*Mathematical Calendar 2016*, Theoni Pappas asked the following question:

**Determine the geometric mean of 8 & 32.**

Solution: let

*g*be the geometric mean. From the definition of geometric mean, 8/

*g*=

*g*/32. Cross multiplying,

*g*^2 = 256. So

*g*= +/- 16. From the definition of geometric mean,

*g*is positive; so the geometric mean of 8 and 32 is 16. Notice that yesterday's date was in fact the 16th.

I don't post questions from the Pappas calendar very often, but this is the second time this month that I've done so. And of course, I post this question because once again, it lines up almost perfectly with today's Geometry post. Lesson 14-2 of the U of Chicago text is on lengths in right triangles -- specifically those that involve the geometric mean. Of course, this isn't a perfect match, since I used

*yesterday's*question with

*today's*lesson, but it's close enough. Had it been the other way around, the Pappas question could have served as a great warm-up question.

This is what I wrote last year about today's lesson:

Lesson 14-2 of the U of Chicago text is on lengths in right triangles -- specifically, those lengths that are related to the altitude and involve the geometric mean.

Geometric Mean Theorem:

The geometric mean of the positive numbers

*a*and

*b*is sqrt(

*ab*).

(Note: This may sound like a

*definition*, but actually the U of Chicago defines

*geometric mean*to be the number

*x*such that

*a*/

*x*=

*x*/

*b*, so we need a theorem to get the geometric mean as sqrt(

*ab*).)

Right Triangle Altitude Theorem:

In a right triangle:

a. The altitude of the hypotenuse is the geometric mean of the segments dividing the hypotenuse.

b. Each leg is the geometric mean of the hypotenuse and the segment adjacent to the leg.

In this lesson, I give the proof of the Pythagorean Theorem based on similarity, but this time I gave the proof in the book, which mentions the geometric mean. Let's look at the proof -- as usual, with an extra step for the Given:

Given: Right triangle

Prove:

*a*^2 +

*b*^2 =

*c*^2

Proof:

Statements Reasons

1. Right triangle 1. Given

2.

*a*geometric mean of

*c*&

*x*, 2. Right Triangle Altitude Theorem

*b*geometric mean of

*c*&

*y*

3.

*a*= sqrt(

*cx*),

*b*= sqrt(

*cy*) 3. Geometric Mean Theorem

4.

*a*^2 =

*cx*,

*b*^2 =

*cy*4. Multiplication Property of Equality

5.

*a*^2 +

*b*^2 =

*cx*+

*cy*5. Addition Property of Equality

6.

*a*^2 +

*b*^2 =

*c*(

*x*+

*y*) 6. Distributive Property

7.

*x*+

*y*=

*c*7. Betweenness Theorem (Segment Addition Postulate)

8.

*a*^2 +

*b*^2 =

*c*^2 8. Substitution (step 6 into step 7)

It is uncertain whether this is the proof that Common Core intends the students to learn, or whether my earlier proof that avoids geometric means suffices.

Actually, since posting this last year, I've decided to check both the PARCC and SBAC released test questions for those related to the proof of the Pythagorean Theorem. There were a few questions that required use of Pythagoras, but none directly related to the

*proof*. Of course, some people lament that there aren't very many proofs on the Common Core tests.

Notice that one of the questions that I included from the text involves the Girl Scouts -- in particular, a girl scout troop leader who calculates the height of a tower using notebook paper (the sole purpose of which is to ensure that the angle is actually 90 degrees). This time of year is Girl Scout Cookie season, so of course I had to include a Girl Scout problem.

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