Question

a pyramid whose altitude is 5ft weighs 800lbs. at what distance from its vertex must it be cut by a plane parallel to its base so that the two solids of equal weight will be formed?

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

A structure which has a square base and four triangular sides meeting at a point is called pyramid.

At distance of 3.97 feet from its vertex , pyramid is cut by plane  so that the two solids of equal weight will be formed.

It is assumed that weight of pyramid is proportional to its volume.

So,   $w = k V$ , where V is volume of original pyramid and v is volume of small pyramid and k is constant.

Let us consider that at h  distance,  pyramid is cut from its vertex. So a small pyramid is also formed.

Assume that base area of original pyramid is A and base of small pyramid is a.

Volume of original pyramid is,   $V= \frac{1}{3} *A* 5$

So, weight of original pyramid,  $W = k *(\frac{1}{3} *A* 5) = 800$

Volume of small pyramid is,  $v = \frac{1}{3}* a* h$

So, weight of small pyramid, $w = k*(\frac{1}{3}* a* h)=400$

Since, base and height of small pyramid and original pyramid are in proportion.

So,  $\frac{a}{A} = (\frac{h}{5}) ^{2}$

       $a = (\frac{h}{5} )^{2}A$

Substituting value of a in  equation $k*(\frac{1}{3}* a* h)=400$

So, $k*(\frac{1}{3}* \frac{h^{2} }{25}A * h)=400$

       $(k*\frac{1}{3}* A*5)*\frac{1}{5} *\frac{h^{2} }{25} * h)=400$

Since, $(k *\frac{1}{3} *A* 5) = 800$, substitute in above equation.

    So, $800*\frac{h^{3} }{125}=400\\\\h^{3}=\frac{125}{2}\\\\h=\sqrt[\frac{1}{3} ]{62.5}\\\\h=3.97 feet$

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