## Tuesday, July 24, 2007

### Are all triangles in the universe similar to 3-4-5?

(a) AB = 8, BC = 6 in rectangle ABCD. Find the lengths of all segments shown in the diagram above.
Specifically: BD, CF,DE, BE, FE, CE

Comments: This is another in a series of rectangle investigations. To deepen student understanding of triangle relationships and to provide considerable practice with these ideas, the question asks for more than just one result. Students should be encouraged to first draw ALL of the triangles in the diagram separately and recognize why they are all similar! Using ratios, students should be able to find all the segments efficiently. One could also demonstrate the altitude on hypotenuse theorems as well!

(b) In case, students need a bit more of a challenge, have them derive expressions for all of the above segments given that AB = b and BC = a. To make life easier, assume b > a. This should keep your stronger students rather busy! This algebraic connection is powerful stuff. We want our students to appreciate that algebra is the language of generalization.

Totally_clueless said...

It may be more efficient to first find CE via areas.

Clueless

Dave Marain said...

tc--
I 'totally' agree! Areas are usually so much easier to work with than altitude on hypotenuse theorems. Students often overlook that approach. However, our students also need to develop more of a comfort level with similar triangles in general and that's the main reason I developed this problem. All of the segments can be found readily once the student matches up corresponding sides of pairs of similar triangles all of which are readily seen to be multiples of 3-4-5 (this enables the lengths to be determined without a calculator!). This question is very similar to others on this and related blogs.
Dave

Mike said...

We can deduce that all of the triangles are similar by use of the angle-angle theorem (postulate?). Then just set up ratios. I'm not sure how to use the "area" method ...

mathmom said...

Mike, it took me a while, but the area method just hit me. If you look at triangle BCD you can calculate the area by either multiplying the two legs, or the hypotenuse by the altitude. (If you didn't know that as a theorem, just think about adding together the 2 smaller triangles.)

Since we know the lengths of all 3 sides of that triangle, the only unknown is the length of the altitude.

I'd have set up the ratios too, but that is quite a nice approach.

Here's a pointer back to a related post from April as well: Ratio of Areas

mathmom said...

well, actually I should have said "twice the area" in both cases above. And instead of thinking about adding up the triangles, it's probably easier to think about finding the area of the surrounding rectangle, which is of course twice the area of the triangle.