Properties of Similar Triangles Part 2

September 15, 2011 GB High School Geometry, High School Mathematics

This is the third and the conclusion of the Triangle Similarity Series. The two prequels are 1. Introduction to Similarity and 2. Properties of Similar Triangles (Part 1).

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In the previous post, we have investigated the properties of similar triangles. We have learned that corresponding angles of similar triangles are congruent. In this post, we are going to discuss more about the properties of similar triangles. If you have not performed the investigation in the previous post, you can use the applet below.

[iframe http://mathandmultimedia.com/wp-content/uploads/2011/09/propertiessimilartriangles.html 539 464]

You would have realized from your exploration of the applet that aside from the angles, there is also something unique about the side lengths of the corresponding sides of the triangles (check the Show/Hide Side Length check box above). We can verify they have the same ratio. That is, if triangle ABC is similar to triangle DEF, then the following relationships hold:

$\displaystyle\frac{DE}{AB} = \frac{EF}{BC} = \frac{DF}{AC}$ .

Now, let us examine the relationship of the areas (check the Show/Hide Area check box). Suppose we let the ratio of the sides of triangle $ABC$ and $DEF$ be $k$ , then if the base and height (altitude) of a triangle $ABC$ is $b$ and $h$ , then the base and height of the triangle $DEF$ is $kb$ and $kh$ (Can you see why?). Therefore, the areas of the two triangles are

$A_1 = \displaystyle\frac{bh}{2}$ and $A_2 = \displaystyle\frac{(kb)(kh)}{2} = \frac{k^2(bh)}{2}$ .

The equations above show that if the ratio of the sides of two similar triangles is $k$ , then the ratio of their area is $k^2$ .

It can also be observed that if we connect the vertices of two similar triangles with lines (check the Show/Hide Lines check box), the lines meet at a point (Is this always true? If not, what should be the condition(s) to make it true?). This point is called the point of similarity or the point of dilation.

If we let $O$ be the point of dilation, and $k$ be the ratio of the side lengths, then the ratio of the distance of from $O$ to the vertices is also $k$ . This can be easily verified from by downloading the GGB file of the construction and using the distance tool. That is, if triangle $ABC$ and $DEF$ are similar, and the ratio of their side lengths is $k$ , then,

$\displaystyle\frac{OD}{OA} = \frac{OE}{OB} = \frac{OF}{OC}=k$

The ratio of similarity can be extended to 3 dimensional figures. That is, if two solids are similar and the ratio of their corresponding edge lengths is $k$ , then the ratio of the their volumes is $k^3$ .

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Properties of Similar Triangles Part 2

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