circumference; and A B, carried round five times, will form the pentagon. Also the arc A B, bisected in S, will give AS, the tenth part of the circumference, or the side of the decagon. Another Method. Inscribe the isosceles triangle ABC, having each of the angles ABC, ACB, double the angle BA C. Then D bisect the two arcs ADB, A E C, in the points D, E; and draw the chords AD, DB, AE, EC; so shall ADB CE be the inscribed pentagon required. And the decagon is thence obtained as before. NOTE. Tangents, being drawn through the angular points, will form the circumscribing pentagon or decagon. PROBLEM XXXVII. To divide the Circumference of a given Circle into twelve equal Parts, each being 30 Degrees. Or to inscribe a Dodecagon by another Method. quired, dividing it into 12 equal parts at the points marked with the numbers. PROBLEM PROBLEM XXXVIII. To divide a given Circle into any proposed Number of Parts by equal Lines, so that those Parts shall be mutually equal, both in Area and Perimeter. Divide the diameter AB into the proposed number of equal parts at the points a, b, c, &c. Thenon Aa, Ab, Ac, &c. as diameters, describe semicircles on one side of the diameter A B; and on Bd, Bc, Bb, &c. describe semicircles on the other side of the diameter. So shall the corresponding joining semicircles divide the given circle in the manner proposed. And in like manner we may proceed, when the spaces are to be in any given proportion. As to the perimeters, they are always equal, whatever may be the proportion of the spaces. PROBLEM XXXIX. On a given Line AB to describe the Segment of a Circle, capable of containing a given Angle. Draw AC and BC, making the angles BAC and ABC each equal to the given angle. Draw AD perpendicular to AC, and BD perpendicular to BC. With centre D, and radius D A, or D B, describe the segment AEB. Then any angle, as E, made in that segment, will be equal to the given angle. D E PROBLEM PROBLEM XL. To cut off a Segment from a given Circle, that shall contain a given Angle C. Draw any tangent AB to the given circle; and a chord AD, to make the angle DAB equal to the given angle C; then DEA will be the segment required, any angle E made in it being equal to the given angle C. A PROBLEM XLI. To make a Triangle similar to a given Triangle ABC. Let ab be one side of the required triangle. Make the angle a equal to the angle A, and the angle b equal to the angle B; then the triangle abc will be similar to A B C, as proposed. 1 B NOTE. If ab be equal to A B, the triangles will also be equal, as well as similar. PROBLEM PROBLEM XLII. To make a Figure similar to any other given Figure ABCDE. To make a Triangle equal to a given Trapezium ABCD. Draw the diagonal DB, and D CE parallel to it, meeting AB produced in E. Join DE; so shall the triangle ADE be equal to the trapezium AB CD. A B PROBLEM PROBLEM XLIV. To make a Triangle equal to the Figure ABCDEA. Draw the diagonals DA, DB, and the lines EF, CG, parallel to them, meeting the base AB, E both ways produced, in F and G. Join DF, DG; and DFG will be the triangle required. D NOTE. Nearly in the same manner may a triangle be made equal to any right-lined figure whatever. PROBLEM XLV. To make a Rectangle, or a Parallelogram, equal to a given Triangle ABC. |