Amazing Mathematics Geometric Mean Maze Answer Key
Video Tutorial
The mean proportion is any value that can be expressed just the way that 'x' is in the proportion on the above on the left.
In the proportion above on the left 'x', is the geometric mean, we could solve for x by cross multiplying and going from there (more on that later)
In the proportion above on the left, '4', is the geometric mean
So what does this have to do with right similar triangles?
It turns out the when you drop an altitude (h in the picture below) from the the right angle of a right triangle, the length of the altitude becomes a geometric mean.
This occurs because you end up with similar triangles which have proportional sides and the altitude is the long leg of 1 triangle and the short leg of the other similar triangle .
Diagram 1
Interactive Demonstration
To better understand how the altitude of a right triangle acts as a mean proportion in similar triangles, look at the triangle below with sides a, b and c and altitude H.
$ \frac{\class{side1}{side1}}{\class{altitude}{altitude}} = \frac{\class{altitude}{altitude}}{\class{side2}{side2}} \\ \\ \frac{\class{side1}{BD}}{\class{altitude}{AD}} = \frac{\class{altitude}{AD}}{\class{side2}{CD}} \\ \\ \frac{\class{side1 side1-v}{6.19}}{\class{altitude altitude-v}{6.19}} = \frac{\class{altitude altitude-v}{6.19}}{\class{side2 side2-v}{6.19}} $
$ \frac{\class{hyp}{hyp}}{\class{leg1}{leg1}} = \frac{\class{leg1}{leg1}}{\class{side1}{side1}} \\ \frac{\class{hyp}{BC}}{\class{leg1}{AB}} = \frac{\class{leg1}{AB}}{\class{side1}{BD}} \\ \frac{\class{hyp hyp-v}{12.37}}{\class{leg1 leg1-v}{8.75}} = \frac{\class{leg1 leg1-v}{8.75}}{\class{side1 side1-v}{6.19}} $
$ \frac{\class{hyp}{hyp}}{\class{leg2}{leg2}} = \frac{\class{leg2}{leg2}}{\class{side2}{side2}} \\ \frac{\class{hyp}{BC}}{\class{leg2}{AC}} = \frac{\class{leg2}{AC}}{\class{side2}{CD}} \\ \frac{\class{hyp hyp-v}{12.37}}{\class{leg2 leg2-v}{8.75}} = \frac{\class{leg2 leg2-v}{8.75}}{\class{side2 side2-v}{6.19}} $
Drag Points And Click Button To Start
Examples: 2 Types of Problems
Students usually have to solve 2 different core types of problems involving the geometric mean.
Problem Type 1
The altitude and hypotenuse
As you can see in the picture below, this problem type involves the altitude and 2 sides of the inner triangles ( these are just the two parts of the large outer triangle's hypotenuse) . This lets us set up a mean proportion involving the altitude and those two sides (see demonstration above if you need to be convinced that these are indeed corresponding sides of similar triangles .)
What is the length of the altitude below?
This problem is just example problem 1 above (solving for an altitude using the parts of the large hypotenuse)
Problem Type 2. Hypotenuse, Leg and Side
Involves the hypotenuse of the large outer triangle, one its legs and a side from one of the inner triangles.
Find the value of x in the triangle below:
This problem is just example problem 2 because it involves the outer triangle's hypotenuse, leg and the side of an inner triangle.
Practice Problems
Thanks to the HHS Math deptarment for how to think about this topic!
Amazing Mathematics Geometric Mean Maze Answer Key
Source: https://www.mathwarehouse.com/geometry/similar/triangles/geometric-mean.php
Posted by: goddardpaptur.blogspot.com
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