How (m^2)/(n^2)=(m/n)^2 is Fundamental to Geometry!

OVERVIEW
The Common Core stresses the importance of students developing a deeper understanding of fundamental concepts and to discover/uncover the interrelatedness of mathematics. The discussion below can be used to demonstrate how a basic law of exponents is tied to the geometry of similar figures.

THE PROBLEM/INVESTIGATION
1) If the sides of 2 squares are in the ratio 2:1, show that their areas are in the ratio 4:1
(a) visually
(b) numerically by examining particular cases
(b)  algebraically

2) If the sides are in the ratio 3:1, do you think the areas will be in the ratio 6:1 or 9:1? Now do parts a-c as in 1).

3) If the ratio of the sides is 3:2 show algebraically that the ratio of the areas is 9:4.

4) Show algebraically that if the ratio of the sides of 2 squares is m:n then the ratio of their areas is (m/n)^2.

Note: How does this result connect to the idea that the area of a square varies directly as the square of its side length?

4) If squares are replaced by circles using radii or diameters in place of "sides" show that the results of questions 1-4 are the same.

How does this result connect to the idea that the area of a circle varies directly as the square of its radius or diameter (or circumference)?

REFLECTIONS
• Squares and circles are of course special cases of similar figures. Beyond this investigation lies the BIG IDEA:

The areas of 2-dim similar figures are proportional to the squares of their linear dimensions.

Note: In 3 dim, we can replace 'area' by what?

• Do you see this as one of the fundamental theorems of Euclidean geometry? Is it sufficiently stressed in textbooks and in the standards? Of course you may not feel as I do about all this!

• So what is the geometry connection to
(m/n)^3 = (m^3)/(n^3)...

'.

If a hen and a half can lay an egg and a half in a day and a half...

The full version in one of its many many variations:

If a hen and a half can lay an egg and a half in a day and a half, how many eggs can three hens lay in three days? Assume that all hens are a-laying at the same rate.

Putting aside the silliness of the riddle, there really is some serious mathematics going in these kinds of rate/ratio/proportion problems. Rather than solve the "hen" problem for you, I'll leave it to my readers to solve it by their own favorite methods. By the way, the answer to this riddle is in the description of the video below on my YouTube channel. Sorry 'bout that!!

Instead, the video below, which appears on my YouTube channel, MathNotationsVids, presents a developmental approach to a more complicated ratio problem for middle schoolers and beyond. I'm far more interested in your thoughts about the teaching strategies than I am about the problem itself. Please understand, further, that I am not suggesting the method shown in the video is efficient nor would it make much sense for the upper level math or science student. See comments below the video for further discussion of this.


The Problem in the Video Below:


If 10 workers can build 3 houses in 60 days, how many workers are needed to build 5 houses in 40 days? Assume all workers build at the same rate.




More Advanced and Efficient Algorithms


(1) We assume from the "constant rate" assumption in  the problem that the number of houses (H) which can be built varies jointly as the number of workers (W) and the number of days (D).
Thus, H = kWD.

Substituting, H=3, W=10 and D=60, we obtain:
3 = k(10)(60) or k = 1/200. Note that the units of k are Houses/(Workers x Days).
We can interpret k to mean that 1/200 of a house can be built by 1 worker in 1 day. Thus, k is not only a constant but actually represents a rate. Another way of expressing this rate is
(1 House)/(200 Worker-Days) or the reciprocal version:
(200 Worker⋅Days)/(1 House)

Substituting the new set of values into the relationship H = (1/200)WD, we obtain:
5 = (1/200)(W)(40) or W = 25 workers.

(2) This can be made even more efficient using the "factor-label" (dimensional analysis, etc.) format:

(200 Worker⋅Days)/(1 House)) x (5 Houses)/(40 Days) = 25 Workers!

(3)  I could also exploit the inverse variation between W and D, but that's for my readers to bring up or for another video!

I see these efficient methods as "black box" methods for some students. Developing a deeper understanding of direct and inverse variation is far more important for the younger student.



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"All Truth passes through Three Stages: First, it is Ridiculed... Second, it is Violently Opposed... Third, it is Accepted as being Self-Evident." - Arthur Schopenhauer (1778-1860)