 | Education - 1902 - 880 pages
...parallelepiped, b) a pyramid. Derive one of these formulas. 5 Complete and demonstrate the following: the lateral area of a frustum of a cone of revolution is equal to ... 6 Prove that the sum of the angles of a spheric triangle is greater than 180° and less than 540°.... | |
 | George Albert Wentworth - Geometry, Solid - 1902 - 248 pages
...Tis ' - ~° _ ^trKH _M,Ji,_tf_ll ,L 1 -J 7T-£i -Tz .41 -TZ Jt JI Lt PROPOSITION XLII. THEOREM. 727. The lateral area of a frustum of a cone of revolution is equal to half .the sum of the circumferences of its bases multiplied by the slant height. Let S denote the lateral... | |
 | Education - 1902 - 780 pages
...parallelepiped, b) a pyramid. Derive one of these formulas. 5 Complete and demonstrate the following: the lateral area of a frustum of a cone of revolution is equal to ... 6 Prove that the sum of the angles of a spheric triangle is greater than 180° and less than 540°.... | |
 | Fletcher Durell - Geometry, Solid - 1904 - 232 pages
...R'2 H' R'3 H'3 L'3' 'H'2' (Why?) (Why?) QED BOOK Vm. SOLID GEOMETRY y PROPOSITION XI. THEOREM 726. 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. Given a frustum... | |
 | Fletcher Durell - Geometry - 1911 - 553 pages
...7rirF/- (Why?) ^ __ = ^ (Why?) Vf %nR'*H' R'^'H' Rf* R* L"' (Why?) QEB PROPOSITION XL THEOREM 726. 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. Given a frustum... | |
 | George Albert Wentworth - Geometry - 1904 - 496 pages
...— Ra ~ L'* — H'* V lirR*H _ R* H _ R° _ 1Is _ L' ' a I '3 '2 K PROPOSITION XLII. THEOREM. / 727. The lateral area of a frustum of a cone of revolution is equal to half the sum of the circumferences of its bases multiplied by the slant height. Let S denote the lateral... | |
 | Education - 1912 - 942 pages
...regular pyramid equals one half the product of its slant height and the perimeter of its base. 748 Definition. Conical surface. Directrix. Generatrix....Corollary 3. The lateral area of a frustum of a cone equals the product of the height and the circumference of a circle whose radius equals the sect of... | |
 | George Clinton Shutts - Geometry - 1905 - 412 pages
...their altitudes or the squares of the radii of their bases. See method of § 581. 609. COROLLARY III. The lateral area of a frustum of a cone of revolution is equal to one-half the product of its slant height by the sum of the perimeters of its bases. Suggestion. Circumscribe... | |
 | Daniel Alexander Murray - 1906 - 466 pages
...frustum, and MN is its slant height, then lateral area of frustum = .J. MN (pi + p2) = MN • P. b. The lateral area of a frustum of a cone of revolution is equal to the product of the slant height of the frustum and half the sum of the circumferences of its bases. [Suggestion for proof:... | |
 | Isaac Newton Failor - Geometry - 1906 - 440 pages
...£ irR2H = R2 x H _ R3 = H8 _L! V'~ATrR'2H'~R'2 H'~R'3~H'3~L13' § 739 PROPOSITION XL. THEOREM 741 The lateral area of a frustum of a cone of revolution is equal to half the sum of the circumferences of its bases multiplied by its slant height. HYPOTHESIS. S is the... | |
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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... 
