Frustum of a Pyramid

If a pyramid is cut through by a plane parallel to its base portion of the pyramid between that plane and the base is called frustum of the pyramid.

Volume of Frustum of a Pyramid:

A general formula for the volume of any pyramid can be derived in terms of the areas of the two bases and the height of the frustum.

Consider any frustum of a pyramid KE in figure with the lower base {A_1}, upper base {A_2} and the altitude h. Complete the pyramid O - HD of which the frustum KE is a part.

Denote by p, the volume of the small pyramid O - QM, whose altitude is a. Then the altitude of O - HD is a  + h.

Let V and P respectively, represents the volume of the frustum KE and the pyramid O - HD.


frustum-pyramid

\therefore From the figure it is easily seen that V = P - p. Expressing this equality in terms of the dimensions, we may write.

\begin{gathered} V = \frac{1}{3}{A_1}\left( {h +  a} \right) - \frac{1}{3}{A_2}a = \frac{1}{3}\left[ {{A_1}h + {A_1}a - {A_2}a}  \right] \\ \Rightarrow V =  \frac{1}{3}\left[ {{A_1}h + a\left( {{A_1} - {A_2}} \right)} \right]\, -  -  -  \left( i \right) \\ \end{gathered}

The pyramid O - HD may be considered as cut by the two parallel planes HD and QM. Hence

\frac{{{a^2}}}{{{{\left( {h + a} \right)}^2}}} =  \frac{{{A_2}}}{{{A_1}}}

(as if a pyramid is cut by two parallel planes, the areas of the sections are proportional to the squares of their distances from the vertex).

Taking the square root of both sides, we have

\begin{gathered} \frac{a}{{h + a}} =  \frac{{\sqrt {{A_2}} }}{{\sqrt {{A_1}} }} \\ \Rightarrow a\sqrt {{A_1}}  = h\sqrt {{A_2}}  + a\sqrt {{A_2}} \\ \end{gathered}

Transposing a\sqrt {{A_2}} to the L.H.S. of this equation and factorizing,

\begin{gathered} a\left( {\sqrt {{A_1}}  - \sqrt {{A_2}} } \right) = h\sqrt  {{A_2}} \\ \Rightarrow a = \frac{{h\sqrt  {{A_2}} }}{{\sqrt {{A_1}}  - \sqrt  {{A_2}} }} \\ \end{gathered}

Substituting the value of a in (1), we have

\begin{gathered} V = \frac{1}{3}\left[ {{A_1}h +  \frac{{h\sqrt {{A_2}} }}{{\sqrt {{A_1}}   - \sqrt {{A_2}} }}\left( {{A_1} - {A_2}} \right)} \right] \\ \Rightarrow V =  \frac{1}{3}\left[ {{A_1}h + \frac{{h\sqrt {{A_2}} }}{{\sqrt {{A_1}}  - \sqrt {{A_2}} }}\left( {{A_1} + {A_2}}  \right)\left( {{A_1} - {A_2}} \right)} \right] \\ \Rightarrow V =  \frac{1}{3}\left[ {{A_1}h + h\sqrt {{A_2}} \left( {{A_1} + {A_2}} \right)}  \right] \\ \Rightarrow V =  \frac{1}{3}\left[ {{A_1}h + h\sqrt {{A_1}{A_2}}   + {A_2}h} \right] = \frac{1}{3}h\left[ {{A_1} + {A_2} + \sqrt  {{A_1}{A_2}} } \right] \\ \end{gathered}

i.e. The volume of a frustum of a pyramid is equal to the one-third the product of the altitude and the sum of the upper base, the lower base and the square root of the product of two bases.

Lateral Surface Area of Frustum of a Pyramid:

Each of the faces such as CDML, of the frustum of a pyramid is a trapezium and the area of each trapezium will be half the sum of the parallel sides, CD and ML, multiplied by the slant distance between them.

In the frustum of pyramid on a square base, let a denote the length of each side of the base, b the length of each side of the other end, l the height of the frustum.

Each face CDML is a trapezium, the lengths of the parallel sides a andb.

\begin{gathered} {\text{Area}}\,{\text{CDML}} =  \frac{1}{2}\left( {a + b} \right)l \\ Lateral\,area =  \frac{1}{2}\,\left( {sum\,of\,perimeters\,of\,bases \times slant\,height}  \right) \\ \end{gathered}

Example:

A frustum of a pyramid has rectangular ends, the sides of the base being 25dm and 36dm. If the area of the top face is 784sq.dm and the height of the frustum is 60dm, find its volume.

Solution:

Here

{A_1} = 25 \times 36 = 900\,{\text{sq}}{\text{.dm}}, {A_2} = 784\,{\text{sq}}{\text{.dm}}

\begin{gathered} V = \frac{1}{3}h\left[ {{A_1} +  {A_2} + \sqrt {{A_1}{A_2}} } \right] = \frac{1}{3} \times 60\left[ {900 + 784 +  \sqrt {900 \times 784} } \right] \\ V = 20 \times \left( {1684 +  840} \right) = 20 \times 2524 = 50500\,{\text{sq}}{\text{.}}\,{\text{dm}} \\ \end{gathered}

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