Dedication: Thanks to Professor Harald Jensen (18981994), Physics Department, Lake Forest College, who originally worked this idea with the high school physics teachers at several summer institutes during the 1970s. This fine example of a phenomenological presentation would not exist if it were not for him.
Introduction
This is a lesson plan for teachers.
If you want other teachers to understand the phenomenological approach to teaching/learning, this is a good way to begin.
Objectives: 

For teachers of teachers (K  Phd):  To model the phenomenological approach. 
For teachers of students (6 and above):  To enable each student to prove the Pythagorean Theorem on his own. 
For teachers of students (K  5):  To enable students to identify, and/or define rectangle, square, triangle, and the concept of area (a measure of the amount of surface). 
Materials Needed:
Strategy:
For teachers: Ask: "Who has seen this before; anyone?
Raise your
hands."
If any hands are raised, then announce  "I need your help!
If you
have seen or done this before, please do not give it away to those
who have not.
Please don't spoil their fun."
Next, ask people to form pairs or partners by holding
their hands up together. Tell them to remember who their partner
is.
Have the overhead projector prepared ahead of time by placing a blank transparency centered on its projection area. Then begin by placing the pieces of the puzzle on the overhead and viewing the image on a screen so all can see and participate. Challenge teachers to tell you how to assemble the pieces into a solid rectangle using all the pieces  they must tell what to do; cannot show.
NOTE: Invariably, teachers  or almost anyone for that matter  will find it difficult to tell you what to do. e.g. They might say, "Move the piece on top next to the gray one." And you will move the piece, but not place them in contact; or you will move the wrong piece, etc. You will not automatically do what they want you to do, but rather only and literally what they tell you to do. They will laugh to see how "stupid" you seem to be, but they will see that you are doing only what they told you to do.
After 5 minutes or so, somebody might use the word "triangle" or "square" or "rectangle" to describe the piece they wish you to move. As soon as one of these words is used, repeat the word several times, (e.g. "triangle") and ask a volunteer to define the word. Ask them to name the other pieces and get their definition for each piece until everyone agrees and understands correctly the names of the pieces. For triangles, make sure everyone agrees to the meanings of "hypotenuse", "altitude", and "base".
This brings out the need for a common vocabulary, and the need to be able to express one's thoughts with precision. If, then, someone asks you to move a green triangle adjacent to the large, orange square, you will do so, but again, the result is not what the person intended for you to do. You then might ask, "Do you mean that you want me to move a green triangle so that its hypotenuse is in continuous contact with an entire side of the large orange square?" If they express agreement, then do it. Then see if others can express their thoughts with precision by telling you what to do next. But once the point is made, do not belabor it; go on to the next step.
See if they can direct you to the point where you have placed each of the four identical triangles so that each has its hypotenuse congruent with one of the four sides of the largest square, thus forming a single, solid square. Once this is done, solving the puzzle will proceed rapidly. But if more than 10  13 minutes have passed (aside from the digressions into the need for vocabulary, etc.), then pass out a sandwich bag of puzzle pieces to each pair. Tell them they have just 5 minutes to solve and win a $100 Grand prize.
Then challenge each pair to complete the puzzle to form a solid rectangle using their pieces. NOTE: If no pair succeeds within 5 minutes, then give a hint: Using the pieces on the overhead, show them how to form the single, solid square mentioned in the previous paragraph. Then let them take it from there with their paper puzzles. (See Sketch 1.)
The first pair to complete their puzzle should come up and show the rest of us how, using the plastic pieces already on the overhead projector. (After appropriate applause, etc. award them each a $100 Grand candy bar, which you have kept out of sight.)
Then say: Thanks! Now all pairs complete your puzzles!
Everyone complete? OK!
Can you arrange your puzzle to form two squares of equal area, using all the pieces? (There is an alternate solution to the first part where there is an extra rectangle, in which case you omit the phrase 'using all of the pieces'  but, until they ask, do not tell them that they do not need to use one of the rectangles. See Sketch 2.
Please do so now! (This will happen quickly for the pieces from Sketch 1.) Then  ask a pair to show their solution using the puzzle pieces already on the overhead projector.
For any right triangle, the square of the hypotenuse is equal to the sum of the squares of the two sides. In other words:
If C is the length of the hypotenuse, and A is the length of its altitude and B is the length of its base, then C^{2} = A^{2} + B^{2}.
With the two equal squares projected on the overhead for all to see, show that they have equal areas by laying them on top of each other. Make sure that the squares lie on the blank transparency. Now place them along side each other, and using a felt marker pen, draw an equal sign between them.
Next, remove two of the four triangles from one square, and one the rectangles from the other. Show that the two triangles and the one rectangle have equal areas (superposition is one easy way). Since we have subtracted an equal amount of area from each of the originally equal squares, the remaining areas must be equal on the left and right sides of the equal sign.
Again, remove two more triangles from one side of the equal sign, and a rectangle from the other side. Again, the remaining area on the left side must equal the area remaining on the right side.
But on one side there will be a small square with the length of its side equal to the base of a triangle, and a midsize square with the length of its side equal to the altitude of the same triangle. On the other side will be a single largest square with the length of its side equal to the hypotenuse of the same triangle. This is easily seen by placing the three squares on the three appropriate sides of any one of the triangles.
Performance Assessment:
OK  there are four steps:
References:
Write a letter [c/o BCPS Department: IIT], telephone me
at IIT [13125673384], or
email me. NOTE:
the puzzle
should be scaled so that the diagonal square is 3 inches on a side.
Sketch 1
In order to draw the puzzle on your own, use 2 sheets of 8.5 x 11 paper, a pencil, a ruler, a straight edge and a scissors.
Draw a square three inches on a side. (This is easily done by starting at one corner of one of the papers and measuring 3 inches down each edge.)Cut out the square.
Place the square so that its edges lie along the bottom right corner edges of the second sheet of paper. Now raise and tilt the square so that its right bottom corner has moved up the right edge of the page by about 1.5 inches; its left bottom corner should lie at the bottom edge of the page, about 2.5 inches from the right bottom corner of the page. The now tilted bottom of the square will be the hypotenuse of a right triangle, and the right bottom edges of the page will be the altitude and base of the triangle. Use some tape to hold the square in place on the sheet.
Next, draw a horizontal line across the page so that it passes through the topmost corner of the tilted square. Then draw a vertical line so that it passes through the leftmost corner of the tilted square. The square will now be circumscribed within a larger square formed by the horizontal and vertical lines drawn on the sheet. This also leaves the original tilted square surrounded by four identical triangles; the hypotenuses of the triangles are the four sides of the tilted square.
For the upperleft triangle, draw a square using one of the triangles sides as one side of the square. Draw another square using the other side of the triangle.
Now draw vertical lines through the vertical sides of the smallest square (on the left of the triangle).
Then draw horizontal lines through the horizontal sides of the midsize square (on the top of the triangle).
You should now have formed three identical rectangles with long sides vertical (and equal in length to the altitude of the triangles), and short sides horizontal (and equal in length to the base of the triangles).
Your puzzle is now complete. Cut it out and play with it. Enjoy!
Sketch 2
Who was Pythagoras?