## M4005 Geometry

### February 15, 2006

Student activities related to the Stillwell chapter 2.

Activity 1: Triangle area from Rectangles

Activity 2: Parallelogram Area

During the first 15 minutes of class, groups reviewed their work. Then representatives gave 5-minute present ions, as follows:

 Group Leader Other members. (* = joined group Wednesday) Presentation 1 Brealand Deano Shannon Harris*, Latasha Moore, Sam Pruitt, Allison Poche, Allison Ward To group 2 by Shannon Harris 2 Danielle Miller Laura Cancienne, Julie Dunn, Laurie Randall, Sarah Rihner To group 3 by Laurie Randall 3 Patrick McClain Mark Christofferson, Garrett Ford*, Daniel Hotard, Gary VanVracken To group 4 by Patrick Mcclain 4 Josh Hogan Hongyi Chen, Marcus Mann, Danica Robinson, Melba Steuber* To group 5 by Melba Steuber 5 Jen Tassin Ben Alberty, Sean Carroll, Karly Ford, Valerie Menter To group 6 by Jen Tassin 6 Kelly Minor Kathryn Gianfala, Gabe Jacobs, Paige LeBlanc To group 7 by Kelly Minor 7 Katie Sheffield Ashley Duncan, Jalayne Kling, Brandon McDonald*, Vimla Rupnarain To group 1 by Katie Sheffield

During the remainder of the class, groups worked on Activity 2.

## Homework

Hand in Wednesday, February 22: Activity 3: Cutting triangles.

On Monday, you should have read and absorbed all of Chapter 2 of Stillwell.

There will be a quiz on Monday, with one or two of the following problems:

Problem 1. Show that if A and B are the ends of the diameter of a circle and C is any point on the circle, then angle ACB is right.

Solution. Let O be the center of the circle and divide ABC into two isosceles triangles by the segment OC. The sum of all the angles in ABC is the same as the sum of the four base angles of the two isosceles triangle. Each isosceles triangle has one base angle at C, so angle ACB includes exactly half the angle measure of the whole triangle. But the angles in a triangle sum to two right angles. So ACB is right.

Problem 2. Suppose ABC is a right triangle with right angle at C. Let D be the base of the perpendicular from C to AB. Prove the Pythagorean theorem using the fact that the triangles ADC, CDB and ACB are similar.

Solution. By similarity, |AC| / |AD| = |AB| / |AC| so |AC| |AC| = |AD| |AB|. Also by similarity, |BC| / |BD| = |AB| / |BC| so |BC| |BC| = |BD| |AB|. Thus,

|AC| |AC| + |BC| |BC| = |AD| |AB| + |BD| |AB| = ( |AD| + |BD| ) |AB| = |AB| |AB|.

Thus, the square of |AC| plus the square of |BC| equals the square of |AB|.

Problem 3. Suppose ABC is a right triangle with right angle at C. Show how to cut the square on the hypotenuse AB into two rectangles, one with area equal to the square on leg AC and the other with area equal to the square on leg BC. (In this problem, I do not demand a proof.)

Solution. Let m be the line through C that is perpendicular to AB. It cuts the square on AB into two rectangles. The rectangle on the same side of m as A has area equal to the area of the square on AC and the rectangle on the same side of m as B has area equal to the area of the square on BC.