Chapter 8.4, Problem 16WE

To determine

To complete: the given table.

Answer to Problem 16WE

d	e	f
$3\sqrt{3}$	$9$	$6\sqrt{3}$

Explanation of Solution

Given:

The figure:

And the table:

d	e	f
?	?	$6\sqrt{3}$

Concept Used:

Theorem 3-11: The addition of measures of all angles of a triangle is 180.

Theorem 8-7: In a $30°-60°-90°$ triangle, the longer Side is $\sqrt{3}$ times as long as the shorter Side and the hypotenuse of right triangle are two times as long as the shorter side of triangle.

Label the given figure as shown below:

Apply theorem 3-11 on the triangle $△ABC$ ,

  $\begin{array}{c}m\angle A+m\angle B+m\angle C=180\\ m\angle A+90+60=180\\ m\angle A+150=180\end{array}$

  $\begin{array}{c}m\angle A=180-150\\ m\angle A=30\end{array}$

Thus, the given triangle is a $30°-60°-90°$ . Here f is the side opposite to the right angle, so it is the hypotenuse and d is the shorter side and e are the longer side.

Now, by theorem 8-7, the hypotenuse is two times as long as the shorter side, so

  $\begin{array}{c}f=2d\\ 6\sqrt{3}=2d\\ \frac{6\sqrt{3}}{2}=d\\ 3\sqrt{3}=d\end{array}$

Also, by theorem 8-7, the longer side is $\sqrt{3}$ times as long as the shorter side.

Thus,

  $\begin{array}{c}e=\sqrt{3}d\\ e=\sqrt{3}\left(3\sqrt{3}\right)\\ e=9\end{array}$

Thus, the complete table is:

d	e	f
$3\sqrt{3}$	$9$	$6\sqrt{3}$