 | Thomas Dix (of Oundle.) - 1799 - 242 páginas
...is the fide of a Decagon. PROBLEM. XXVII. To infer ibe any regular Polygon, in a given circle. Draw the diameter AB, and divide it into as many equal parts as the Polygon is to have fides. With the radius AB, and A as a centre, defcribe an arc. With the fame radius, and В as a centre,... | |
 | Peter Nicholson - 1809 - 426 páginas
...describe two arcs intersecting each other at F. 2. From B, draw BC perpendicular, and divide the arc AC into as many equal parts as the polygon is to have sides. 3. Through the second division D, draw BG, make FE equal to FD, and through E, draw AG, meeting BG... | |
 | Pierce Morton - 1830 - 584 páginas
...circumscribed about the circle, viz. by dividing the circumference (or conceiving it to be divided) into as many equal parts as the polygon is to have sides, and drawing tangents through the points of division. Then, because the area of the polygon is equal... | |
 | Edward Shaw - 1832 - 402 páginas
...two arcs, intersecting each other at F. 2. From B, draw В С perpendicular, and divide the arc А С into as many equal parts as the polygon is to have sides. 3. Through the second division D, draw BG, make FE equal to FD, and through E, draw AG, meeting BG... | |
 | 1835 - 684 páginas
...circumscribed about the circle, viz. bv dividing the circumference (or conceiving it to be divided) into as many equal parts as the polygon is to have sides, and drawing tangents through the points of division. Then, because the area of the polygon is equal... | |
 | Edward Shaw - 1836 - 438 páginas
...two arcs intersecting each other at F. 2. From B, draw В С perpendicular, and divide the arc А С into as many equal parts as the polygon is to have sides. 3. Through the second division D, draw BG, make FE equal to FD, and through E, draw AG, meeting BG... | |
 | Chambers W. and R., ltd - 1842 - 744 páginas
...circle about equal in size to the required size of the polygon ; then the circumference is divided into as many equal parts as the polygon is to have sides ; all that then remains to complete the figure is to draw straight lines or chords between each two... | |
 | William Pease - 1843 - 80 páginas
...AB any length on either side as A C. From A as a centre, with AB as a radius, describe a semicircle. Divide it into as many equal parts as the polygon is to have sides (say five.) 3. Draw lines through all the divisions, (minus one,) as AD, A 3 E, A 4 F. 4. From D and... | |
 | Robert Griffith Hatfield - 1845 - 326 páginas
...acdb will be the square required. Fig- 61. Fig. C2. Fig. 63. AMERICAN HOUSE-CARPENTER. feb ; divide ac into as many equal parts as the polygon is to have sides, and extend those divisions from c towards d ; from the second point of division counting from c towards... | |
 | George Roberts Perkins - 1847 - 326 páginas
...polygon of a certain number of sides, in a given circle, we must be able to divide the circumference into as many equal parts as the polygon is to have sides ; and then, by joining these successive points of division, we shall form the polygon desired. For... | |
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AB of the circle into as many equal parts as the polygon is to have sides. With the points A and B as centers and radius AB, describe arcs cutting each other at C. 
