 | Adrien Marie Legendre - Geometry - 1838 - 382 pages
...Decagon .... 10 ... 7.6942088 Undecagon ... 11 ... 9.3656399 Dodecagon ... 12 .... 11.1961524 Now. since the areas of similar polygons are to each other as the squares of their homologous sides (Book IV. Prop. XXVII.), we shall have I- : tabular area : : any side squared : area. Or, to find the... | |
 | Charles Davies - Geometrical drawing - 1840 - 262 pages
...Undecagon, 9,3656404 1,2028437 i 12 Dodecagon, 11,1961524 1,8660254 Mensuration of Surfaces. Now, since the areas of similar polygons are to each other as the squares described on their homologous side (see Part I. ยง VII. Art. 3). we have I2 : tabular area : : any... | |
 | Adrien Marie Legendre - Geometry - 1841 - 288 pages
...FGHIK, as one antecedent ABC is to its consequent FGH, or as AB is to FG (219). Therefore the surfaces of similar polygons are to each other as the squares of their homologous sides. 222. Corollary. If three similar figures be constructed, whose homologous sides are equal to the three... | |
 | J. M. Scribner - Measurement - 1844 - 130 pages
...of each is equal to 1 : it also shows the length of the Radius of the inscribed circle. Now, since the areas of similar polygons are to each other as the squares of their homologous sides, if the square of the side of a polygon be multiplied by the multiplier of the like figure, the product... | |
 | Charles Davies - Geometrical drawing - 1846 - 254 pages
...radius of the inscribed circle. 50. How do you find the area of any polygon from the above table ? Since the areas of similar polygons are to each other as the squares described on their homologous sides, we have 1* : tabular area : : any side squared : area. Hence,... | |
 | George Roberts Perkins - Geometry - 1847 - 326 pages
...GHKLM, as any one antecedent ABC, is to its corresponding consequent GHK, or as AB' is to GH'. Hence the areas of similar polygons are to each other as the squares of their homologous sides. Cor. If three similar rectilineal. figures are constructed on the three sides of a right-angled triangle,... | |
 | Benjamin Peirce - Geometry - 1847 - 204 pages
...as the squares of their homologous altitudes, and as the squares of their perimeters. 268. Theorem. Similar polygons are to each other as the squares of their homologous sides. Proof. In the similar polygons AB CD &c., A'B'C'D' &c. (fig. 108), the triangles ABC, A'B C', which... | |
 | J. M. Scribner - Mechanical engineering - 1849 - 286 pages
...showing the multipliers of the ten regular polygons, when the sides of each are eqflal to 1 : Now, since the areas of similar polygons are to each other as the squares of their homologous sides, if the square of a side of a polygon be multiplied by the multiplier of the like figure, the product... | |
 | Charles Davies - Trigonometry - 1849 - 372 pages
...FGHIK, as one antecedent ABC, is to its consequent FGH, or as AB 2 is to FG 2 (Prop. XXV.) ; hence the areas of similar polygons are to each other as the squares described on the homologous sides. Cor. If three similar figures were constructed, on the three sides... | |
 | Daniel Adams - Arithmetic - 1849 - 142 pages
...feet, and the altitude of one of its equal triangles is 8'660254 feet ? Ans. 259>80762 sq. ft. IT 54. The areas of similar polygons are to each other as the squares of one of their sides. H 52. Hence, the areas of regular polygons may be more readily found by the... | |
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Now, since the areas of similar polygons are to each other as the squares of their homologous sides... 
