Question 1165476
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The volume of a truncated prism with an equilateral triangle as its horizontal base is equal to 3600 cm3. 
The vertical edges at each corner are 4, 6, and 8 cm. respectively. Find one side of the base.
a. 22.37 b. 25.43 c. 37.22 d. 17.89
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<pre>
First, let's look at the position of the point at the sloped section, 
which is exactly over the center of the regular triangle at the base.


Vertical coordinate of this point is the arithmetic mean of the numbers 4, 6 and 8

    z = {{{(4+6+8)/3}}} = 6 cm  over the base plane.


Now, let's accept without a proof, that the volume of the truncated prism is the same 
as the volume of the regular prism with the same base and the height of 6 sm.


If the side of the base is 'a', then the area of the base is  {{{a^2*(sqrt(3)/4)}}} cm^2,

and the volume of the regular prism is  {{{a^2*6*(sqrt(3)/4)}}} cm^3 = {{{a^2*(3*sqrt(3)/2)}}} cm^3.


It gives us an equation for 'a'

    {{{a^2*(3*sqrt(3)/2)}}} = 3600.


From this equation

    {{{a^2}}} = {{{(3600*2)/(3*sqrt(3))}}} = {{{2400/sqrt(3)}}} = {{{(2400*sqrt(3))/3}}} = {{{800*sqrt(3)}}},

    a = {{{sqrt(800*sqrt(3))}}} = 37.22 cm  (rounded).


<U>ANSWER</U>.  The size of the base is about 37.22 cm ,  which is option (c)  in the answer list.
</pre>

Solved.