Chapter 16.3, Problem 4PS

To determine

To find the measure of the third side of the triangle.

Answer to Problem 4PS

  $BC=21$

Explanation of Solution

Given:

The following is the given figure-

  

Calculation:- The angle bisector divides the opposite side in segments proportional to the angle's sides.

  $\begin{array}{l}\frac{AB}{AC}=\frac{BD}{DC}\\ \frac{9}{18}=\frac{BD}{DC}\\ \frac{1}{2}=\frac{BD}{DC}\\ DC=2BD\end{array}$

Using Stewart’s theorem-

  $A{B}^2\cdot DC+A{C}^2\cdot BD=BC\cdot A{D}^2+BC\cdot BD\cdot DC$

Now, substituting the values-

  $\begin{array}{l}{9}^2\cdot DC+{18}^2\cdot BD=BC\cdot 8^2+BC\cdot BD\cdot DC\\ 81(2BD)+324BD=(BD+2BD)\cdot 64+(BD+2BD)\cdot BD(2BD)\\ 162BD+324BD=192BD+6B{D}^3\\ 6B{D}^3-294BD=0\\ 6BD(B{D}^2-49)=0\\ BD=0\\ BD=\pm 7\end{array}$

The only positive solution is $BD=7$ ,

Therefore, the length of the side BC is $BC=DC+BD=3BD=3\cdot 7=21$