Chapter 11.7, Problem 21WE

To determine

To Find: The name of similar triangles and give the ratio of their area.

Answer to Problem 21WE

  $\Delta ABC:\Delta CAD=1:1$

  $\Delta ABF:\Delta DEF=9:4$

  $\Delta ABG:\Delta CEG=9:25$

  $\Delta ABF:\Delta CEB=9:25$

Explanation of Solution

Given:

  

Observe the figure

  $\begin{array}{l}\angle A=\angle C(opp.angle)\\ \angle B=\angle D(opp.angle)\\ \angle DAC=\angle BCA\\ \angle B=\angle D\\ \angle DAC=\angle BCA\\ thus\\ \Delta ABC\sim \Delta CAD\end{array}$

Now in triangles $\Delta ABCand\Delta CAD$

The value of two sides is $6$

Thus the ratio of two sides $ABandCD$ is

  $6:6=1:1$

Thus

  $\Delta ABC:\Delta CAD=1:1$

Area of similar triangles $\Delta ABCand\Delta CAD$

  $\Delta ABC:\Delta CAD=1^2:1^2=1:1$

Now

  $\begin{array}{l}\angle ABG=\angle CEG(alternateangle)\\ \angle GAB=\angle ECG(alternateangle)\\ thus\\ \Delta ABG\sim \Delta CEG\end{array}$

Now in triangles $\Delta ABGand\Delta CEG$

The ratio of two sides $ABandCE$ is

  $6:10=3:5$

Thus

  $\Delta ABG:\Delta CEG=3:5$

Area of similar triangles $\Delta ABGand\Delta CEG$

  $\Delta ABG:\Delta CEG=3^2:5^2=9:25$

Similarly

In triangle $\Delta ABFand\Delta DEF$

  $AB:CE=6:4=3:2$

  $\Delta ABF:\Delta DEF=3^2:2^2=9:4$

Similarly

  $\Delta ABF:\Delta CEB=9:25$