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Session 5: Solutions
 
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Solutions for Session 5 Homework

See solutions for Problems: H1 | H2 | H3 | H4 | H5


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.

<< back to Problem H1


 

Problem H2

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.

<< back to Problem H2


 

Problem H3

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.

<< back to Problem H3


 

Problem H4

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.

<< back to Problem H4


 

Problem H5

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.

<< back to Problem H5


 

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