and B as centers, with the radius AB, describe arcs cutting each other in g. Again, from g, the point of intersection, with the same radius, describe the circle ABC, which will contain the given side AB six times when applied to its circumference, and will be the hexagon required.

Problem 17.-To describe a regular octagon upon a given straight line: Let AB (Fig. 40) be the given line. From the extremities A and B erect the perpendiculars AE and BF; extend the given line both ways to k and 1, forming external right angles with the lines AE and BF. Bisect these external right angles, making each of the bisecting lines AH and BC equal to the given line AB. Draw HG and CD parallel to AE or BF, and each equal in length to AB. Draw from G, GE parallel to BC, and intersecting AE in E, and from D draw DF parallel to AH, intersecting BF in F. Join EF, and ABCDFEGH is the octagon required. Or from D and G as centers, with the given line AB as radius, describe arcs cutting the perpendiculars AE and BF in E and F, and join GE, EF, FD, to complete the octagon.

Otherwise, thus:-Let AB (Fig. 41) be the given straight line upon which the octagon is to be described. Bisect it in a, and draw the perpendicular ab equal to Aa or Ba. Join Ab, and produce ab to c, making bc equal to Ab; join also Ac to Bc, extending them so as to make cE and CF each equal to Ac or Bc. Through c draw Ccg, at right angles to AE. Again, through the same point c, draw DH at right angles to BF,

making each of the lines cC, CD, cG, and cH equal to Ac and Bc, and, consequently, equal to each other. Lastly, join BC, CD, DE, EF, FG, GH, HA; ABCDEFGH will be the regular octagon described upon AB, as required.

Problem 18.-In a given square to inscribe a given octagon: Let ABCD (Fig. 42) be the given square. Draw the diagonals AC and BD, inter

H

FIG. 41.

m

FIG. 42.

secting each other in e; then from the angular points ABC and D as centers, with a radius equal to half the diagonal, viz.: Ae or Ce, describe arcs cutting the sides of the square in the points f, g, h, k, l, m, n, o, and the straight lines of gh, kl, and mn, joining these points will complete the octagon, and be inscribed in the square ABCD, as required.

Problem 19.-To find the area of any regular polygon: Let the given figure be a hexagon; it is required to find its area. Bisect any two adjacent angles, as those at A and B (Fig. 43), by the straight lines AC and BC, intersecting in C, which

will be the center of the polygon. Mark the altitude of this elementary triangle by the dotted line drawn from C perpendicular to the base AB; then multiply together the base and altitude thus found, and this product by the number of sides: one-half of the product gives the area of the whole figure.

Or otherwise thus: Draw the straight line DE, equal to six times; that is, as many times

AB, the base of the elementary triangle, as there are sides in the given polygon. Upon DE describe an isosceles triangle, having the same altitude as ABC, the elementary triangle of the given polygon; the triangle thus constructed is equal in area to the given hexagon; consequently, by multiplying the base and the altitude of this triangle together half the product will be the area required. The rule may be expressed in other words as follows: The area of a regular polygon is equal to its perimeter, multiplied by half the radius of its inscribed circle, to which the sides of the polygon are tangents.

Problem 20.-To describe the circumference of a circle through three given points: Let A, B, C (Fig. 44) be the given points not in a straight line. Join AB and BC; bisect each of the straight lines AB and BC by perpendiculars meeting in D; then A, B and C are all equi-distant from D; therefore a circle described from D, with the radius DA, DB, or DC, will pass through all the three points, as required.

Problem 21.-To divide a given circle into any number of equal or proportional parts by concentric divisions: Let ABC (Fig. 45) be the given

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D

B

FIG. 45.

FIG. 46..

circle, to be divided into five equal parts. Draw the radius AD, and divide it into the same number of parts as those required in the circle; and upon the radius thus divided, describe a semicircle; then from each point of division on AD, erect perpendiculars to meet the semi-circumference in e, f, g and h. From D, the center of the given circle, with radii extending to each of the different points of intersection on the semicircle,

describe successive circles, and they will divide the given circle into five parts of equal area, as required; the center part being also a circle, while the other four will be in the form of rings.

Problem 22.-To divide a circle into three concentric parts, bearing to each other the proportion of one, two, three, from the center: Draw the radius AD (Fig. 46), and divide it into six equal parts. Upon the radius thus divided, describe a semicircle: from the first and third points of division, draw perpendiculars to meet the semicircumference in e and f. From D, the center of

the given circle, with radii extending to e and f, describe circles which will divide the given circle into three parts, bearing to each other the same proportion as the divisions on AD, which are as 1, 2 and 3. In like manner circles may be divided in any given ratio by concentric divisions.

Problem 23.-To draw a straight line equal to any given arc of a circle: Let AB (Fig. 47)