Chapter 8.4, Problem 39WE

To determine

To find: the value of $AB$ .

Answer to Problem 39WE

The required length of $AB$ is $4\sqrt{2}$ .

Explanation of Solution

Given Information:

The six edges of the solid shown are 8 units long. $A$ and $B$ are mid points of two edges as shown.

Given figure,

  

Calculation:

Consider the following figure,

  

The law of cosines can be stated as follows,

  $a^2=b^2+c^2-2bc\cos \left(A\right)$

Suppose that $\theta$ is the angle the edge containing $B$ makes with the bottom face.

Let $a=8$ be the length of each face.

Then, it implies that,

  $\begin{array}{c}\cos \theta =\frac{{\left(\frac{\sqrt{3}}{2}a\right)}^2+a^2-{\left(\frac{\sqrt{3}}{2}a\right)}^2}{2\left(\frac{\sqrt{3}}{2}a\right)a}\\ =\frac{1}{\sqrt{3}}\end{array}$

So the length of $AB$ is given by,

  $\begin{array}{c}A{B}^2={\left(\frac{\sqrt{3}}{2}a\right)}^2+{\left(\frac{a}{2}\right)}^2-2\left(\frac{\sqrt{3}}{2}a\right)\left(\frac{a}{2}\right)\cos \theta \\ ={\left(\frac{\sqrt{3}}{2}a\right)}^2+{\left(\frac{a}{2}\right)}^2-2\left(\frac{\sqrt{3}}{2}a\right)\left(\frac{a}{2}\right)\frac{1}{\sqrt{3}}\\ =\frac{4}{4}{a}^2-\frac{1}{2}{a}^2\\ =\frac{1}{2}{a}^2\end{array}$

Simplify further to get,

  $\begin{array}{c}AB=\frac{1}{\sqrt{2}}a\\ =\frac{1}{\sqrt{2}}8\\ =4\sqrt{2}\end{array}$

Therefore, the required length of $AB$ is $4\sqrt{2}$ .