Problem of the Week Problem E and Solution Across the Prism

Problem

A trapezoidal prism is a prism in which opposite parallel ends are congruent trapezoids.

For the trapezoidal prism shown, the opposite parallel sides of each trapezoid are $$36$$ cm and $$16$$ cm in length. The non-parallel sides of each trapezoid are $$16$$ cm and $$12$$ cm in length. The prism is $$40$$ cm long. The volume of the prism is $$9984$$ cm$$^3$$.

A body diagonal of a prism is a line connecting two vertices that are not on the same face. Find the length of the body diagonal $$AB$$ (indicated in the diagram), accurate to $$1$$ decimal place.

Solution

Let $$h$$ represent the height of the trapezoid. We will determine the height using two different methods.

Method 1: Finding the Height Using the Given Volume

To find the volume, $$V$$, of a prism, we multiply the area of one of the congruent bases by the perpendicular distance, $$d$$, between the two bases. Since the bases are trapezoids, we can calculate the area of the base using the formula $$A=\frac{h(a+b)}{2}$$, where $$h$$ is the perpendicular distance between the two parallel sides $$a$$ and $$b$$. We know $$V=9984\text{ cm}^3$$, $$d=40\text{ cm}$$, $$a=16\text{ cm}$$, and $$b=36\text{ cm}$$.

We can find $$h$$. \begin{align} V&=\frac{h(a+b)}{2}\times d\\ 9984&=\frac{h(16+36)}{2}\times 40\\ 9984&=26h\times 40\\ 9984&=1040h\\ h&=9.6\text{ cm} \end{align}

Method 2: Finding the Height Without Using the Given Volume

We start by breaking the trapezoids into two right-angled triangles and a rectangle. This can be done by drawing a line from each of the two vertices on the shorter parallel side to meet the longer parallel side at a right angle. The longer parallel side of the trapezoid breaks into pieces with lengths $$a$$ cm, 16 cm, and $$36-16-a=20-a$$ cm. This is shown in the diagram.

Using the Pythagorean Theorem, we can find two different expressions for $$h$$: $$h=\sqrt{16^2-(20-a)^2}$$ and $$h=\sqrt{12^2-a^2}$$.

Since $$h=h$$, $$\sqrt{16^2-(20-a)^2}=\sqrt{12^2-a^2}$$.
Squaring both sides and expanding, we get $$256-400+40a-a^2=144-a^2$$.
Simplifying further, we get $$40a=288$$, and so $$a=7.2$$ cm.

Substituting $$a=7.2$$ into $$h=\sqrt{144-a^2}$$, we find that $$h=\sqrt{144-7.2^2}=9.6$$ cm.

With both methods, we find that $$h=9.6$$ cm. We now want to find the length of $$AB$$. Let $$DE$$ represent the length of the longer parallel side of the back trapezoid, with $$C$$ on $$DE$$ such that $$DC\perp BC$$. It follows that $$BC=h=9.6$$ cm.

From Method $$2$$, we know that $$CE=a=7.2$$ cm. Then $$DC=DE-CE=36-7.2=28.8$$ cm.

The sides of the prism are perpendicular to the ends, so $$AD\perp DC$$ and $$\triangle CDA$$ is a right-angled triangle. Using the Pythagorean Theorem in $$\triangle CDA$$, \begin{align} AC^2&=DC^2+AD^2\\ &=28.8^2+40^2\\ &=2429.44 \end{align}

To find the required length, $$AB$$, we note that $$AC$$ is a line segment drawn across the top of the prism. $$BC$$ is a line segment perpendicular to the top and bottom of the prism. It follows that $$\angle ACB=90^\circ$$ and $$\triangle ACB$$ is a right-angled triangle. We will use the Pythagorean Theorem in $$\triangle ACB$$ to find the length $$AB$$. \begin{align} AB^2&=AC^2+BC^2\\ &=2429.44+9.6^2\\ &=2521.6 \end{align} Since $$AB > 0$$, then we have $$AB = \sqrt{2521.6}\approx 50.2$$ cm.

Therefore, length of the body diagonal, $$AB$$, is approximately $$50.2$$ cm.