305 : but the ratio of 7 to 22 is greater than the ratio of 1000 to 3142.7305: therefore the diameter has a less ratio to the circumference than 7 has to 22; or the circumference is less than 22 of the parts of which the diameter contains 7. The circumference exceeds 3 and 10-7 1st parts of the diameter. It has been proved (p. 8 of this), that the perpendicular froin the centre on the chord of an arc, is a mean preportional between half the radius and the sum of the radius and a perpendicular from the centre on the chord of A HC twice the arc. CG, P, Q, S and T were used, in this order, as perpendiculars on the chords of 1-6th, 1-12th, 1-24th, 1-48th and 1-96th part of the circumference. CG2, was made equal to AHXAF=500X1500=750000, and CG= 866.02545— : therefore AC+CG=1866.02545--. P2. was made equal to AHX(AC+CG)=500X1866.02545—, and P=965.92585—; therefore ACP=1965.92585– Q?, was made equal to AHX(AC+P)=500X 1965.92585%, and Q=991.44495—: therefore AC+Q=1991.44495— S.% was made equal to AHX(AC+Q)=500X1991.444954, and S =997.85895 -: therefore AC+8=1997.85895. It is also proved (p. 8 of this), that the chord of the arc is a mean proportional between the diameter and the excess of the radius above the perpendicular: therefore (the chord of 1-96o=ABX(AC-8)= 2000X2.14105+=4282.1 t, and the chord=65.4377+. Now this E chord is the side of an inscribed polygon of 96 sides, the perimeter of which is 96 X65.4377+=6282.019+: But the circumferenec of the circle exceeds the perimeter of the inscribed polygon; therefore the circumference exceeds 6282.019+ of those parts of which the radius contains 1000; or 3141.0095 of those parts of which the radius contains 500, or the diameter contains 1000. Now 1000 has a greater ratio to 3141.0095 than 1 has to 3 10-71st parts; that is, the excess of the circumference above three times the diameter is greater than 10 of those parts of which the diameter contains 71. Cor. 1. Hence the diameter of a circle being given, the circum ference may be found nearly, by making as 7 to 22, so the given diameter to a fourth proportional, which shall be greater than the circumference. And, if as 1 to 3 10-71, or 71 to 223, so the given diameter to a fourth proportional, this will be nearly equal to the cir. cumference, but will be less than it. Cor. 2. Because the difference between 1-7th and 10-71st is 1-497th, therefore the lines found by these proportionals differ by 1-497th of the diameter; therefore the difference of either of them from the circumference must be less than the 497th part of the diameter. Cor. 3. As 7 to 22 so the square of the radius to the area of the circle nearly. For it has been shown (cor. 1 p. 5 of this), that the diameter of a circle is to its circumference as the square of the radius to the area of the circle; but the diameter is to the circumference nearly as 7 to 22, therefore the square of the radius is to the area of the circle nearly in that same ratio. N. B. In the preceding calculations, for .461-read.47+. THE INTERSECTION OF PLANES: Definitions. 1. A straight line is perpendicular to a plane when it makes right angles with every straight line it meets in the plane. 2. A plane is perpendicular to a plane, when the straight lines drawn in one of the planes at right angles to their common section are perpendicular to the other plane. 3. The declination of a straight line from a plane is the acute angle of a right angled triangle opposite to the side perpendicular to the plane. 4. The declination of two intersecting planes is the angle made by two straight lines, one in each plane, which meet in the line of section, at right angles to that line. 5. Two planes have a declination equal to that of other two, when the angles of declination are equal. 6. A straight line is parallel to a plane, when it does not meet the plane, however produced. 7. Planes are parallel to each other, which do not meet, however far produced. 8 A solid angle is made by the meeting of more than two plane angles in the same point. Propositions. 1 Th. One part of a straight line cannot be in a plane and another part above it. Recite Pos. 1; cor. def. 4, 1, 2 Th. Any three straight lines which meet each other not in the same point, are in one plane. Recite def. 7,1; p. 1 of this. 3 Th. If two planes cut one another, their common section is a straight line. Recite def. 7, 1. 4 Th. If a straight line stand at right angles to each of two straight lines, at their point of intersection, it will also be at right angles to the plane in which those straight lines are. Recite p 47 and 48 of b. 1; def. 1 of this. 5 Th. If three straight lines meet in a point, and a straight line stand at right angles to each of ihem in that point, these three straight lines shall be in the same plane. Recite p. 3 and 4 of this. |