Chapter 8.2, Problem 22WE

To determine

To find: x

Answer to Problem 22WE

  $x=4\sqrt{7}$

Explanation of Solution

Given:

The figure:

Concept Used:

Theorem4-1 corollary-3: vertex angle bisector of isosceles triangle is perpendicular bisector of the base.

Theorem 8-2: In right triangle, square of hypogenous is equal to the square of base and perpendicular.

Label the given figure as shown below:

m:math display='block'>△ACD

It can be observed that two sides of the triangleare congruent, so it is an isosceles triangle.

So, by Theorem4-1 corollary-3, it can be said that,

  $\begin{array}{c}AB=\frac{1}{2}AC\\ =\frac{1}{2}\left(x\right)\\ =\frac{x}{2}\end{array}$

Consider the right triangle $△ABD$ .

Now, $\overline{AD}$ is the hypotenuse and has length 8 and the lengths of the legs are $AB=\frac{x}{2}$ and $BD=6$ .

So, by theorem 8-2,

  $\begin{array}{c}A{D}^2=A{B}^2+B{B}^2\\ 8^2={\left(\frac{x}{2}\right)}^2+6^2\\ 64=\frac{x^2}{4}+36\end{array}$

  $\begin{array}{c}\frac{x^2}{4}=64-36\\ \frac{x^2}{4}=28\\ x^2=28\times 4\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{28\times 4}\\ x=\sqrt{4\times 4\times 7}\\ x=4\sqrt{7}\end{array}$