...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 off...

...the number of whose faces is indefinitely increased. PROPOSITION V.— THEOREM. 23. The lateral area of a cone of revolution is equal to the product of the circumference of its base by half its slant height. Let S-MNPQ be a cone generated by the revolution...

...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 off...

...the number of whose faces is indefinitely increased. PROPOSITION V.— THEOREM. 23. The lateral area of a cone of revolution is equal to the product of the circumference of its base by half its slant height. Let S-MNPQ be a cone generated by the revolution...

...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,...

...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 and...

...revolution. Draw Cff±, and AD II, to Y Y'. Then area AB = AB X 2 тг С Н, §662 (the lateral arca 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 and...

...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. Candidates...

...basal radii rl and r2 find the radius of the circle in which the two cones cut. 329 THEOREM III. 832. The lateral area of a frustum of a cone of revolution is the product of the projection of the frustum s slant height on the axis by twice TT times a perpendicular...

...basal radii r, and r, find the radius of the circle in which the two cones cut. 329 THEOREM III. 832. The lateral area of a frustum of a cone of re-volution is the product of the projection of the frustums slant height on the axis by twice ?r times a perpendicular...