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by the sum and difference of the hypothenuse, and one of the sides is equal to the square of the other side.

2

4. Since by cor. 2. AC+ CB.AC-CB-AB22.AB.BD, and AC+CB.AC—CB=AC2—CB 2 (5. 2.) ·· AC2—CB)2=AB÷F 2.AB.BD, and AC2=AB)2+CB2F2.AB.BD. Or the square of the side AC is less or greater than the sum of the squares of AB and CB, by twice the rectangle contained by the base, and the segment CB, according as the angle ABC is acute or obtuse. This is the same as 12 and 13. 2 Euclid.

250. B. The chord of one sixth part of the circumference being given, to find the chord of half that arc, and thence to inscribe within the circle a polygon of a great number of sides.

Let ABD be a semicircle, C its centre, draw the chord DA=AC (1.4), bisect the arc DA in E (30. 3.), and join EA; EA will be the side of a regular polygon of 12 sides. Bisect EA, and draw a straight line from A to the point of section, and it will be the side of a polygon of 24 equal sides; and by continually bisecting, we obtain the sides of polygons of 48, 96, 192, 384, &c. equal sides.

D

E

B

251. To find the circumference and area of a circle, having the diameter given r.

RULE. First. Since there is no geometrical method for determining accurately, the length of the whole, or any part of the circumference, we must be content with an approximation; which however, may be obtained to such a degree of exactness, as to differ from the truth by a line less than any given line.

Secondly. If two similar polygons of a great number of sides, be one inscribed in, and the other circumscribed about a circle,

? This problem will serve to shew by what laborious methods Wallis, Romanus, Metius, Snellius, Van Ceulen, and others, obtained approximations to the circles periphery; the same thing may however be performed with much more expedition and ease, by the method of fluxions, infinite series, &c. See Simpson's Doctrine and Application of Fluxions, part. 1. sect. 8.

the circumference will be greater than the sum of the sides of the former, but less than the sum of the sides of the latter; and therefore, if the numbers expressing these sums agree in a certain number of figures, those figures may be considered as expressing (as far as they go) the length of the circumference which lies between the two polygons; and if half the difference of the remaining figures be added to the less number, or subtracted from the greater, the result will afford a still more accurate expression for the length of the circumference.

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Let r=AC=1, c=AB=1, the arc AE÷ED=DB, and t=AE the chord of one third of the arc AB; then since the arc EB is double the arc AE, the angle EAH=ACE (20.3.) and AEC is common, . the triangles AEC and AEH are equiangular (32.1.) and CA: AE :: AE: EH (4.6.); that is, r:

xx

x:x: EH; also CE: AE:: AH : EH ··· AE=AH; in

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like manner it is shewn that BD=BK, ·: AĤ=BK, ·: AH+ BK=2x, and HK=(AB~AH—BK=) c−2x; but CE: ED

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:: CH: HK (4.6.), or r: x :: r--:c-2x; whence, multiply.

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tion, &c. (since c and r each =1,) becomes r3-3x=-1, the root of which is the chord of AE, or of T's part of the whole circumference.

Next to trisect the arc AE, let 3y-y=x, the chord of AE,

we shall have r3 27 y3 —27 y3 +9 y1 — y9

and -3x=-9y+3 y3.

and +1 =

5

+1

Their sum x3-3x+1=-9y+30 y3 — 27 y3 +9 y1—y°+1=0, the root of which is the chord of part of the whole circumference.

Again, to trisect the arc of which y is the chord; let 3 z— z3=y, and if this value be substituted for y in the last equation, we shall obtain an expression in which the value of z will be the chord of the part of the whole circumference. Proceeding in this manner after sixteen trisections, the chord of 2557376 part of the circumference (the radius being unity) will be found to be .000000024326999289832033, which number being multiplied by 258280326 (or the number of sides of the polygon, of which the above number expresses the length of one side) the product will be 6.283185307179585968482758 =the perimeter of the inscribed polygon.

