824. A Conical Surface is generated by a straight line moving so as always to pass through a fixed point called the apex. X 825. A Cone is a solid bounded by a conical surface and a plane. A 826. A right circular cone may be generated by revolving a right triangle about one pependicular. All the elements are equal, and each is called the Slant Height of the cone. 827. The Frustum of a pyramid or cone is the portion included between the base and a parallel section. A 828. To find the lateral area of a right circular cone. A B│C B RULE. Multiply the circle of its base by half the slant height. FORMULA. K K = Ich = arh. PROOF. If the curved surface be slit along a slant height, and spread out flat, it becomes a sector of a circle, with slant height as radius, and the circle of cone's base as arc; therefore, by 811, its area is ch. 829. COROLLARY. Since, by 810, the base = cr, therefore the curved surface is to the base as the slant height is to the radius of base, as the length of circle with radius is to circle with radius r, as the perigon is to the sector angle. 830. To find the lateral area of the frustum of a right circular RULE. Multiply the slant height of the frustum by half the sum of the circles of its bases. FORMULA. F = 1⁄2h (c, + c2) = h (r1 + r2). PROOF. Completing the cone, and slitting it along a slant height, the curved surface of the frustum develops into the difference of two similar sectors having a common angle, the arcs of the sectors being the circles of the bases of the frustum. By 813, the area of this annular sector, F = 1⁄2h(c, + C2). 831. The axis of a right circular cone, or cone of revolution, is the line through its vertex and the center of its base. EXERCISES. 116. Given the two sides of a right-angled triangle. Find the area of the surface described when the triangle revolves about its hypothenuse. HINT. Calling a and b the given altitude and base, and x the perpendicular from the right angle to the hypothenuse, by 828, the area of the surface described by a is πxa, and by b is 7xb. Also a : x :: √a2 + b2 : b. 117. In the frustum of a right circular cone, on each base stands a cone with its apex in the center of the other base; from the basal radii r, and r2 find the radius of the circle in which the two cones cut. 2 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 π times a perpendicular erected at the mid-point of this slant height, and terminated by the axis. PROOF. By 830, the lateral area of the frustum whose slant height is PR and axis MN is F= (PM+ RN)PR. But if is the mid-point of PR, then PM + RN = 2QO, .. F = 2′′ × PR × QO. But ▲ PRL ~ ▲ QCO, since the three sides of one are drawn perpendicular to the sides of the other; . F EXERCISES. PR × QO = PL × QC, = 2π(LP × QC) = 2′′(MN × CQ). 118. Reckon the mantel from the two radii when the inclination of a slant height to one base is half a right angle. 119. If in the frustum of a right cone the diameter of the upper base equals the slant height, reckon the mantel from the altitude a and perimeter p of an axial section. RULE. Multiply four times its squared radius by PROOF. In a circle inscribe a regular polygon of an even number of sides. Then a diameter through one vertex passes through the opposite vertex, halving the polygon symmetrically. Let PR be one of its sides; draw PM, RN, perpendicular to the diameter BD. From the center C the perpendicular CQ bisects PR. Drop the perpendiculars PL, QO. Now, if the whole figure revolve about BD as axis, the semicircle will generate a sphere, while each side of the inscribed polygon, as PR, will generate the curved surface of the frustum of a cone. By 832, this F = 2π(MN × CQ); and the sum of all the frustums, that is, the surface of the solid generated by the revolving semi-polygon, equals 27CQ into the sum of the projections, BM, MN, NC, etc., which sum is BD. .. Sum of surfaces of frustums 2π CQ X BD. |
