Chapter 8.2, Problem 19WE

To determine

To find: x

Answer to Problem 19WE

  $x=3$

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 $△ABD$ are congruent, so it is an isosceles triangle.

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

  $\begin{array}{c}BC=\frac{1}{2}CD\\ =\frac{1}{2}\left(8\right)\\ =4\end{array}$

Consider the right triangle $△ABC$ .

Now, $\overline{AB}$ is the hypotenuse and has length 5 and the lengths of the legs are $AC=x$ and $BC=4$ .

So, by theorem 8-2,

  $\begin{array}{c}A{B}^2=A{C}^2+B{C}^2\\ 5^2=x^2+4^2\\ 25=x^2+16\end{array}$

  $\begin{array}{c}25-16=x^2\\ 9=x^2\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{9}\\ x=3\end{array}$

That is,

  $AO=OC=8$

And thus,

  $\begin{array}{c}AC=AO+OC\\ =8+8\\ =16\end{array}$

The length of the other diagonal of rhombus is 16 cm.