PROBLEM XXXIV. In a given Circle, to infcribe a Trigon, a Hexagon, or a Dodecagon. The radius is the fide of the hexagon. Therefore from any point A in the circumference, with the distance of the radius, describe the arc BOF. Then is AB the fide of the hexagon; and therefore carrying it fix times. round will form the hexagon, or divide the circumference into fix equal parts, each containing 60 degrees.-The second of these c, will give AC the fide of the trigon or equilateral triangle; and the arc AC one-third of the circumference, or 120 degrees.-Alfo the half of AB, or an is one-12th of the circumference, or 30 degrees, and gives the fide of the dodecagon. Note. If tangents to the circle be drawn through all the angular points of any infcribed figure, they will form the fides of a like circumfcribing figure. PROBLEM XXXV. In a given Circle to infcribe a Pentagon, or a Decagon. Draw the two diameters AP, mn perpendicular to each other, and bifect the radius on at q.-With the center q and distance qA, defcribe the arc Ar; and with the center A, and radius ar, describe the arc rB. Then is AB one-fifth of the circumference; and AB B m A ૧ n P carried five times over will form the pentagon.-Alfo the arc AB bifected in s, will give As the tenth part of the circumference, or the fide of the decagon. Note. Note. Tangents being drawn through the angular points, will form the circumfcribing pentagon or decagon. PROBLEM XXXVI. To divide the Circumference of a given Circle into 12 equal Parts, each of 30 Degrees. Or to infcribe a Dodecagon by another Method. Draw two diameters 17 and 4 10 perpendicular to each other. Then with the radius of the circle, and the four extremities 1, 4, 7, 10, as centers, defcribe arcs through the center of the circle; and they will cut the circumference in the points required, dividing it into 12 equal parts, at the points marked with the numbers. 3 21 12 II 10 9 678 PROBLEM XXXVII. To draw a right Line equal to the Circumference of a given Circle. Take III 1 equal to 3 times the diameter and part more; and it will be equal to the circumference, very nearly. PRO To find a Right Line equal to any given Arc AB of a Circle, Through the point A and the center draw Am, making mn equal to of the radius no.-Alfo draw the indefinite tangent AP perpendicular to it. Then through m and B, draw mp; fo fhall AP be equal to the arc AB very nearly. Otherwife. Divide the chord AB into 4 equal parts.-Set one part AC on the arc from в to D.Draw CD, and the double of it will be nearly equal to the arc ADB. m B P A PROBLEM XXXIX. To divide a given Circle into any propofed Number of Parts by equal Lines, fo that thofe Parts fhall be mutually equal, both in Area and Perimeter. Divide the diameter AB into the propofed number of equal parts at the points a, b, c, &c,-Then on Aa, Ab, A AC, &c, as diameters, describe femicircles on one fide of the diameter AB; and on вd, BC, Bb, &c, defcribe femicircles B on the other fide of the diameter. So fhall the correfponding joining femicircles divide the given circle in the manner propofed. 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 the proportion of the spaces is. PROBLEM XL. On a given Line AB to defcribe the Segment of a Circle capable of containing a given Angle. Draw AC and Bc making the angles BAC and ABC each equal the given angle.Draw AD perpendicular to AC, and BD perpendicular to BC. With center D, and radius DA or DB, describe the fegment AEB. Then any angle, as E, made in that fegment, will be equal to the given angle. D B To make a Triangle fimilar to a given Triangle ABC. Let AB be one fide 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 fimilar to ABC as propofed. Note. If ab be equal to AB, the triangles will also be equal, as well as fimilar. C A B a PRO |