between the front string and level facia, and in many other instances. The curve is not circular, but of the form of the parabola, (Fig. 93;) yet in large angles the difference is not perceptible. This problem can be applied to describing the

C

Fig. 49.

curve for door heads, window-heads, &c., to rather better advantage than Art. 87. For instance, let a b, (Fig. 49,) be the width of the opening, and cd the height of the arc. Extend c d, and make de equal to c d; join a and e, also e and b; and proceed as directed above.

с

Fig. 50

90.-To describe a circle within any given triangle, so that the sides of the triangle shall be tangical. Let a bc, (Fig. 50,) be the given triangle. Bisect the angles a and b, according to Art. 77; upon d, the point of intersection of the bisecting lines, with the radius, de, describe the required circle.

91.-About a given circle, to describe an equi-lateral tri angle. Let a db c, (Fig. 51.) be the given circle. Draw the diameter, cd; upon d, with the radius of the given circle, describe the arc, a eb; join a and b; draw ƒ g, at right angles to de; make fc and c g, each equal to a b; from ƒ, through a, draw ƒ h, also from g, through b, draw g h; then f g h will be the triangle required.

92.-To find a right line nearly equal to the circumference of a circle. Let a b c d, (Fig. 52,) be the given circle. Draw the diameter, a c; on this erect an equi-lateral triangle, a e c, according to Art. 93; draw gf, parallel to a c; extend e c to f, also e a to g; then gf will be nearly the length of the semi-circle, a d c; and twice gf will nearly equal the circumference of the circle, a b c d, as was required.

Lines drawn from e, through any points in the circle, as o, o and o, to P, P and P will divide gf in the same way as the semi-circle, a d c, is divided. So, any portion of a circle may be transferred to a straight line. This is a very useful problem, and should be well studied; as it is frequently used to solve problems on stairs, domes, &c.

92, a.—Another method. Let a bf c, (Fig. 53,) be the given circle. Draw the diameter, a c; from d, the centre, and at right angles to a c, draw db; join b and c; bisect be at e; from d, through e, draw df; then ef added to three times the diameter, will equal the circumference of the circle sufficiently near for

b

C

Fig. 53.

many uses. The result is a trifle too large, If the circumfer erence found by this rule, be divided by 648-22, the quotient will be the excess. Deduct this excess, and the remainder will be the true circumference. This problem is rather more curious than useful, as it is less labor to perform the operation arithmetically simply multiplying the given diameter by 3-1416, or where a great degree of accuracy is needed by 3.1415926.

93.-Upon a given line to construct an equi-lateral triangle. Let a b, (Fig. 54,) be the given line. Upon a and b, with a b for radius, describe arcs, intersecting at c; join a and c, also c and b; then a cb will be the triangle required.

94.-To describe an equi-lateral rectangle, or square. Let a b, (Fig. 55,) be the length of a side of the proposed square. Upon a and b, with a b for radius, describe the arcs a d and bc; bisect the arc, a e, in f; upon e, with e f for radius, de

scribe the arc, cf d; join a and c, c and d, d and b; then a o db will be the square required.

95.- Within a given circle, to inscribe an equi-lateral tri angle, hexagon or dodecagon. Let a b c d, (Fig. 56,) be the given circle. Draw the diameter, b d; upon b, with the radius of the given circle, describe the arc, a e c; join a and c, also a and d, and c and d-and the triangle is completed. For the hexagon: from a, also from c, through e, draw the lines, a f and c g; join a and b, b and c, c and ƒ, &c., and the hexagon is completed. The dodecagon may be formed by bisecting the sides of the hexagon.

Each side of a regular hexagon is exactly equal to the radius of the circle that circumscribes the figure. For the radius is equal to a chord of an arc of 60 degrees; and, as every circle is supposed to be divided into 360 degrees, there is just 6 times 60, or 6 arcs of 60 degrees, in the whole circumference. A line drawn from each angle of the hexagon to the centre, (as in the figure,) divides it into six equal, equi-lateral triangles.

96.- Within a square to inscribe an octagon. Let a b c d,

(Fig. 57,) be the given square. Draw the diagonals, a d and bc; upon a, b, c and d, with a e for radius, describe arcs cutting the sides of the square at 1, 2, 3, 4, 5, 6, 7 and 8; join 1 and 2, 3 and 4, 5 and 6, &c., and the figure is completed.

In order to eight-square a hand-rail, or any piece that is to be afterwards rounded, draw the diagonals, a d and b c, upon the end of it, after it has been squared-up. Set a gauge to the distance, a é, and run it upon the whole length of the stuff, from each corner both ways. This will show how much is to be chamfered off, in order to make the piece octagonal. (Art. 159.)

97.- Within a given circle to inscribe any regular polygon. Let a b c 2, (Fig. 58, 59 and 60,) be given circles. Draw the diameter, a c; upon this, erect an equi-lateral triangle, a e c, according to Art. 93; divide a c into as many equal parts as the polygon is to have sides, as at 1, 2, 3, 4, &c.; from e,