Chapter 10.3, Problem 14WE

To determine

To calculate:The value of $z$ and $PW$ .

Answer to Problem 14WE

  $z=7,PW=17$

Explanation of Solution

Given information: $CW=2z^2-5z-12$ and $CP=z^2-15$

Calculation:

  

As AU, BV and CW are the medians of triangle ABC, P is the centroid of triangle.

(Centroid is the intersection point of medians of the triangle)

As centroid divides the median in ratio $2:1$ ,

  $\begin{array}{l}\frac{CP}{PW}=\frac{2}{1}\\ \Rightarrow \frac{PW}{CP}=\frac{1}{2}\end{array}$

Add 1 on both sides.

  $\begin{array}{l}\frac{PW}{CP}+1=\frac{1}{2}+1\\ \Rightarrow \frac{PW+CP}{CP}=\frac{1+2}{2}\\ \Rightarrow \frac{CW}{CP}=\frac{3}{2}\end{array}$

Put $CW=2z^2-5z-12$ and $CP=z^2-15$

  $\begin{array}{l}\frac{2z^2-5z-12}{z^2-15}=\frac{3}{2}\\ \Rightarrow 4z^2-10z-24=3z^2-45\\ \Rightarrow z^2-10z+21=0\\ \Rightarrow z^2-7z-3z+21=0\\ \Rightarrow z\left(z-7\right)-3\left(z-7\right)=0\\ \Rightarrow \left(z-3\right)\left(z-7\right)=0\\ \Rightarrow z=3,7\end{array}$

  $\begin{array}{l}PW=CW-CP\\ =2z^2-5z-12-z^2+15\\ =z^2-5z+3\end{array}$

At $z=3;PW=3^2-5\left(3\right)+3=9-15+3=-3$

But length cannot be negative.

So,

  $z\ne 3$

So,

  $z=7$

At $z=7;PW=7^2-5\left(7\right)+3=49-35+3=49-32=17$