 | George Albert Wentworth - Mathematics - 1896 - 68 pages
...equal to one-half the sum of the circumferences of its bases multiplied by the slant height. 676. Cor. 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. 677. The volume... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry - 1896 - 554 pages
...cone of revolution is equal to the slant height of the frustum of a cone. PROPOSITION VI. THEOREM 002. The lateral area of a frustum of a cone of revolution is equal to Iialf the sum of the circumferences of -its bases multiplied by its slant height. GIVEN a frustum of... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry - 1896 - 570 pages
...radius of a section half-way between the bases of the frustum, R+r Hence K=S=2irKx L. That is, thc lateral area of a frustum of a cone of revolution is equal to the circumference of a section half-way between its bases multiplied by its slant height. PROPOSITION VII.... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry - 1897 - 374 pages
...is the radius of a section half-way between the bases of the frustum, 2 Hence S—2irKxL. That is, the lateral area of a frustum of a cone of revolution is equal to the product of the circumference of a section half-way between its bases and its slant height. 8lS6. Exercise. — Find... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry - 1896 - 574 pages
...If K is the radius of a section half-way between the bases of the frustum, rr R + r Hence That is, the lateral area of a frustum of a cone of revolution is equal to the circumference of a section half-way between its bases multiplied by its slant height. PROPOSITION VII.... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry - 1897 - 376 pages
...cone of revolution is equal to the slant height of the frustum of a cone. PROPOSITION VI. THEOREM 883. The lateral area of a frustum of a cone of revolution is equal to one-half the product of the sum of the circumferences of its bases and its slant height. GIVEN a frustum... | |
 | George Washington Hull - Geometry - 1897 - 408 pages
...variables which are always equal. Hence lim. S'— lim. l(P+p)L', §140 Or S=l(C+c)L. QED 532. COR. 1.— The lateral area of a frustum of a cone of revolution is S = n(R + r)L, in which R and r are the radii of the bases. For S = \(1nR + 27tr)L = n(R + r)L. §... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry, Modern - 1898 - 324 pages
...is the radius of a section half-way between the bases of the frustum, Hence S=2TrKxL. That is, thc lateral area of a frustum of a cone of revolution is equal to the circumference of a section half-way between its bases multiplied by its slant height. PROPOSITION VII.... | |
 | James Howard Gore - Geometry - 1898 - 232 pages
...revolution is equal to the circumference of its base multiplied by one-half its slant height. 461. COR. 3. The lateral area of a frustum of a cone of revolution is equal to one-half the sum of the circumferences of its bases multiplied by its slant height. If R and R' denote... | |
 | George Albert Wentworth - Geometry, Solid - 1899 - 248 pages
...L as a limit. § 279 But $' = %(P + j>) XL, always. § 644 .'. S = $(C + c)X L. §284 QBD 728. COR. 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. BOOK VII. SOLID... | |
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The lateral area of a frustum of a cone of revolution is equal to one-half the sum of the circumferences of its bases multiplied by its slant height. Hyp. S is the lateral area, C and C... 
