Chapter 10.3, Problem 5CE

To determine

To find: the length of EP and PR. If $FT=9$ , find PT and FP.

Answer to Problem 5CE

  $EP=10, PR=4,\text{ and }FP=6.$

Explanation of Solution

Given:

Medians of triangle DEF are given; $FT=9$ .

Calculation:

a.

From the figure, and the theorem 10.4,

  $\begin{array}{l}PS=\frac{1}{3}ES,\\ \Rightarrow ES=EP+PS.\\ \text{Now, if }EP=x,\\ ES=x+5,\\ \Rightarrow 5=\frac{1}{3}(x+5),\\ \Rightarrow 15=x+5\\ \Rightarrow x=10\\ \Rightarrow EP=10.\end{array}$

b.

From the figure, and the theorem 10.4,

  $\begin{array}{l}DP=\frac{2}{3}DR,\\ \Rightarrow DR=DP+PR.\\ \text{Now, if }PR=y,\\ DR=8+y,\\ \Rightarrow 8=\frac{2}{3}(8+y),\\ \Rightarrow 24=16+2y,\\ \Rightarrow 2y=8,\\ \Rightarrow y=4,\\ \Rightarrow PR=4.\end{array}$

c.

From the figure, and the theorem 10.4,

  $\begin{array}{l}FP:PT:FT=2:1:3.\\ \text{And it is also given that }FT=9,\\ \Rightarrow PT:9=1:3,\\ \Rightarrow 3PT=9,\\ \Rightarrow PT=\frac{9}{3},\\ \Rightarrow PT=3.\\ \text{Since }FP:9=2:3,\\ \Rightarrow FP=\frac{2}{3}\times 9,\\ \Rightarrow FP=\frac{18}{3},\\ \Rightarrow FP=6.\end{array}$