Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Let the straight line X be equal to half the circumference of the circle ABD, and the straight line Y to half the circumference of the

X

Y

circle GHL: And because the rectangles AO.X and GP. Y are equal to the circles ABD and GHL (5.1. Sup.); therefore AO.X: GP.Y:: AD2: GL :: AO: GP; and alternately, AO. X: AO :: GP.Y:GP2; whence, because rectangles that have equal altitudes are as their bases (1. 6.), X: AO:: Y: GP, and again alternately, X: Y:: AO: GP; wherefore, taking the doubles of each, the circumference ABD is to the circumference GHL as the diameter AD to the diameter GL.

S

COR. 2. The circle that is described upon the side of a right angled triangle opposite to the right angle, is equal to the two circles described on the other two sides. For the circle described upon SR is to the circle described upon RT as the square of SR to the square of RT; and the circle described upon TS is to the circle described upon RT as the square of ST to the square of RT. Wherefore, the circles described on SR and on ST are to the circle described on RT as the squares of SR and of ST to the square of RT (24. 5.). But the squares of RS and of ST are equal to the square RT(47. 1.); therefore the circles described on RS and ST are equal to the circle described on RT.

of

R

T

PROP. VII. THEOR.

Equiangular parallelograms are to one another as the products of the numbers proportional to their sides.

Let AC and DF be two equiangular parallograms, and let M, N, P and Q be four numbers, such that AB: BC::M:N; AB: DE:: M: P; and AB: EF:: M: Q, and therefore ex æquali, BC: EF:: N:Q. The parallelogram AC is to the parallelogram DF as MN to PQ.

Let NP be the product of N into P, and the ratio of MN to PQ will be compounded of the ratios (def. 10. 5.) of MN to NP, and of NP to PQ. But the ratio of MN to NP is the same with that of M

C

F

A

B D

E

to P (15. 5.), because MN and NP are equimultiples of M and P; and for the same reason, the ratio of NP to PQ is the same with that of N to Q; therefore the ratio of MN to PQ is compounded of the ratios of M to P, and of N to Q. Now, the ratio of M to P is the same with that of the side AB to the side DE (by Hyp.); and the ratio of N to Q the same with that of the side BC to the side EF. Therefore, the ratio of MN to PQ is compounded of the ratios of AB to DE, and of BC to EF. And the ratio of the parallelogram AC to the parallelogram DF is compounded of the same ratios (23. 6.); therefore, the parallelogram AC is to the parallelogram DF as MN, the product of the numbers M and N, to PQ, the product of the numbers P and Q. Therefore, &c. Q. E. D.

COR. 1. Hence, if GH be to KL as the number M to the number N; the square described on GH

will be to the square described G

on KL as MM, the square of the

H K L

number M to NN, the square of the number N.

num

COR. 2. If A, B, C, D, &c. are any lines, and m, n, r, s, &c. ber's proportional to them; viz. A : B :: m: n, A : C :: m :r, A : D::m:s, &c.; and if the rectangle contained by any two of the lines be equal to the square of a third line, the product of the numbers proportional to the first two, will be equal to the square of the num ber proportional to the third; that is, if A. C-B3, mxr=nxn, or

=n2.

For by this Prop. A.C: B3:: mxr: n3; but A.C=B', therefore mxr=n3. Nearly in the same way it may be demonstrated, that whatever is the relation between the rectangles contained by these lines, there is the same between the products of the numbers proportional to them.

So also conversely if m and r be numbers proportional to the lines A and C; if also A.C=B3, and if a number n be found such, that n3 =mr, then A:B::m: n. For let AB::m: q, then since, m, q, rare proportional to A, B, and C, and A.C=B3; therefore, as has just been proved, q2=mxr; but n2=qxr, by hypothesis, therefore n3=q3, and n=q; wherefore A:B::m:n

SCHOLIUM.

In order to have numbers proportional to any set of magnitudes of the same kind, suppose one of them to be divided into any number, m rts, and let H be one of those parts. Let H be found я

times in the magnitude B,r times in C,s times in D, &c., then it is evident that the numbers m, n,r,s are proportional to the magnitudes A, B, C and D. When therefore it is said in any of the following propositions, that a line as A= a number m, it is understood that A=m XH, or that A is equal to the given magnitude H multiplied by m, and the same is understood of the other magnitudes, B, C, D, and their proportional numbers, H being the common measure of all the magnitudes. This common measure is omitted for the sake of brevity in the arithmetical expression; but is always implied, when a line, or other geometrical magnitude, is said to be equal to a number. Also, when there are fractions in the number to which the magnitude is called equal, it is meant that the common measure H is farther subdivided into such parts as the numerical fraction indicates. Thus, if A=360.375, it is meant that there is a certain magnitude H, 375

such that A=360x H+ 1000H, or that A is equal to 360 times H,

together with 375 of the thousandth parts of H. And the same is true in all other cases, where numbers are used to express the relations of geometrical magnitudes.

[blocks in formation]

The perpendicular drawn from the centre of a circle on the chord of any arch is a mean proportional between half the radius and the line made up of the radius and the perpendicular drawn from the centre on the chord of double that arch: And the chord of the arch is a mean proportional between the diameter and a line which is the difference between the radius and the aforesaid perpendicular from the centre.

