Tangrams Part 1
First I traced the small triangle. The relationship between the angles and sides are identified.
When using the small red triangle, I formed perfect squares along each side of the red triangle using small red triangles. For each of the two legs it took 2 small red triangles to form a perfect square, and for the hypotenuse it took 4 small red triangles. If you add the triangles formed on the legs (2+2=4 red triangles), it is equal to the number of triangles formed on the hypotenuse (4 red triangles).
I then used the medium yellow triangle. I again formed perfect squares on each side of the yellow triangle using small red triangles. For each of the two legs it took 4 small red triangles, and for the hypotenuse it took 8 small red triangles. If you add the triangles formed on the legs (4+4=8 red triangles), it is equal to the trangles formed on the hypotenuse (8 red triangles).
I repeated this again using the large blue triangle. I formed perfect squares on each side of the blue triangle using small red triangles. For each of the legs it took 8 small red triangles, and for the hypotenuse it took 16 small red triangles. If you add the triangles formed on the legs (8+8=16 red triangles), it is equal to the triangles formed on the hypotenuse (16 red triangles).
- “Discuss the relationship between the areas of the squares along each leg of the right triangle to the area of the square along the hypotenuse.” (from Step 5 from “The Pythagorean Theorem with Tangrams” – linked above)
All the pictures above show the relationship between the squares of the legs and the square of the hypotenuse. The area of the square from leg ‘a’ + the area of the square from leg ‘b’ = the area of the square from the hypotenuse ‘c’. If you use the area formula A = l × w, the area for the square with side length ‘a’ is a × a, or a². The area for the square with side length ‘b’ is b × b, or b². The area for the square with the side length ‘c’ (hypotenuse) is c × c, or c². So they would discover the formula a² + b² = c², which is Pythagorean Theorem.
- What connection can students make between the numbers of shapes needed to create the sides of the triangle?
The students can experiment with the various shapes and make a connection between them. They will make the connection that 1 medium yellow triangle is the same as 2 small red triangles. Or 1 large blue triangle is the same as 2 medium yellow triangles (or 4 small red triangles). Experimenting with this, they can create other models that would represent the same relationship and compare it to the models above working with only small red triangles.
- In your own words explain why this activity is a good introduction to square roots and rational numbers.
This would be a good activity to lead into square roots and rational numbers. After going over finding squares, the natural step is to explain square roots. Especially when finding a missing ‘leg’ variable for the Pythagorean Theorem, they will get the connection to finding the square root and how value is associated to a rational number. It would be a good time to review the basics of rational numbers.
- How would you present this activity to your students?
I think this is a wonderful activity to do with students. So often, the Pythagorean Theorem is taught and the students memorize the forumla but they don’t really know what the fomula means or how it was found. I would first do this activity like the steps above and have them compare their findings with a partner. (Did the partner come up with the solution? Was the same, or different when comparing with a partner?) After the developing the squares with the small triangle, have them predict what would happen with the medium and the large triange before continuing. (Were the predictions true?) I would have them experiment and challenge them to use different shapes to form the squares. (What can we conclude about the different shapes in relation to the small red squares?) I would also have them journal their findings.