Question 796887
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Drawing a little picture, you can see that, on the bottom block, you can see 4 complete faces of the cube presuming you would actually be able to walk around it.  If we let *[tex \Large e_1] represent the measure of the edge of the larger cube, then the surface area of each of the faces that can be seen in their entirety is *[tex \Large e_1^{\,2} ], so the total of those areas is *[tex \Large 4e_1^{\,2}].  The top block, which edge we will refer to as *[tex \Large e_2], has the four sides and the top visible, so that part is represented by *[tex \Large 5e_2^{\,2}].  That leaves the partially exposed top face of the bottom block to account for.  If the top cube wasn't there, the exposed area of the top face of the bottom block would be simply *[tex \Large e_1^{\,2} ], but since it is partly covered by an area that amounts to *[tex \Large e_2^{\,2} ], the actual visible area is *[tex \Large e_1^{\,2}\ -\ e_2^{\,2} ].  Putting the whole thing together:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(e_1^{\,2}\ -\ e_2^{\,2}\right)\ +\  4e_1^{\,2}\ +\ 5e_2^{\,2}\ =\ \ 61].


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5e_1^{\,2}\ +\ 4e_2^{\,2}\ =\ \ 61].


Finally, since the total height of the stack is 5 feet:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e_1\ +\ e_2\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e_2\ =\ 5\ -\ e_1 ]



Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5e_1^{\,2}\ +\ 4\left(5\ -\ e_1\right)^{\,2}\ =\ \ 61].


The only thing left for you to do is to expand the binomial, collect like terms, put the quadratic in *[tex \Large e_1] into standard form, and solve.  Since *[tex \Large e_1] represents the edge of the larger cube and the height is 5, the edge of the larger cube has to be larger than 2.5 feet.  Hence, if you get a root that is smaller than 2.5 feet (and you will get one given correct arithmetic), discard it.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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