The problem is as follows:

For the architecture class Joan and Peter must each make a right regular prism of equal volume. Joan built a triangular prism and Peter constructed a quadrangular prism whose height is $12\sqrt{3}$ inches. If their bases from both prisms have the same perimeter. How tall is the prism Joan built?

Apparently the problem lies in the interpretation of the word "regular prism" by looking on this source, a regular prism is a prism with bases that are regular polygons with the latter meaning from this other source being a polygon for which all sides are congruent and all angles are congruent. In other words, for this problem would meant that the base of $\textrm{Prism A}$ is a equilateral triangle and $\textrm{Prism B}$ is a square.

From the previous information I made a sketch of how I thought to solve the problem.

Sketch of the problem

Prism $\textrm{A}: a=b=c$ and $\textrm{Prism B: e=f}$. Hence reducing the formulas for area and volume to this:

$$V_{1}=h_{1} \times \sqrt{s(s-a)(s-a)(s-a)}$$ $$V_{2}=12\sqrt{3}\times e\times e$$



Therefore both volumes are the same,

$$h_{1} \times \sqrt{s(s-a)(s-a)(s-a)}=12\sqrt{3}\times e^{2}$$

$$h_{1} \times \sqrt{\frac{3a}{2}(\frac{3a}{2}-a)^{3}}=12\sqrt{3}\times e^{2}$$

$$h_{1} \times \sqrt{\frac{3a}{2}(\frac{a}{2})^{3}}=12\sqrt{3}\times e^{2}$$

$$h_{1} \times \sqrt{\frac{3a^{4}}{2^{4}}}=12\sqrt{3}\times e^{2}$$

$$h_{1} \times \frac{a^{2}}{4}\sqrt{3}=12\sqrt{3}\times e^{2}$$

Replacing the $\textrm{e}$ by $e=\frac{3a}{4}$

$$h_{1} \times \frac{a^{2}}{4}\sqrt{3}=12\sqrt{3}\times \frac{3^{2}a^{2}}{4^{2}}$$

$$h_{1} = \frac{12\times 3^{2}}{4}$$

$$h_{1} = 27$$

Therefore the result would become into $\textrm{27 inches}$. By following this way of writing Heron's formula:

$$A=\frac{1}{4}\times \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$

I got the same result.

Overall, does the method I used is right?.

  • $\begingroup$ It seem that we need some other given, for the triangle indeed for a same perimeter we can have different area for the base (see Eron formula) and dofferent hieghts, or am I wrong? $\endgroup$ – user Mar 4 '18 at 23:05
  • $\begingroup$ @gimusi There will be different areas and heights. But I could be naive and just think that $12\sqrt{3}$ would be the height. But we cannot prove that. $\endgroup$ – Chris Steinbeck Bell Mar 4 '18 at 23:09

You don't have enough information to answer the question. The heights have to be in inverse ratio to the area of the bases. The area of Prism B has area $ef$ and perimeter $2e+2f$. You are given that $a+b+c=2e+2f$ to make the perimeters equal, but the area of the triangular base can be anything from $0$ up to $\frac {\sqrt 3}{36}(2e+2f)^2$

  • $\begingroup$ As you mentioned i'm adding the missing information. I also noticed it. However can you tell me how did you got to $\frac{\sqrt{3}(2e+2f)^{2}}{36}$? $\endgroup$ – Chris Steinbeck Bell Mar 4 '18 at 23:41
  • $\begingroup$ I got it by deriving the area of an equilateral triangle given its perimeter. If the perimeter is $p$, the side is $\frac p3$, the altitude is $\frac {\sqrt 3}6p$ $\endgroup$ – Ross Millikan Mar 4 '18 at 23:54

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