Chapter 6.9, Problem 8PPS

a.

To determine

To find : Measure of segment GD

Answer to Problem 8PPS

  $GD=12m$

Explanation of Solution

Given information : Triangles in the given figure are similar

Calculation :

  

  $\Delta AGD\sim \Delta CED$

As sides of similar triangles are proportional, therefore

  $\begin{array}{l}\frac{GD}{ED}=\frac{AG}{CE}\\ \Rightarrow \frac{GD}{6}=\frac{9}{4.5}\\ \Rightarrow GD=\frac{9}{4.5}(6)=12m\end{array}$ .

b.

To determine

To find : Measure of segment BF

Answer to Problem 8PPS

  $BF=6.85m$

Explanation of Solution

Given information: $GF\cong FE$

  $\begin{array}{l}GD=12m\\ \Rightarrow GF+FE+6=12\\ \Rightarrow GF+FE=12-6=6\\ \Rightarrow 2GF=6\left\{\because GF\cong FE\right\}\\ \Rightarrow GF=3m\\ \Rightarrow GF=FE=3m\\ \Rightarrow DF=FE+6=3+6=9m\end{array}$

Also, $\Delta BFD\sim \Delta CED$

As sides of similar triangles are proportional,

  $\begin{array}{l}\frac{BF}{CE}=\frac{FD}{ED}\\ \Rightarrow \frac{BF}{4.5}=\frac{9}{6}\end{array}$

  $\Rightarrow BF=\frac{9}{6}(4.5)=6.75m$

c.

To determine

To find : Measure of segments $BC,CD$

Answer to Problem 8PPS

  $BC=3.75m,CD=7.5m$

Explanation of Solution

Given information: $AD=15m$

  $\Delta AGD\sim \Delta CED$

As sides of similar triangles are proportional,

  $\begin{array}{l}\frac{AG}{CE}=\frac{AD}{CD}\\ \Rightarrow \frac{9}{4.5}=\frac{15}{CD}\\ \Rightarrow CD=\frac{15}{9}(4.5)=7.5m\end{array}$

Also, $\Delta BFD\sim \Delta CED$

As sides of similar triangles are proportional,

  $\begin{array}{l}\frac{BF}{CE}=\frac{BD}{CD}\\ \Rightarrow \frac{6.75}{4.5}=\frac{BD}{7.5}\\ \Rightarrow BD=\frac{6.75}{4.5}(7.5)=11.25m\\ BC=BD-CD=11.25-7.5=3.75m\end{array}$.