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You can make your own tangram set from construction paper. Start with a large square of construction paper and follow the directions below:

**Step 1:** Fold the square in half along the diagonal; unfold and cut along the crease. What observations can you make about the two pieces you have? How could you prove that your observations are correct?

**Step 2:** Take one of the triangles you have, and find the midpoint of the longest side. Connect this to the opposite vertex and cut along this segment.

**Step 3:** Take the remaining half and lightly crease to find the midpoint of the longest side. Fold so that the vertex of the right angle touches that midpoint, and cut along the crease. Continue to make observations.

**Step 4:** Take the trapezoid and find the midpoints of each of the parallel sides. Fold to connect these midpoints, and cut along the fold.

**Step 5:** Fold the acute base angle of one of the trapezoids to the adjacent right base angle and cut on the crease. What shapes are formed? How do these pieces relate to the other pieces?

**Step 6:** Fold the right base of the other trapezoid to the opposite obtuse angle. Cut on the crease. You should now have seven tangram pieces. Are there any other observations you can make?

Try this with a very long and skinny parallelogram.

When people work on the dissection problems, they often create a figure that looks like a rectangle, but they can’t explain why the process works. Or sometimes they perform a cutting process that they think should work, but the result doesn’t look quite right. Reasoning about the geometry of the process allows you to be sure. A cutting process is outlined below. Your job is to analyze if the cuts really work. Does this algorithm turn any parallelogram into a rectangle? If so, provide the justifications. If not, explain what goes wrong.

First, cut out the parallelogram. Then fold along both diagonals. Cut along the folds, creating four triangles as shown.

Slide the bottom triangle (number 4) straight up, aligning its bottom edge with the top edge of triangle 2. Slide the left triangle (number 1) to the right, aligning its left edge with the right edge of triangle 3.

Here are some new cutting problems. When you find a process that you think works, justify the steps to be sure.

Form two angles that are smaller than the smallest one in the original triangle, and a third angle that is larger than the largest angle in the original.

Start with a scalene triangle. Find a way to dissect it into pieces that you can rearrange to form a new triangle, but with three different angles. That is, no angles of the new triangle have the same measure as any angle of the original triangle. You should be able to demonstrate the following:

a. |
The final figure is a triangle. (It has exactly three sides.) |

b. |
The two joined edges match. |

c. |
All three angles are different from those of the original triangle. |

d. |
As a challenge, can you solve this with just one cut? |

Start with a scalene triangle. Find a way to dissect it into pieces that you can rearrange to form a new triangle, but with three different sides. That is, no sides of the new triangle have the same length as any side of the original triangle. You should be able to demonstrate the following:

a. |
The final figure is a triangle. (It has exactly three sides.) |

b. |
The two joined edges match. |

c. |
All three sides are different from those of the original triangle. |

Remember from Session 3 that you can divide any quadrilateral into two triangles.

Start with an arbitrary quadrilateral. Find a way to dissect it into pieces that you can rearrange to form a rectangle. Test your method on quadrilaterals like these. Try to justify why it will always work.

Problems H3-H5 adapted from *Connected Geometry,* developed by Educational Development Center, Inc. pp.169, 171. © 2000 Glencoe/McGraw-Hill. Used with permission. www.glencoe.com/sec/math

**Senechal, Marjorie (1990). Shape. In On the Shoulders of Giants: New Approaches to Numeracy. Edited by Lynn Arthur Steen (pp. 148-160). Washington, D.C.: National Academy Press.**

Reproduced with permission from the publisher. Copyright © 1990 by National Academy Press. All rights reserved.

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**Problem H1**

Follow the instructions.

**Step 1:** The two shapes are congruent isosceles right triangles since all three pairs of corresponding sides are congruent, and two of the square’s right angles are preserved, one in each triangle.

**Step 2:** Again, we get two congruent isosceles right triangles since the two legs of each are half the length of the original diagonal of the square, and since the angle between the two legs is the right angle (SAS congruence).

**Step 3:** We get an isosceles right triangle and a trapezoid. The base of the triangle is half the length of the longer base of the trapezoid (midline theorem), and is of same length as the shorter base of the trapezoid.

**Step 4:** The two shapes are congruent right trapezoids, each one of which has two right angles, one angle of 45° and one of 135°.

**Step 5:** We get a square and an isosceles right triangle. These shapes are similar to those encountered in previous steps; i.e., their corresponding sides are proportional.

**Step 6:** The final two shapes are a parallelogram and an isosceles right triangle. The triangle is congruent to the one created in Step 5. The parallelogram has one pair of sides congruent to the leg of the triangle created, and one pair of sides congruent to the hypotenuse of the triangle just created.

The construction does not work in general. To see this, consider triangle number 4: In the final arrangement, one of its angles is also an angle of the final shape, which is supposed to be a rectangle. The angle in question is formed by the intersecting diagonals, so the construction only works if the diagonals of the original parallelogram intersect at an angle of 90°. This happens when the original parallelogram is a rhombus.

Start with a scalene triangle ABC. Find the midpoint D of the side opposite the smallest angle.

Cut along the segment DB, then rotate the triangle DCB about the vertex D by 180° (see picture). The final figure has three sides. Sides AD and DC match since they are of equal length, and points B, D, and B’ are collinear since B’D is obtained by rotating DB by 180°. The angle at A is larger than any of the angles in the original triangle (since it is the sum of the two largest angles of the original triangle). The angles at B and B’ are smaller than (hence different from) any of the angles in the original triangle since they are both smaller than the smallest angle in the original triangle. We only used one cut.

Let’s say AB is the longest side of the original triangle.

Draw the altitude CD (from vertex C to the side AB). Notice that it is shorter than any of the three sides of the original triangle.

Make the triangle into a rectangle using one of the methods we found in the session. Use side AB for the base. The other side will have length 1/2 CD (half the altitude).

Now, cut along a diagonal of the rectangle. (Note that the diagonal is longer than side AB, and so it is longer than all three sides of the original triangle, since AB was longest.) This creates two small right triangles; call them FAB and BEF (they share two vertices; the right angles are at A and E).

Form an isosceles triangle by flipping and translating triangle BEF so that angles A and E (the right angles) are adjacent and sides FA and EB align, thus forming a new shortest side of the new triangle.

The shortest side of this triangle is twice the smaller side of our rectangle. That means it’s the same as the altitude of the original triangle, and hence smaller than any of the original sides. The other two sides are the same, and since they are diagonals of the rectangle, they are longer than any of the original three sides.

Split the quadrilateral in half along an interior diagonal. You have two triangles that have a side in common. For each triangle, use the common side as the base, and turn it into a rectangle using your algorithm from Problem B4. Now stack the two rectangles on top of each other. They match up because they have a common base (the diagonal of the original quadrilateral).

**Quadrilateral 1:**

**Quadrilateral 2:**

Note that in quadrilateral 2, there is only one interior diagonal to choose from, while quadrilateral 1 has two interior diagonals.