Frustum of a Cone

If a cone is cut by a plane parallel to its base, the portion of a solid between this plane and the base is known as frustum of a cone.

The volume denoted by ABCD in figure is a frustum of the cone ABE.


frustum-cone-01

Volume of Frustum of a Cone:

Since, we know that cone is a limit of a pyramid therefore; frustum of a cone will be the limit of frustum of a pyramid. But volume of a pyramid is
                       

V = \frac{1}{3}h\left[ {{A_1} + {A_2} + \sqrt  {{A_1}{A_2}} } \right]


Where              {A_1} = \pi {R^2}
                        {A_2} = \pi {r^2}
\therefore                         = \frac{h}{3}\left( {\pi {R^2} + \pi {r^2} + \pi  \sqrt {{R^2}{r^2}} } \right)
                         = \frac{{\pi h}}{3}\left( {{R^2} + {r^2} + Rr}  \right)


frustum-cone-02

Example:

A cone 12cm high is cut 8cm from the vertex to form a frustum with a volume of 190cu.cm. Find the radius of the cone.

Solution:

Given that:
            Height of cone             = 12{\text{cm}}
            Height of frustum       h  = 12 - 8 = 4{\text{cm}}
            Volume of frustum      =  190{\text{cu}}.{\text{cm}}

            Now volume of frustum cone
                         = \frac{{\pi h}}{3}\left( {{R^2} + {r^2} + Rr}  \right)
            or         190 = \frac{{3.1415 \times 4}}{3}\left[ {\left(  {\frac{2}{3}{r^2}} \right) + {r^2} + \frac{2}{3}{r^2}} \right] = 4.19 \times  \frac{{19}}{9}{r^2}
            or         {r^2} = \frac{{190 \times 9}}{{19 \times 4.19}} =  \frac{{1710}}{{79.61}} = 21.48
                        r = \sqrt {21.40}  = 4.63

            Hence required radius of cone  = 4.63

Curved Surface Area of a Frustum of a Cone:

Since, a cone is the limiting case of a pyramid, therefore the lateral surface of frustum of a cone can be deduced from the slant surface of frustum of a pyramid, i.e., curved (lateral) surface of frustum of cone.


frustum-cone-03

             = \frac{1}{2}\left( {{\text{sum of circumferences  of bases}}} \right) \times \left( {{\text{slant height}}} \right)
             = \frac{1}{2}\left( {2\pi R + 2\pi r}  \right)l\,\,\, = \pi \left( {R + r} \right)l
l, being the slant height of frustum, R and r being two radius of bases.

Note:
            (1)        Total surface area of frustum of a cone
                                   

 = \pi {R^2} + \pi {r^2} + \pi \left( {R + r}  \right)l


            (2)        To find the slant height of the cone, use Pythagorean theorem.

Example:

A material handling bucket is in the shape of the frustum of a right circular cone as shown in figure. Find the volume and the total surface area of the bucket.

Solution:


frustum-cone-04

            Slant height                 l = \sqrt {{h^2}  + {r^2}} \,\,\, = \sqrt {{{\left( {17} \right)}^2} + {{\left( {2.5}  \right)}^2}} \,\,\, = 17{\text{cm}}
            Lateral surface area      =  \pi \left( {R + r} \right)l
                                                 = \pi \left( {7.5 + 5} \right)\left( {17}  \right)\,\,\,\, = 670\,{\text{sq}}{\text{.cm}}
\therefore            Base areas                    = \pi  {R^2} + \pi {r^2}\,\,\,\, = \pi \left( {{R^2} + {r^2}} \right)
                                                 = \pi \left( {56.25 + 25} \right)\,\,\,\, = 177 +  79
\therefore            Total surface area          = \pi {R^2} + \pi {r^2} + \pi \left( {R + r} \right)l
                                                 = 177 + 79 + 670
                                                 = 926\,{\text{sq}}{\text{.cm}}

            Volume            = \frac{{\pi  h}}{3}\left( {{R^2} + {r^2} + Rr} \right)
                                     = \frac{{17h}}{3}\left( {{{\left( {7.5}  \right)}^2} + {{\left( 5 \right)}^2} + \left( {7.5} \right)\left( 5 \right)}  \right)
                                     = 2114\,{\text{cu}}{\text{.cm}}

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