Chapter 8.2, Problem 21WE

To determine

To find: x

Answer to Problem 21WE

  $x=3\sqrt{5}$

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:

It can be observed that two sides of the triangle $△ACD$ are congruent, so it is an isosceles triangle.

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

  $\begin{array}{c}AB=\frac{1}{2}AC\\ =\frac{1}{2}\left(6\right)\\ =3\end{array}$

Consider the right triangle $△ABD$ .

Now, $\overline{AD}$ is the hypotenuse and has length x and the lengths of the legs are $AB=3$ and $BD=6$ .

So, by theorem 8-2,

  $\begin{array}{c}A{D}^2=A{B}^2+B{B}^2\\ x^2=3^2+6^2\\ x^2=9+36\\ x^2=45\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{45}\\ x=3\sqrt{5}\end{array}$