The front of a rectangular prism has an area of $12 \mathrm{~cm}^{2}$, the side has an area of $6 \mathrm{~cm}^{2}$, and the top has area $8 \mathrm{~cm}^{2}$. The volume of the prism in $\mathrm{cm}^{3}$, is
(A) 24
(B) 26
(C) 48
(D) 72
(E) 52

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-087.jpg?height=241&width=336&top_left_y=259&top_left_x=1296)

## Part B: Each correct answer is worth 6.
Solution 1

Suppose that the height is $h \mathrm{~cm}$, the width is $w \mathrm{~cm}$, and the depth is $d \mathrm{~cm}$.

Using the given areas of the faces, $w h=12, d w=8$ and $d h=6$.

Multiplying these together, we get $w h d w d h=12(8)(6)$ or $d^{2} h^{2} w^{2}=576$.

Thus, $(d h w)^{2}=576$, so $d h w=\sqrt{576}=24$, since $d h w>0$.

The volume is $d h w \mathrm{~cm}^{3}$, so the volume is $24 \mathrm{~cm}^{3}$.

## Solution 2

We try to find dimensions that work to give the three face areas that we know.

After some trial and error, we see that if the width is $4 \mathrm{~cm}$, the height is $3 \mathrm{~cm}$, and the depth is $2 \mathrm{~cm}$, then the given face areas are correct.

Thus, the volume is $4 \mathrm{~cm} \times 3 \mathrm{~cm} \times 2 \mathrm{~cm}=24 \mathrm{~cm}^{3}$.

## Solution 3

Suppose that the height is $h \mathrm{~cm}$, the width is $w \mathrm{~cm}$, and the depth is $d \mathrm{~cm}$.

Using the given areas of the faces, $w h=12, d w=8$ and $d h=6$.

From the first equation, $w=\frac{12}{h}$.

Substituting into the second equation, we obtain $\frac{12 d}{h}=8$.

Multiplying this equation by the equation $d h=6$, we obtain $12 d^{2}=48$ which gives $d^{2}=4$ and so $d=2$, since $d>0$.

Since $d h=6$, then $2 h=6$ so $h=3$.

Since $d w=8$, then $2 w=8$ so $w=4$.

Thus, the volume is $4 \mathrm{~cm} \times 3 \mathrm{~cm} \times 2 \mathrm{~cm}=24 \mathrm{~cm}^{3}$.

ANSWER: (A)