Chapter 16.5, Problem 4PS

To determine

To find: The length of the side AM.

Answer to Problem 4PS

The length of the side AMis $\frac{12+2\sqrt{3}}{3}\text{units}$ .

Explanation of Solution

Given information:

The triangle $\Delta \text{ABC}$ is given.

Assign the ratios and the weights as shown below,

  

The length of the side PMis calculated as,

  $\begin{array}{l}{w}_1\left(AP\right)=w_2\left(PM\right)\\ \left(\frac{1}{\sqrt{3}}\right)\left(4\right)=\left(2\right)\left(PM\right)\\ PM=\frac{2}{\sqrt{3}}\\ PM=\frac{2\sqrt{3}}{3}\text{unit}\end{array}$

The length of the side AMis calculated as,

  $\begin{array}{c}AM=AP+PM\\ =4+\frac{2\sqrt{3}}{3}\\ =\frac{12+2\sqrt{3}}{3}\text{units}\end{array}$

Thus, the length of the side AM is $\frac{12+2\sqrt{3}}{3}\text{units}$ .