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" The lateral area of a frustum of a cone of revolution is equal to the circumference of a section equidistant from its bases multiplied by its slant height. "
Elements of Geometry: Geometry of space - Page 426
by Andrew Wheeler Phillips, Irving Fisher - 1898
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Elements of Geometry: Including Plane, Solid, and Spherical Geometry

George Washington Hull - Geometry - 1807 - 408 pages
...PROPOSITION XI. THEOREM. 531. The 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. Given — S the lateral area, C and c the circumferences of the bases, and L the slant height of a frustum...
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A Treatise on Elementary Geometry: With Appendices Containing a Collection ...

William Chauvenet - Geometry - 1871 - 380 pages
...altitudes, or as the squares of the radii of their bases. /"r\ -.vA-.y PROPOSITION VI.— THEOREM. >.. 26. The lateral area of a frustum of a cone of revolution is equal to the half sum of the circumferences of its bases multiplied by its slant height. The plane which cuts...
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Elements of Geometry and Trigonometry from the Works of A.M. Legendre ...

Charles Davies - Geometry - 1872 - 464 pages
...and its slant height is equal to that of the cone: hence, the convex surface of the frustum of a cone is equal to half the sum of the circumferences of its bases multiplied by the slant height; which was to be proved. Scholium. From the extremities A and Z>, and from the middle...
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A Treatise on Elementary Geometry: With Appendices Containing a Collection ...

William Chauvenet - Geometry - 1872 - 382 pages
...of their altitudes, or as the squares of the radii of their bases. PROPOSITION VI.— THEOREM. 26. The lateral area of a frustum of a cone of revolution is equal to the half sum of the circumferences of its bases multiplied by its slant height. The plane which cuts...
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Elements of Geometry and Trigonometry: From the Works of A.M. Legendre

Adrien Marie Legendre - Geometry - 1874 - 500 pages
...its slant height is equal to that of the cone : hence, the convex surface of the frustum of a cone is equal to half the sum of the circumferences of its bases multiplied by the slant height ; which was to be proved. Scholium. From the extremities A and D, and from the middle...
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Annual Statement, Volumes 11-20

1876 - 646 pages
...are mutually equiangular they are also mutually equilateral ; and are either equal or symmetrical. 8. The lateral area of a frustum of a cone of revolution is equal to the half sum of the circumferences of its bases multiplied by its slant height. ENGLISH GRAMMAR. JUNE,...
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Manual of Geometry and Conic Sections: With Applications to Trigonometry and ...

William Guy Peck - Conic sections - 1876 - 376 pages
...a frustum of a right pyramid, whose bases are regular polygons having an infinite number of sides. half the sum of the circumferences of its bases multiplied by its slant height, (P. 4, Cor., B. 7). Cor. 3. The volume of a conic frustum is equal to the sum of its bases ph1s a mean...
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Elements of Plane and Solid Geometry

George Albert Wentworth - Geometry - 1877 - 416 pages
...frustum of a cone of revolution. Draw CHI-, and AD II, to YY'. Then area AB = AB X 2 jr СH, §662 (the lateral area of a frustum of a cone of revolution is equal to the slant height multiplied 'by the circumference of a section equidistant from its bases). The AABD...
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Yale Examination Papers

F. B. Stevens - Examinations - 1884 - 202 pages
...mutually equiangular, they are also mutually equilateral ; and are either equal or symmetrical. 8. The lateral area of a frustum of a cone of revolution is equal to the half sum of the circumferences of its bases multiplied by its slant height. June, 1882. NOTE 1....
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Elements of Geometry and Trigonometry from the Works of A.M. Legendre ...

Charles Davies, Adrien Marie Legendre - Geometry - 1885 - 538 pages
...its slant height is equal to that of the cone : hence, the convex surface of the frustum of a cone is equal to half the sum of the circumferences of its bases multiplied by its slant height ; which was to be proved. • Scholium. From the extremities A and D, and from the middle point I,...
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