Solving simultaneous equations

Here is a link to practice problems for simultaneous equations. The coin problems all use elimination instead of substitution. Here are the same coin problems using substitution, with the answers in the comment below.

a) 100 coins, all quarters and pennies, total = $13.72

b) 100 tickets, all children ($6) and adult ($12), total =$912

c) 100 coins, all quarters and nickels, total = $8.20

d) 100 coins, all dimes and pennies, total = $3.79

e) 100 coins, all dimes and nickels, total = $7.85

f) 100 coins, all quarters and dimes, total = $14.65

Two points on the plane define a slope (or a slope is undefined)

Here are previous posts dealing with the definition of slope and practice problems for finding the definition of a line that passes through two points (x1, y1) and (x2, y2).

Tiliing the plane

Here is a link to a previous blog I wrote and pictures of some of the regular tilings, either single shape or mix and match.

And here is some fun I had with the Penrose tiles back when I bought them in 2011.

Notes for the week of October 12 through 15

Notes for area of a triangle defines by three points and the distance formula.

Notes for interior angles sums of polygons and the measure of regular angles.

More practice on the interior angles sums and regular angle measurements.

Classifying triangles defined by three vertices in the plane.

In class, we have worked with three points in the plane where one is the origin (0, 0) and the other two are labeled (a, b) and (c, d).  In the notes here on the blog, the points are given coordinates with subscripts such as (x1, y1) and (x2, y2). The formulas are the same, only the letters used in the formulas are different.

The area for the triangle define by (0, 0), (a, b) and (c, d) is given by the formula


Area = ½| ad - bc |

This formula is relatively simple compared to Heron's formula, with no square roots necessary. On the other hand, we will need to do three square roots to get the lengths of the three line segments for both classification methods.  Let's look at an example.

Example # 1: (0, 0), (6, 3), (7, 1)

Area: ½| (6)(1) - (3)(7)| = ½| 6 - 21| = ½| - 15 | = ½(15) = 7.5

Distances:

Between (0, 0) and (6, 3): When (0, 0) is involved the subtractions are easy and we get

Distance = sqrt((0-6)² + (0-3)²) = sqrt(6² + 3²) = sqrt(36 + 9) = sqrt(45)

Usually, I'd ask to have this simplified, but this problem just leave the answer as the square root of 45, it will make the next steps easier.

Between (0, 0) and (7, 1) 

Distance = sqrt((0-7)² + (0-1)²) = sqrt(7² + 1²) = sqrt(49 + 1) = sqrt(50)

Between (6, 3) and (7, 1)   

Distance = sqrt((6-7)² + (3-1)²) = sqrt((-1)² + 2²) = sqrt(1 + 4) = sqrt(5)

Now we have the distances, the classifications can be done.

Equilateral, isosceles or scalene?  The numbers are all different sqrt(45), sqrt(50) and sqrt(5), so the triangle is scalene.

Obtuse, right or acute? Here we need to square the distances, which is easy because they are all written as square roots, so the squares are 45, 50 and 5.  We need to compare the sum of the two short sides squared versus the long side squared and in this case

45 + 5 = 50.

This is a right triangle.

Final answer: Area of 7.5 square units, a scalene right triangle.

Example #2: (3, 2), (4, 1), (1, 7)

Here, we don't have a point at the origin, so we will subtract a value of one given vertex from all of the vertices.  That will give us three new points, but the triangle defined by our new points will have the same area and classifications.  I will choose to subtract (3, 2), but using (4, 1) or (1, 7) would also work.

New vertex #1: (3, 2) - (3, 2) = (0, 0)
New vertex #2: (4, 1) - (3, 2) = (1, -1)
New vertex #4: (1, 7) - (3, 2) = (-2, 5)

Area = ½| (1)(5) - (-1)(-2)| = ½| 5 - 2| = ½| 3 | = ½(3) = 1.5


Between (0, 0) and (1, -1):

Distance = sqrt((0-1)² + (0-(-1))²) = sqrt(1² + 1²) = sqrt(1 + 1) = sqrt(2)

Usually, I'd ask to have this simplified, but this problem just leave the answer as the square root of 45, it will make the next steps easier.

Between (0, 0) and (-2, 5) 

Distance = sqrt((0-(-2))² + (0-5)²) = sqrt(2² + 5²) = sqrt(4 + 25) = sqrt(29)

Between (1, -1) and (-2, 5)   

Distance = sqrt((1-(-2))² + ((-1)-5)²) = sqrt(3² + (-6)²) = sqrt(9 + 36) = sqrt(45)

Now we have the distances, the classifications can be done.

Equilateral, isosceles or scalene?  The numbers are all different sqrt(2), sqrt(29) and sqrt(45), so the triangle is scalene.

Obtuse, right or acute? Here we need to square the distances, which is easy because they are all written as square roots, so the squares are 2, 29 and 45.  We need to compare the sum of the two short sides squared versus the long side squared and in this case

2 + 29 = 31 < 45.

This is an obtuse triangle.


Final answer: Area of 1.5 square units, a scalene obtuse triangle.

Notes for the week of October 5 through 8

Classification of triangles, including the triangle inequality and Heron's formula, which will give us the area of a triangle defined by side lengths, here using the letters u, v and w.

Finding a line segment of the length of the square root of any number.

The easiest thing is to find the square root of any given odd number. To find the square root of any given even number, find the square root of the odd number one less than it. For example, if I want to find sqrt(44), it will be a two step process, where first we find the sqrt(43).



Step 1: 43/2 = 21.5, so the whole numbers closest to splitting 43 in half are 22 and 21. It's easy to prove that 22² - 21² = 43, so we make a triangle with the hypotenuse = 22 and the long leg = 21.  The short leg will be sqrt(43).

Step 2: Now we make a new right triangle by extending the leg of length 21 by 1. Now we have a new right triangle with legs of length sqrt(43) and 1, and the hypotenuse will be sqrt(44).

Notes for the October 1 class

The new topic covered on Thursday night was the Pythagorean Theorem. Here is a link to the posts on the topic.