 | William Elwood Byerly - Calculus, Integral - 1881 - 220 pages
...the area of the surface of a frus— tum of a cone of revolution is, by elementary Geometry, one-half the sum of the circumferences of the bases multiplied by the slant height. The frustum generated by the chord in the figure will have an area differing by infinitesimals of higher... | |
 | William Elwood Byerly - Calculus, Integral - 1882 - 218 pages
...of the surface of a frus— turn of a cone of revolution is, by elementX l±X an" Geometry, one-half the sum of the circumferences of the bases multiplied by the slant height. The frustum generated by the chord in the figure will have an area differing by infinitesimals of higher... | |
 | William Elwood Byerly - Calculus, Integral - 1888 - 424 pages
...and the area of the surface of a frustum of a cone of revolution is, by elementary Geometry, one-half the sum of the circumferences of the bases multiplied by the slant height. The frustum generated by the chord in the figure will have an area differing by infinitesimals of higher... | |
 | George Washington Hull - Geometry - 1897 - 408 pages
...II L 3 H 3 ff PROPOSITION XI. THEOREM. 531. ITie lateral area of a frustum of a cone of revolution is equal to one half of the sum of the circumferences of its bases multiplied by its slant height. the bases, and L the slant height of a frustum of a cone... | |
 | Wooster Woodruff Beman, David Eugene Smith - 1903 - 158 pages
...respectively, of F', then if the number oj faces of F' increases indefinitely, PROPOSITION III. 441. Theorem. The lateral area of a frustum of a right circular cone equals one-half the product of the slant height and the sum of the circumferences of its bases. Given... | |
 | Wooster Woodruff Beman, David Eugene Smith - Geometry, Solid - 1900 - 160 pages
...respectively, of F', then if the number of faces of F' increases indefinitely, PROPOSITION III. 441. Theorem. The lateral area of a frustum of a right circular cone equals one-half the product of the slant height «nd the sum of the circumferences of its bases. Given... | |
 | William Elwood Byerly - Fourier series - 1902 - 474 pages
...and the area of the surface of a frustum of a cone of revolution is, by elementary Geometry, one-half the sum of the circumferences of the bases multiplied by the slant height. The frustum generated by the chord in the figure will have an area differing by infinitesimals of higher... | |
 | George Bruce Halsted - Geometry - 1904 - 324 pages
...A frustum of a cone is the portion included between the base and a plane parallel to the base. 455. Theorem. The lateral area of a frustum of a right circular cone is half the product of its slant height by the sum of the lengths of Us bases. Proof. It is the difference... | |
 | George Bruce Halsted - Geometry - 1904 - 313 pages
...A frustum of a cone is the portion included between the base and a plane parallel to the base. 455. Theorem. The lateral area of a frustum of a right circular cone is half the product of its slant height by the sum of the lengths of its bases. Proof. It is the difference... | |
 | Education - 1911 - 946 pages
...pyramid is one half the product of the slant height and the sum of the perimeters of the bases. [Q7-] 8. The lateral area of a frustum of a right circular cone is one half the product of the slant height and the sum of the circumferences of the bases. [Qg (a).]... | |
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The lateral area of a frustum of a right circular cone is one half the product of the slant height and the sum of the circumferences of the bases. 
