Question 1043186
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A {{{highlight(cross(frustrum))}}} frustum of pyramid consists of square base of length 10 cm and a top square of 7 cm. 
The heights of the {{{highlight(cross(frustrum))}}} frustum is 6 cm calculate
(A) the surface area
(B) volume
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Volume


<pre>
See <A HREF=https://en.wikipedia.org/wiki/Frustum>this</A> Wikipedia article or <A HREF=>this</A> WEB-page.

The volume formula of a frustum of a square pyramid was introduced by the ancient Egyptian mathematics 
in what is called the Moscow Mathematical Papyrus, written ca. 1850 BC.:

V = {{{(1/3)*h*(a^2+ab+b^2)}}} 

where a and b are the base and top side lengths of the truncated pyramid, and h is the height. 
The Egyptians knew the correct formula for obtaining the volume of a truncated square pyramid, 
but no proof of this equation is given in the Moscow papyrus.
</pre>

By applying the formula, &nbsp;we get &nbsp;&nbsp;V = {{{(1/3)*6*(10^2 + 10*7 + 7^2)}}} = 438 {{{cm^3}}}.


<pre>
By the way, the proof is easy.
As you know, the volume of the larger pyramid is  {{{V[1]}}} = {{{(1/3)*a^2*H[1]}}}, where {{{H[1]}}} is its height.
The volume of the smaller pyramid is  {{{V[2]}}} = {{{(1/3)*b^2*H[2]}}}, where {{{H[2]}}} is its height.

Obviously, from similarity {{{H[1]}}} = {{{const*a}}}  and  {{{H[2]}}} = {{{const*b}}}, where const is a constant value independent of "a" and "b".

Then  {{{V[1]}}} = {{{(1/3)*const*a^3}}},  {{{V[2]}}} = {{{(1/3)*const*b^3}}}.

The volume of the frustum is the difference 

V = {{{V[1] - V[2]}}} = {{{(1/3)*const*a^3 - (1/3)*const*b^3)}}} = {{{(1/3)*const*(a^3 - b^3)}}} = {{{(1/3)*const*(a-b)*(a^2 + ab + b^2)}}}.

But const*(a-b) = {{{H[1]-H[2]}}} = h, the height of the frustum.

Therefore,  the volume of the frustum is  V = {{{(1/3)*h*(a^2+ab+b^2)}}}.   QED.
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