F

H

E B

252. Next, we are to find the length of the side of a circumscribed polygon of the same number of sides, in order to which, let AB=the side D of the inscribed polygon as found above, A DE the side of a similar circumscribed polygon; bisect AB in H, join CH and produce it to F. Then CAP-AH}2 = √ H2 or 12-.000000012163499644916)2= 1— .000000000000000147950723611871658

084647056 = .9999999999999998520492 76388128342, &c.=CH2, the square root of which number is .99999999999999999

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.000000012163499644916, &c. DF, which number multiplied by 2 gives .00000002432699929832, &c.==DE

But

.00000002432699928983, &c.= AB

and since these two numbers agree as far as the 16th place of decimals, and the arc AFB lies between DE and AB, it follows, that those 16 decimal places will express the length of the arc

AFB very nearly; that is, the above number will differ from the truth by a very small decimal, whose highest place is 17 places below unity. Whence .0000000243269992842=the length of the arc AFB or of the part of the whole circumference extremely near. Now if the length of the arc AFB as above determined be multiplied into the denominator of this fraction, the product will be 6.2831853061898472=the circumference of a circle whose diameter is 2, very nearly

253. Having found the circumference of a circle, we can readily find the area, if not with strict accuracy, at least sufficiently near the truth for any practical purpose, in order to which, let us suppose an indefinite number of straight lines drawn from the centre to the circumference, these will divide the circle into as many sectors, the bases of which will be indefinitely small arcs, and their common altitude the radius of the circle; now since these small arcs coincide indefinitely near with the sides of a circumscribed or inscribed polygon of the same number of sides as there are sectors, these sectors may evidently be considered as triangles, the bases of which are the above small arcs, and their common altitude the radius; but half the base of a triangle, multiplied into the altitude, will give the area (42. 1.) wherefore, (half the sum of the bases, that is) half the circumference of the circle, multiplied into the radius, will give the area of the triangles, that is, the area of the circle; thus 6.2831853, &c. × 1

2

diameter is 2.

=3.1415926, &c.—the area of a circle, whose

n

254. Having found the circumference of a circle, whose diameter is 2, we are by means of it enabled to find the circumference of any other circle, whatever its diameter may be; for let the inscribed polygon (whose sides coincide indefinitely near with the circumference) have a sides, the length of each being r; and let a similar polygon be inscribed in any other circle having the length of its sides, then will nr=the periphery of the first polygon, and ns that of the second. Let the radius of the former circle, t=that of the latter; then if lines be drawn from each centre to the points of division in the respective circumferences, we shall have 1:r::t: s, (4. 6.) whence (16.5.) 1: t::r: s, and consequently (15. 5.) 1: t:: nr ns, that is, the peripheries of the similar polygons are to each

other as the radii of their circumscribed cricles; but these polygons coincide indefinitely near with their circumferences: wherefore the circumferences of circles are as their radii.

255. The area of one circle being known, that of another circle having a given diameter, may be found; let D=the diameter of a circle, A=its area, and d=the diameter of another circle, whose area x is required; then (2.12.) D2 : d2 : : A ; x, d2 A the area required. D2

whence r¬

PRACTICAL GEOMETRY.

255. Practical Geometry teaches the application of theoretical Geometry, as delivered by Euclid and other writers, to practical uses".

256. To draw a straight line from a given point A, to represent any length; in yards, feet, inches, &c.

RULE. I. Let each of the divisions on any convenient scale of equal parts represent a yard, foot, inch, or other unit of the measure proposed.

II. Extend the compasses on that scale until the number of units proposed be included exactly between the points.

III. With this distance in the compasses, and one foot on A, describe a small arc at B; lay the edge of a straight scale or ruler from A to B, and draw the line AB with a pen or pencil, and it will be the line required.

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EXAMPLES.-1. To draw a straight line from the point A to represent 12 inches.

a The following problems are intended as an introduction to the practical. application of some of the principal propositions in the Elements of Euclid, and likewise to assist the student in acquiring a knowledge of the use of a case of mathematical instruments. From a great variety of problems usually given by writers on Practical Geometry, we have selected such as appear most necessary, and likewise such methods of solving them as appear most simple and obvious; to a learner well acquainted with Euclid, other methods will occur, and he should be encouraged to exercise his ingenuity in discovering and applying them. The best elementary treatises on Practical Geometry and Mensuration, are those of Mr. Bonnycastle and Dr. Hutton.

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