Periods 2 and 3
Using what we know about right triangles, we drew right triangles on dot paper. We were given the lengths of the legs. We were asked to draw these as the sides of the right angle and then join them to make a right triangle. This final, "slanty" side is the hypotenuse. We then measured the length of the hypotenuse to the nearest tenth of a centimeter. We recorded the results.
Following that we found the area of squares that would have the leg measures that were given. For instance, if the leg measure was 2, the area of a square with side lengths of 2 would be four. If the other leg was 3, the area of that square would be 9. We then added those two numbers together. In this case, it would be 4 + 9 which equals 13. We then took the square root of 13. It is about 3.6. We compared this answer to the measurement we got earlier. In most cases the measurements were the same.
This is really what the Pythagorean Theorem tells us that C²(the hypotenuse squared) = a²(leg 1 squared) + b² (leg 2 squared). By finding the square root of c² we find the hypotenuse of the triangle.
We will continue this tomorrow.
Assignment: Pythagorean Theorem worksheet, #1 -4
We built on what we learned yesterday about the degree of a polynomial and adding and subtracting like terms to begin multiplying polynomials. The method we discussed today is using a table. In this case we set up a table that is like an area model. Above each horizontal section is one of the terms from one of the polynomials we are multiplying. Along each vertical section is one of the terms from the other polynomial. The products of the terms are written in the cells of the table. From there we combine like terms and find the final solution.
Tomorrow we will begin to discuss FOIL.
Assignment: worksheet 9.2A, #1 -13.
Today we worked solving quadratic equations using square roots. Through the use of an activity we were able to see graphs of three different kinds of equations. Each of these represent different numbers of possible solutions. If you have a problem like x² = 0, there is one solution; it is zero.
For problems like x² = 4, there are two solutions; they are ±2. For ones like x² = -4, there are no solutions because you cannot find the square root of a negative number.
We generalized these ideas and will continue this tomorrow.
Assignment; none tonight.