Chapter 8.2, Problem 31WE

a.

To determine

To find: the length of median to the hypotenuse.

Answer to Problem 31WE

  $5$

Explanation of Solution

Given:

The right triangle has legs of length 6 and 8.

Draw a right triangle with legs of length 6 and 8 also draw a median to the hypotenuse:

Let $AC=a$ and $\overline{BD}$ be the median to the hypotenuse $\overline{AC}$

Consider the right triangle $△ABC$ .

Now, $\overline{AC}$ is the hypotenuse and has length a and the lengths of the legs are $AB=6$ and $BC=8$ .

So, by theorem 8-2,

  $\begin{array}{c}A{C}^2=B{C}^2+A{B}^2\\ a^2=8^2+6^2\\ a^2=64+36\\ a^2=100\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{a^2}=\sqrt{100}\\ a=10\end{array}$

Now $\overline{BD}$ be the median to the hypotenuse $\overline{AC}$

So, D is the midpoint of $\overline{AC}$ . That is,

  $\begin{array}{c}CD=\frac{1}{2}AC\\ =\frac{1}{2}\left(10\right)\\ =5\end{array}$

Now, by theorem 5-15, the midpoint D of hypotenuse $\overline{AC}$ of a right triangle $△ABC$ is equidistant from the three vertices.

Thus,

  $BD=CD=5$

Hence, the length of the median to the hypotenuse is 5 units.

b.

To determine

To find: the length of altitude to the hypotenuse.

Answer to Problem 31WE

  $4.8$

Explanation of Solution

Given:

The right triangle has legs of length 6 and 8.

Draw a right triangle with legs of length 6 and 8 also draw an altitude to the hypotenuse:

Let $AC=a$ and $\overline{BD}$ be the altitude to the hypotenuse $\overline{AC}$

Consider the right triangle $△ABC$ .

Now, $\overline{AC}$ is the hypotenuse and has length a and the lengths of the legs are $AB=6$ and $BC=8$ .

So, by theorem 8-2,

  $\begin{array}{c}A{C}^2=B{C}^2+A{B}^2\\ a^2=8^2+6^2\\ a^2=64+36\\ a^2=100\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{a^2}=\sqrt{100}\\ a=10\end{array}$

Now $\overline{BD}$ be the altitude to the hypotenuse $\overline{AC}$

So by theorem 8-1 corollary 2, the midpoint D of hypotenuse $\overline{AC}$ of a right triangle $△ABC$ is equidistant from the three vertices.

  $\begin{array}{c}\frac{AC}{AB}=\frac{AB}{AD}\\ \frac{10}{6}=\frac{6}{AD}\\ 10AD=6\times 6\end{array}$

  $\begin{array}{c}10AD=36\\ AD=\frac{36}{10}\\ AD=3.6\end{array}$

Now consider the right triangle $△ABD$ .

Let $BD=x$

Now, $\overline{AB}$ is the hypotenuse and has length 6 and the lengths of the legs are $AD=3.6$ and $BD=x$ .

So, by theorem 8-2,

  $\begin{array}{c}A{B}^2=A{D}^2+B{D}^2\\ 6^2={\left(3.6\right)}^2+x^2\\ 36=12.96+x^2\end{array}$

  $\begin{array}{c}36-12.96=x^2\\ 23.04=x^2\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{23.04}\\ x=4.8\end{array}$

Hence, the length of the altitude to the hypotenuse is 4.8 units.