Applications

When we work with similar triangles, we are harnessing the familiar power of proportions! Similar shapes are scaled versions of one another – so they are proportional.

[latex]\frac{15'}{4.5'}=\frac{X}{6'}[/latex]

The parallel lines include two similar triangles, although they may be hard to see.

Two triangles are similar if the three angles of one triangle have the same measure as the three angles of the second triangle. The lengths of the sides of similar triangles will be in the same proportion. The triangles will have the same shape but the lengths will be scaled up or down.

Because right triangles already have one angle set at 90°, they only need to have one additional angle the same to be similar. This makes them useful for solving applied problems.

We use similar triangles to find an unknown dimension on one triangle by setting up a proportion. The known side on one triangle divided by the similar known side on the other triangle make one side of the proportion. That is set equal to the unknown side divided by the similar known side.

To find h in the diagram above, use the following proportion: [latex]\frac{18}{12} = \frac{h}{8}[/latex]
To find c in the diagram above, use the following proportion: [latex]\frac{42}{36} = \frac{c}{42}[/latex]

On the job, we may find similar triangles useful when dealing with a measurement that is difficult to take. We can use shadows, reflections, and angles sighted with a compass or surveying instruments to build similar triangles and solve for a measurement we are otherwise unable to take.

For example, you might need to get the distance across a body of water without crossing it, and can do so as follows:

Start at Point A and sight your compass to Point C (a prominent object across the lake). Follow that heading and measure the distance you walk along that bearing to the lake shore at Point B. Turn away from the lake and walk to Point D. Record the bearing from Point D to Point B. Turn to walk toward the lake until you can face Point C at the same bearing you took from Point D to Point B. Lines DB and EC will be parallel and you’ve created similar triangles with a known distance from D to B.  You’ve got enough known measurements figure out the width of the lake along either line!

Find BC [latex]\frac{AB}{AB+BC}=\frac{AD}{AE}[/latex]

or

Find EC [latex]\frac{EC}{DB}=\frac{AE}{AD}[/latex]