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.
Prism $\textrm{A}: a=b=c$ and $\textrm{Prism B: e=f}$. Hence reducing the formulas for area and volume to this:
$$V_{1}=h_{1} \times \sqrt{s(s-a)(s-a)(s-a)}$$ $$V_{2}=12\sqrt{3}\times e\times e$$
then:
$$3a=4e$$
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?.
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 $\frac {\sqrt 3}{36}(2e+2f)^2$