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ANOTHER METHOD.

193. Let AC (Jig. 15, pl. II,) be the length or chord, and BD the versed sine. Join AB, and draw BE parallel to AC, making BE of any length, not less than AB. Form a triangular piece of wood, ABE: bring the angular point B, of the triangle, to the point A; and move the triangle, so that the side BA may slide upon A, and the side BE upon B: then if, during the motion, a pencil be held at the angular point B, with its point tracing over the plane, the arc AB will be described by the point of the pencil. The arc AB being described, the arc BC will be described in a similar manner; and, consequently, the whole segment of the circle, as required to be done.

Problem 20.

194. Between two straight lines, E and F, (Jig. 1, pl. Ill,) to find a mean proportional.

Draw the straight line AB. Make AC equal to E, and CB equal to F. Upon AB, as a diameter, describe the semi-circle ADB: from the point C draw CD, perpendicular to AB, and CD will be the mean proportional required.

Problem 21.

195. To find a straight line equal in length, nearly, to the arc of a circle. Let ABC, (Ar. 2, pi. HI,) be the given arc Join AC, which prolong to F.

Bisect the arc ABC in B, and make AE equal to twice AB. Divide CE into three equal parts, and set one of them off from E to F; then the straight line AF is nearly equal in length to the arc ABC.

Problem 22.

196. To describe a triangle, of which the three sides shall be equal to three given straight lines, provided that any two of them are greater than the third.

Let D, E, F, (Jig. 3, pl. HI,) be the three given straight lines. Draw AB, and make AB equal to the straight line D. From the point A, with the distance of the line F, describe an arc; and from the point B, with the extent of the line CE, describe another arc, cutting the former at C, and join AC and

BC: then is ABC the triangle required.

In this manner a triangle may be made equal to another given triangle;

for this is only making the sides of the triangle equal to those of the given

triangle.

Problem 23.

197. To describe a trapezium equal and similar to a given trapezium.

Let it be required to describe a trapezium equal and similar to the given trapezium, ABCD {fig. 4, pl. III).

Draw the straight line, FG, fig. 5, and make FG equal to BC: upon FG describe the triangle FGH, equal to the triangle BCD; and, upon FH, describe the triangle FHE, equal to the triangle BDA, and the whole figure, EFGH, will be equal and similar to the figure ABCD.

Problem 24.

198. To make a rectangle equal to a given triangle.

It is required to make a rectangle equal to the given triangle, ABC (fig. 6, pi III).

Draw CF perpendicular to AB, cutting AB in F. Divide CF into two equal parts in the point G. Through G draw DE parallel to AB, and draw AD and BE perpendicular to AB; then the rectangle ABED will be equal to the triangle ABC, as required to be done.

Problem 25.

199. To make a square equal to a given rectangle.

Let it be required to make a square equal to the given rectangle, ABCD, (fig. 7, pl. III).

Produce the side AB of the rectangle to E, and make BE equal to BC. Draw BG perpendicular to AE; and, on AE, as a diameter, describe the semicircle AGE; and, on the straight line, BG, describe the square BGFH; which is the thing required to be done.

We now see that a triangle may be reduced to a rectangle, and a rectangle may be reduced to a square; therefore a triangle may be reduced to a square.

PROBLEM M.

200. To make a square equal to two given squares {Jig. 8, pl. HI).

Draw the straight line, AB, and BC perpendicular to AB. Make AB equal to the side of one of the given squares, and BC equal to the side of the other, and join AC: then the square described upon AC will be equal to the sum of two other squares described upon AB, BC: Thus, on AB describe the square ABGF; on BC describe the square BHIC; and on AC describe the square ACDE; then will the square ACDE, described upon the hypothenuse, be equal to the two squares ABFG, CBHI, described upon the legs.

Problem 27.

201. To describe a square equal to any given number of squares {figures 9, 10, pl. III).

Let it be required to make a square equal to the three given squares, A, B, C. Make AB, fig. 10, equal to the side of the square A. Prolong BA to F, and draw BG perpendicular to FB, and on BG make BC equal to the side of the square B, and join AC; then a square described on AC is equal to the two squares A and B. Make BG equal to AC, and FB equal to the side of the square C, and join FG, and on FG describe the square FGHI; then the square FGHI is equal to the squares of FB and BG; but the square of FB is equal to the square of C; and, since AC is equal to the squares of A and B, the square of BG will be equal to the squares of A and B; therefore the square FGHI is equal to the three given squares, A, B, C.