Let ADB be a circle, of which the centre is C; DBE any arch, and DB the half of it; let the chords DE, DB be drawn; as also CF and CG at right angles to DE and DB; if CF be produced it will meet the circumference in B: let it meet it again in A, and let AC be bisected in H; CG is a mean proportional between AH and AF; and

[blocks in formation]

BD a mean proportional between AB and BF, the excess of the radius above CF.

Join AD; and because ADB is a right angle, being an angle in a semicircle; and because CGB is also a right angle, the triangles ABD, CBG are equiangular, and, AB: AD :: BC: CG (4. 6.), or alternately, AB: BC:: AD: CG; and therefore, because AB is double of BC, AD is double of CG, and the square of AD therefore equal to four times the square of CG.

But, because ADB is a right angled triangle, and DF a perpendicular on AB, AD is a mean proportional between AB and AF (8.6.), and AD=AB.AF (17.6.), or since AB is=4AH, AD=4AH.AF. Therefore also, because 4CGAD ̊, 4CG®=4AH.AF, and CG2= AH.AF; wherefore CG is a mean proportional between AH and AF, that is, between half the radius and the line made up of the radius, and the perpendicular on the chord of twice the arch BD.

Again, it is evident, that BD is a mean proportional between AB and BF (8. 6.), that is, between the diameter and the excess of the radius above the perpendicular, on the chord of twice the arch DB. Therefore, &c. Q. E. D.

PROP. IX. THEOR.*

The circumference of a circle exceeds three times the diameter, by a line less than ten of the parts, of which the diameter contains seventy, but greater than ten of the parts whereof the diameter contains seventy-one.

Let ABD be a circle, of which the centre is C, and the diameter AB; the circumference is greater than three times AB, by a line less

10

1

10of

than 10,or, of AC, but greater than of AC.

707

In the circle ABD apply the straight line BD equal to the radius

[blocks in formation]

In this proposition, the character+ placed after a number, signifies that something is to be added to it; and the character-,on the other hand, signifies that something is to be taken away from it.

BC: Draw DF perpendicular to BC, and let it meet the circumference again in E; draw also CG perpendicular to BD: produce BC to A, bisect AC in H, and join CD.

It is evident, that the arches BD, BE are each of them one sixth of the circumference (Cor. 15.4.), and that therefore the arch DBE is one third of the circumference. Wherefore, the line (8. 1. Sup.) CG is a mean proportional between AH, half the radius, and the line AF. Now because the sides BD, DC, of the triangle BDC are equal, the angles DCF, DBF are also equal; and the angles DFC, DFB being equal, and the side DF common to the triangles DBF, DCF, the base BF is equal to the base CF, and BC is bisected in F.

Therefore, if AC or BC=1000, AH=500, CF-500, AF-1500, and CG being a mean proportional between AH and AF, CG2= (17. 6.) AH.ÄF=500×1500=750000; wherefore CG=866.0254 +, because (866.0254) is less than 750000. Hence also, AC+CG 1866.0254+.

2

Now, as CG is the perpendicular drawn from the centre C, on the chord of one-sixth of the circumference, if P= the perpendicular from C on the chord of one-twelfth of the circumference, P will be a mean proportional between AH (8.1. Sup.) and AC+CG, and P3 — AH (AC+ CG)=500 x (1866.0254+)=933012.7+. Therefore, P 965.9258+, because (965.9258) is less than 933012.7. Hence also, AC+P=1965.9258+.

=

2

=

Again, if Q=the perpendicular drawn from C on the chord of one twenty-fourth of the circumference, Q will be a mean proportional between AH and AC+P, and Q-AH (AC+P)=500(1965.9258 +) 982962.9+; and therefore Q-991.4449+, because (991. 4449) is less than 982962.9. Therefore also AC+Q=1991.4449+. In like manner, if S be the perpendicular from C on the chord of one forty-eighth of the circumference, S- AH (AC + Q) = 500 (1991.4449+)=995722.45+; and S-997.8589+, because (997. 8589) is less than 995722.45. Hence also, AC+S=1997.8589+. Lastly, if T be the perpendicular from C on the chord of one ninety-sixth of the circumference, T=AH (AC+S)=500(1997.8589 +998929.45+, and T-999.46458+. Thus T, the dicular on the chord of one ninety-sixth of the circumference, is greater than 999.46458 of those parts of which the radius contains

1000.

perpen

But by the last proposition, the chord of one ninety-sixth part of the circumference is a mean proportional between the diameter and the excess of the radius above S, the perpendicular from the centre on the chord of one forty-eighth of the circumference. Therefore, the of the chord of one ninety-sixth of the circumference square AB (AC-S) =2000 × (2.1411-,)=4282.2-; and therefore the chord itself 65.4386-, because (65.4386) is greater than 4282.2. Now, the chord of one ninety-sixth of the circumference, or the side of an equilateral polygon of ninety-six sides inscribed in the circle,

=

« ΠροηγούμενηΣυνέχεια »