Similar Polygons: Practice Problems

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  • 0:03 Similar Polygons
  • 0:51 Problem 1
  • 2:08 Problem 2
  • 4:47 Lesson Summary
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Lesson Transcript
Instructor: Elizabeth Foster

Elizabeth has been involved with tutoring since high school and has a B.A. in Classics.

Similar polygons have the same angles and proportional side lengths. In this lesson, you'll solve some practice problems using their special properties.

Similar Polygons

What do we mean when we say that we have similar polygons? It means that corresponding angles have the same measures, and corresponding sides have proportional lengths.

For example, in this drawing, you can see that the corresponding angles of both polygons, labeled a through e, are all the same.


Each side of the bigger polygon is twice as long as the corresponding side of the smaller polygon, so the sides are proportional in a ratio of 2:1. That means these two polygons are similar. The perimeters of the polygons will also be proportional in the same ratio.

If you know that two polygons are similar, you can use the side lengths and angle measures of one polygon to help you figure out the side lengths and angle measures of the other. In this lesson, we'll look at some practice problems to give you an idea.

Problem 1

Polygon A is similar to Polygon B. The perimeter of Polygon A is 15 meters. The perimeter of Polygon B is 10 meters. x = w, and y = v. Given that the length of side w1 is 3 meters, what is the length of side x?

Problem 1

If you don't know what to do with a problem like this, a good first step is to mark up the diagram with all the information given. The problem tells us that the perimeter of A is 15 meters, and the perimeter of B is 10 meters. If we simplify that, we get a ratio of 1.5 to 1, which means that every side of Polygon A is 1.5 times as long as the corresponding side of Polygon B. We'll use this to simplify the diagram. We know that w1 is equal to 1.5 times w, so instead of w1, we'll write 1.5w. The same goes for all the sides of Polygon A, so we'll replace them all.

The problem also tells us that w1, which we just renamed as 1.5w, is equal to 3. If 1.5w = 3, then w must equal 3/1.5, meaning that w = 2. The problem tells us that x = w, so x must also be 2.

Problem 2

Ready for something a little trickier? You know you got this. The two arrows shown are congruent polygons.


The measure of angle a is 60 degrees, and the measure of angle b is also 60 degrees. The length of side x is 1/3 the length of side y. The length of side y is 12 inches. What's the area of the region shown in green?

All right, let's start by writing everything we know on the diagram. We know that angles A and B are both 60 degrees, so we'll start by marking that on both shapes. The problem tells us that y = 12, so we'll write that on the diagram too. The length of x is 1/3 the length of y, so side x is 4.

Now look at the 'point' of each arrow as a triangle.


Part of one side of the triangle is cut out, but it would be right where the orange line is in the picture. In each triangle, two of the angles are 60 degrees. That means the third angle also has to be 60 degrees, because all the angles in a triangle add up to 180. In other words, the point of each arrow is basically an equilateral triangle with a chunk cut out of one side.

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