PROBLEM 26.

200. To make a square equal to two given squares (fig. 8, pl. III).

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, pl. III,) 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 di

vided 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 Aƒ, Ag, Ah, Ai, cutting DE in the points f, g, h, i. 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 ƒ, 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 (fig. 17, pl. II).—Divide AE and AF 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 1D, 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 f, and fC and ED, cutting each other in g. Bisect g C, by a perpendicular, cutting CD produced in i. Make Ek equal to Ei, and join Fi, cutting AB in 7. Make Em equal to El. Through m draw kp, and iq; and, through 7, draw kr, ih. From i, with the radius i C, describe an arc, qh; and, from k, with the same radius, describe an arc pr. From m, with the radius mp, describe an arc pAq; and, from 7, with the same radius, describe an arc rBh: then will the compound figure, Aq, Ch, Br, Dp, 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

PROBLEM 32.

209. To describe an ellipse with a trammel (fig. 4, pl. IV).

Draw the straight line AB, and make AB equal to the length of the given ellipse. Bisect AB in E, by the perpendicular CD. Make ED and EC each equal to half the lesser axis. Let the trammel-rod be fgh. Suppose the nut f to be fixed, and the nuts g and h to be moveable. Move the nut g so that the distance fg may be equal to EC or ED; and move the nut h, so that the distance fh may be equal to EA or EB: then set g and h in the grooves of the trammel, and, by moving the end of the rod f, so that the pin g, may slide along the line AB, and the pin h along the line CD; the point or pencil will trace out the curve of the ellipse as required.

PROBLEM 33.

210. A rectangle being given, to describe an ellipse, so that the two axes may have the same proportion as the sides of the rectangle (fig. 3, pl. IV).

Let ABCD be the given rectangle; it is required to describe an ellipse that shall pass through the points A, B, C, D, and of which the axes shall have a ratio equal to the sides AB, BC, or to CD, DA, of the rectangle.

Draw the diagonals AC, BD, cutting each other in I; and, through the point I, draw EG, parallel to AB or DC, cutting one of the sides BC in P ; and HF, parallel to AD or BC, cutting one of the sides AB in K. the radius IK, describe an arc, KL, cutting IG in L. Bisect the

From I, with

arc KL in M,

and draw mN, parallel to EG, cutting the diagonal DB in N. Join NP and NK, and draw BF, cutting HF in F, and draw BG, cutting EG in G. Make IE equal to IG, IH equal to IF; then EG and FH are the two axes of the ellipse, which will be described as in the following problem.

PROBLEM 34.

211. To describe an hyperbola, or a figure that may have any curvature at the summit, that we please (fig. 18, pl. II).

Let AC be the base, or what is called a double ordinate: make ED equal to the height, in the middle of AC; then, upon AC, as a side, describe a rectangle,

AFGC, so that the opposite side may pass through D; produce ED to the point B; take the point B, farther or nearer from D, according as the curvature at D is required to be flatter or quicker; and observe that, the quicker the curve is at D the flatter it will be towards A and C. The point B being thus fixed, divide AE into any number of parts, as here into four; also, divide AF into the same number of parts, viz. four; through 1, 2, 3, &c., the points of section in AE, draw lines to B ; and through the points of section, 1, 2, 3, &c. in AF, draw lines to D, cutting the former lines drawn to B, in the points h, i, k, &c.; and, through the points A, h, i, k, &c., draw a curve, which will be the half of an hyperbola, or an hyperbolic curve.

PROBLEM 35.

212. To describe a parabola upon a given ordinate, AE, and a given abscissa, ED* (fig. 19, pl. II).

Make EC equal to EA, and complete the rectangle AFGC; so that the opposite side may pass through D. Proceed as in the two former problems, 33 and 34; excepting that, instead of drawing the lines to B, through the points 1, 2, 3, &c., in AE, to draw them parallel to ED.

PROBLEM 36.

213. To describe the figure of the Sines † (fig. 20, pl. II).

Describe the quadrant FHG, equal to the height of the figure, and divide the arc HG into any number of equal parts; the more of these the more perfect the operation will be; and extend the chords to double the number of parts upon the line AC, which is a continuation of FH, and mark the points of division. Draw the lines 1k, 27, 3m, &c. perpendicular to AC; and, from the points 1, 2, 3, &c. of division in the quadrant, draw lines 1k, 21, 3m, &c. parallel to AC, and through the points A, k, l, m, &c., draw a curve, which will be the figure of the sines, as required.

* The PARABOLA is a figure arising from the section of a cone, when cut by a plane parallel to one of its sides. This, with other terms in Conics, is fully described, under CONIC SECTIONS, hereafter.

The SINE, or right sine, of an arch, is a right line drawn from one end of that arch, perpendicular to a radius, drawn to the other end of the same.-See TRIGONOMETRY, hereafter.