| Etienne Bézout - Calculus - 1824 - 222 pages
...kind of demonstration was called reductio ad absurdum. By this means, having first ascertained that **the areas of similar polygons are to each other as the squares** of their homologous lines, they inferred that circles of different radii are to each other as the square... | |
| Francis Joseph Grund - Geometry, Plane - 1830 - 274 pages
...sides opposite to the equal angles, and also as the squares upon the heights of the triangles. 21. **The areas of similar polygons are to each other, as...SECTION IV. OF THE PROPERTIES OF THE CIRCLE.* QUERY I.** hi how many points can a straight line CD, cut the circumference of a circle ? A, In two points M,... | |
| Adrien Marie Legendre - Geometry - 1836 - 359 pages
...polygon FGHIK, as one antecedent ABC, is to its consequent FGH, or as ABS is to FG2 (Prop. XXV.)j 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... | |
| Charles Davies - Geometrical drawing - 1840 - 252 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. § VII. Art. 3). we have I2 : tabular area : : any... | |
| J. M. Scribner - Measurement - 1844 - 123 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... | |
| 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 - 308 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... | |
| 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... | |
| J. M. Scribner - Mechanical engineering - 1849 - 264 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... | |
| Charles Davies - Trigonometry - 1849 - 384 pages
...Decagon .... 10 ... 7.0942088 Dodecagon ... 12 ... 11.1901524 Undecagon ... 11 ... 9.3050399 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 Or, to find the area of any regular... | |
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