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 = {}h (c, + c2) = πh (r, + r2). F = 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, 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 xa, and by b is xb. 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. 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 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 But if is the mid-point of PR, then PM + RN = 2Q0, :. F = 2′′ × PR × QO. But A PRL ~A QCO, since the three sides of one are drawn perpendicular to the sides of the other; :. PR × QO = PL × QC, :. F = 2′′(LP × QC) = 2π(MN × CQ). EXERCISES. 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 x CQ); and the sum of all the frustums, that is, the surface of the solid generated by the revolving semi-polygon, equals 2′′CQ into the sum of the projections, BM, MN, NC, etc., which sum is BD. Sum of surfaces of frustums = 2π CQ x BD. As we continually double the number of sides of the inscribed polygon, its semi-perimeter approaches the semicircle as limit, and its surface of revolution approaches the sphere as limit, while CQ, its apothem, approaches r, the radius of the sphere, as limit. Representing the sum of the surfaces of the frustums by ΣF, and BD by 2r, we have That is, the variable sum is to the variable CQ in the constant ratio 4; therefore, by 798, Principle of Limits, their limits have the same ratio, 835. The last proof gives also the following rule: To find the area of a zone. RULE. Multiply the altitude of the segment by twice « times the radius of the sphere. FORMULA. Z = 2arπ. CHAPTER IV. SPACE ANGLES. 836. A Plane Angle is the divergence of two straight lines which meet in a point. 837. A Space Angle is the spread of two or more planes which meet in a point. V 838. Symmetrical Space Angles are those which cut out symmetrical spherical polygons on a sphere, when their vertices are placed at its center. 839. A Steregon is the whole amount of space angle round about a point in space. |