# Problem of the Week Problem E and Solution I Want More Volume

## Problem

A rectangular prism is placed on a table. Points $$P$$, $$Q$$, and $$R$$ lie on three different faces of the prism with $$P$$ on the top face and $$Q$$ and $$R$$ on two adjacent side faces. Each point is located where the diagonals of the particular face intersect. Connecting these three points gives us $$\triangle PQR$$.

If $$PQ=4$$ cm, $$QR=5$$ cm, and $$RP=6$$ cm, determine the volume of the rectangular prism.

## Solution

First, we label the top edge of the face containing point $$Q$$ as $$AB$$ and its midpoint as $$M$$. Let $$AM = a$$, $$PM=b$$, and $$MQ = c$$. Since the centre of a rectangle is where its diagonals intersect, $$P$$ and $$Q$$ are at the centres of their respective faces. Further, since the centre of a rectangle is also where the perpendicular bisectors of the sides of the rectangle meet, the rectangular prism has dimensions $$2a$$, $$2b$$, and $$2c$$, as shown in the following diagram.

Since $$\angle{PMQ}=90^\circ$$, it follows that $b^2+c^2=16 \tag{1}$ Similarly, we can conclude the following. $a^2+c^2=36\tag{2}$ $a^2+b^2=25\tag{3}$ Adding equations $$(1)$$, $$(2)$$, and $$(3)$$ gives us the following. \begin{align*} 2a^2+2b^2+2c^2&=77\\ a^2+b^2+c^2&=\frac{77}{2} \tag{4}\end{align*}

Now, we subtract each of equations $$(1)$$, $$(2)$$, and $$(3)$$ from equation $$(4)$$ to obtain $a^2=\frac{45}{2},~b^2=\frac{5}{2},~\text{and}~c^2=\frac{27}{2}$ Multiplying $$a^2,~b^2,$$ and $$c^2$$ together gives the product $a^2b^2c^2=\frac{45}{2}\times\frac{5}{2}\times\frac{27}{2}=\frac{6075}{8}$

Then, taking the positive square root, $abc=\sqrt{\frac{6075}{8}}=\frac{45\sqrt{3}}{2\sqrt{2}}=\frac{45\sqrt{3}}{2\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{45\sqrt{6}}{4}$

To determine the volume of the rectangular prism, we multiply the side lengths together. \begin{aligned} V&=(2a)(2b)(2c)\\ &=8abc\\ &=8\left(\frac{45\sqrt{6}}{4}\right)\\ &=90\sqrt{6}~\text{cm}^3\end{aligned}

Therefore, the volume of the rectangular prism is $$90\sqrt{6}~\text{cm}^3$$.