Right circular cone solid mensuration help!?

Question

The frustum of a right circular cone has a slant height of 9 ft, and the radii of the bases are 5 ft and 7 ft. Find the lateral area and the total area. What is the altitude of this frustum? Find the altitude of the cone that was removed to leave this frustum. Also, find in two ways the volume of the frustum.

*Thank you very much*

Answer 1

The height of the frustum is h= sqrt(9^2-2^2) =sqrt(77)
The lateral area is
M = pi*sqrt[(R-r)^2+h^2)](R+r) =pi*sqrt(4+77)*14= 126*pi sq feet
The total area is A = pi(126+25+81)= 232*pi sq feet
h total/9 =( h total-sqrt(77))/5
5h =9h -9*sqrt(77)
htotal cone =9/4*sqrt(77)
V= 1/3*pih(R^2+Rr+r^2) =1/3*pi*(81+45+25)*sqrt(77)=
151/3*sqrt(77)cubic feet)
V=1/4 h*pi[(R+r)^2+1/3(R-r)^2] make the substitutions

Answer 2

The volume of spherical sector, either open spherical sector or spherical cone, is equal to one-third of the product of the area of the zone and the radius of the sphere. This is similar to the volume of a cone which is, V(cone) = 1/3 * (circular base area = πR²) * h. In spherical sector, replace A(base) with Area of zone and h with R. [ Note : A zone is that portion of the surface of the sphere, which is included between two parallel planes ] V = (1/3) * (2πRh) * R ==> (2/3) πR² h Volume of required spherical cone = (2/3) π * 9² * (9 - √65) {Since . . . . {H = √(9² - 4²) ==> √65 and h = R - H} => 54 * (9 - √65) * π => 159.0842 cubic cm => ~ 159.08 cm^3 (Ans) ______________________________________... Edit : It is true that, Volume of the solid = solid angle of the sector / solid angle of the whole sphere ) x Volume of the sphere Here solid angle of the sector = Ω = 2π (1 - cos θ) and solid angle of the whole sphere = 4π steradian θ = semi apex angle of the cone. => Vol. = (Ω / 4π ) * (4/3πR³ ) => Vol. = (2π (1 - cos θ) / 4π ) * [(4/3)πR³ ] => Vol. = (2π/3) [1 - (√65)/9] * 9³ => Vol. = 54 * (9 - √65) * π => 159.0842 cubic cm P.S. Here, θ / 360 = V / (4/3)πR³ is invalid.