Problem 28. To divide a straight line in the same proportion as another is divided.

202. Method the first.—Let AB, {fig. 11, pi. HI,) be a given straight line, divided into the parts AF, FD, DB; it is required to divide another straight line in the same proportion.

Draw AC, making any angle with AB; and make AC equal to the length of the line proposed to be divided. Join BC, and draw the straight lines DE, FG, parallel to BC, cutting AC in G and E; then will the line AC be divided in the same proportion as the straight line AB; as was required to be done.

203. Method the second.—Let BC, {fig. 1, pl. IV,) be the given line. On BC describe the equilateral triangle BCA. Make, on the sides of this triangle, AD, AE, each equal to the length of the line to be divided, and join DE; then DE will be equal to AD or AE: let the given line BC be divided in the points f, g, h, i. Draw A/, Ag, Ah, At, cutting DE in the points

f, g, h, t. Then the line DE will be divided by these points in the same proportion as the given line BC is divided by the points f, g, h, i.

Problem 29.

204. To describe an octagon whose opposite sides are at a given distance from each other {fig. 2, pl. IV).

Describe a square, ABCD, of which each side is equal to the distance of the opposite sides of the octagon. Draw the diagonals AC and BD, cutting each other in P. From each of the points, ABCD, with a radius equal to AP, BP, CP, or DP, one half of the diagonal, describe the arcs LPF, EPH, GPK, IPM, cutting AB, BC, CD, DA, in the points E, F, G, H, I, K, L, M, and join FG, HI, KL, ME; then will EFGHIKLM be the octagon required.

PROBLEM 30.

To describe an ellipse to any length and breadth.

205. Method the first {fig. 16, pl. II).—Draw the line AC, and make AC equal to the length required; bisect AC by a perpendicular BD, and make EB and ED each equal to half the breadth.

To find any point, g, in the curve; with the difference of ED and EA, as a radius, from any point f, in EB, describe an arc, cutting EC in h. Draw fh, and produce it to g, and make hg equal to EB or ED; then g will be a point in the curve, as required.

In the same manner we may find as many points in the curve as we please.

206. Method the second Chg. 17, pl. II).—Divide AE andAF each into the same number of equal parts, as here into five. Through the points of section 1, 2,3, &c draw the lines Bh, Bi, Bk, &c; and through the points of section, 1, 2,3, &c in AF, draw the lines ID, 2D, 3D, &c, cutting the former lines drawn from B, in the points h, i, k, &c; then through the points A, h, i, k, &c draw a curve, and we shall have the fourth part, or quarter, of the whole curve. In the same manner the other quarter DC may be found.

And, by taking the point D, instead of B, and by describing the rectangle upon AC, so that the opposite side may pass through B, and dividing and drawing lines in the same manner, we shall have the whole curve.

207. Method the third, with a string [fig. 5, pl. IV).—Having placed the axes at right angles, and bisecting each other, from one extremity, C, of the conjugate or shorter axis, with AE or BE, half of the transverse, or longer axis, describe arcs, cutting AB at F, f; then F, f, are the two focal points. Round the points F, f, stretch a string, and fasten the ends of it; carry the point forward, as to H, keeping the string always tense, and the point H, &c will describe the curve as required.

Problem 31.

208. To represent an ellipse by means of the arcs of circles {fig. 6, pl. IV).

Let AB be the length, and CD the breadth, as before. Draw BE perpendicular to CD. Make BF equal to EC. Bisect BF in /, and fC and ED, cutting each other in g. Bisect gC, by a perpendicular, cutting CD produced in i. Make Ek equal to Ei, and join Ft, cutting AB in l. Make Em equal to El. Through in draw hp, and iq; and, through /, drawer, ih. From i, with the radius iC, describe an arc, qh; and, from k, with the same radius, describe an arc pr. From m, with the radius mp, describe an arcpkq; and, from l, with the same radius, describe an arc rBh: then will the compound figure, Aq, CH, Br, DC, represent the ellipse required.

This method is used only for describing an arc on paper: it may also be used for drawing the lines, perpendicular to a real elliptic arch, for the joints of the stones. The first method of describing the curve may be used occasionally, from want of a proper instrument; but, of all the methods, none is so accurate, .in practice, as the trammel, which is next described.

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