Question 1194401
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Find the total area of the frustum of a regular square pyramid which is inscribed
in the frustum of a cone whose upper and lower base diameters are 11 ft and 15 ft
respectively, and whose altitude is 19 ft.
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<pre>
Use the formula for the volume of a regular pyramidal frustum

    V = {{{(1/3)*h*(A[1]+A[2]+sqrt(A[1]*A[2]))}}},


where h is the height,  {{{A[1]}}}  and  {{{A[2]}}}  are the base areas

(see this source https://mathworld.wolfram.com/PyramidalFrustum.html) .


In your case, the bases are the squares with their diagonals 11 ft and 15 ft.


Their areas are  {{{11^2/2}}} square feet  and  {{{15^2/2}}} square feet, respectively;
h = 19 ft.  Therefore, the volume of the frustum is


    V = {{{(1/3)*19*(121/2 + 225/2 + sqrt(((11^2)/2)*((15^2)/2)))}}} = {{{(1/3)*19*(121/2 + 225/2 + (11*15)/2)}}} = 

      = {{{(1/3)*19*((121+225+165)/2)}}} = {{{(1/3)*19*(511/2)}}} = {{{9709/6}}} = 1618.167  cubic feet  (rounded).


<U>ANSWER</U>.  The volume is  {{{9709/6}}} = 1618.167  cubic feet  (rounded).
</pre>

Solved.