Chapter 8.4, Problem 34WE

To determine

To find: As many lengths as possible

Answer to Problem 34WE

  $\begin{array}{l}MN=4\sqrt{2}\\ NL=4\sqrt{2}\\ KL=8\sqrt{2}\end{array}$

  $\begin{array}{l}KN=4\sqrt{6}\\ MK=8\sqrt{2}\end{array}$

Explanation of Solution

Given:

The figure:

Concept Used:

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

Theorem 8-2: in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.

Theorem 8-6: in a $45°-45°-90°$ triangle, the hypotenuse is $\sqrt{2}$ times as long as a leg.

Theorem 8-7: In a $30°-60°-90°$ triangle, the hypotenuse is twice as long as the shorter leg, and the longer leg is $\sqrt{3}$ times as long as the shorter leg.

Consider the given figure:

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

  $\begin{array}{c}m\angle MLN+m\angle LNM+m\angle NML=180\\ m\angle MLN+90+45=180\\ m\angle MLN+135=180\end{array}$

  $\begin{array}{c}m\angle MLN=180-135\\ m\angle MLN=45\end{array}$

Similarly, Apply theorem 3-11 on the triangle $△KNL$ ,

  $\begin{array}{c}m\angle LKN+m\angle KNL+m\angle NLK=180\\ m\angle LKN+90+60=180\\ m\angle LKN+150=180\end{array}$

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

It can be observed that the triangle $△MNL$ is a $45°-45°-90°$ .

Her $\overline{ML}$ is the side opposite to the right angle, so it is the hypotenuse and $\overline{MN}$ and $\overline{NL}$ are the legs.

Now, a $45°-45°-90°$ is an isosceles triangle and thus both its legs are congruent. So,

  $MN=NL$

Now, by theorem 8-6, the hypotenuse is $\sqrt{2}$ times as long as a leg, so

  $\begin{array}{c}ML=\sqrt{2}\left(NL\right)\\ 8=\sqrt{2}\left(NL\right)\\ \frac{8}{\sqrt{2}}=NL\\ 4\sqrt{2}=NL\end{array}$

Thus,

  $MN=NL=4\sqrt{2}$

Also the triangle $△KNL$ is a $30°-60°-90°$ . Here $\overline{KL}$ is the side opposite to the right angle, so it is the hypotenuse and $\overline{NL}$ is the shorter leg and $\overline{KN}$ are the longer leg.

Now, by theorem 8-7, the hypotenuse is twice as long as the shorter leg, so

  $\begin{array}{c}KL=2NL\\ =2\left(4\sqrt{2}\right)\\ =8\sqrt{2}\end{array}$

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

Thus,

  $\begin{array}{c}KN=\sqrt{3}NL\\ =\sqrt{3}\left(4\sqrt{2}\right)\\ =4\sqrt{6}\end{array}$

Consider the right triangle $△MNK$ .

Now, $\overline{MK}$ is the hypotenuse and the lengths of the legs are $MN=4\sqrt{2}$ and $NK=4\sqrt{6}$ .

So, by theorem 8-2,

  $\begin{array}{c}M{K}^2=M{N}^2+N{K}^2\\ ={\left(4\sqrt{2}\right)}^2+{\left(4\sqrt{6}\right)}^2\\ =32+96\\ =128\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{M{K}^2}=\sqrt{128}\\ MK=8\sqrt{2}\end{array}$