Chapter 5, Problem 16CT

To determine

To find: $MN$ , $NE$ and $RT$

Answer to Problem 16CT

  $MN=14$

  $NE=15$

  $RT=26$

Explanation of Solution

Given:

The diagram:

$\overline{RE}$ is an altitude of $△RST$ .

Consider the figure shown below:

Consider the triangle $△RST$ .

Here, points M and N are the mid points of $\overline{RT}$ and $\overline{RS}$ .

So, by theorem 5-11 part (2), the side $\overline{MN}$ is half as long as the third side $\overline{ST}$ .

  $\begin{array}{c}MN=\frac{1}{2}ST\\ =\frac{1}{2}\left(SE+ET\right)\\ =\frac{1}{2}\left(18+10\right)\end{array}$

  $\begin{array}{c}=\frac{1}{2}\left(28\right)\\ =14\end{array}$

Consider the triangle $△RST$ .

It is given that $\overline{RE}$ is an altitude of $△RST$ .

So, it can be said that the triangles $△SER$ and $△TER$ are right triangles.

Now, $\overline{RS}$ is the hypotenuse of the right triangle $△SER$ and N is its midpoint.

So, by theorem 5-15, the midpoint N of the hypotenuse $\overline{RS}$ of a right triangle $△SER$ is equidistant from the three vertices.

Thus,

  $NE=NR=15$

Similarly, $\overline{RT}$ is the hypotenuse of the right triangle $△TER$ and M is its midpoint.

So, by theorem 5-15, the midpoint N of the hypotenuse $\overline{RT}$ of a right triangle $△TER$ is equidistant from the three vertices.

Thus,

  $MT=ME=13$

Also, M is the midpoint of $\overline{RT}$ . So, it implies that,

  $RM=MT=13$

Therefore,

  $\begin{array}{c}RT=RM+MT\\ =13+13\\ =26\end{array}$