Section 9.3

Today we learned about the Altitude-on-Hypotenuse theorems. In the first diagram of the picture above, you see a proof. This is proving the statement: if you draw a right triangle, and you draw altitude to hypotenuse, you create three similar right triangles.

You may now use this statement in your future proofs! YAY! :D
http://www.youtube.com/watch?v=KzBT8130TqU <-- this video shows you the different ways you can create these right triangles, changing the size's of the triangle. :)

http://www.youtube.com/watch?v=tuAjSiuG8j0
Watch it. Wait until after the advertisement. It's worth it...

OKAY. About the next diagram. This shows a right triangle with the altitude, the triangles next to it are the similar triangles formed by the altitude. There are many rules that ALWAYS apply to triangles under the "pink" rule (a.k.a the rule above in pink) These rules that are always true are in ORANGE.

You may notice there are also side lengths, found in the ratios, that are highlighted in blue. This was done NOT ONLY to *try* and make the diagram as cool as Mr. Wilhelm's, but for a learning stand point.


Highlighted in blue are the geometric means. This is relevant to Theorem 68- which, in short, says; The altitude to the hypotenuse is the mean proportional between segments of the hypotenuse.

The theorem also states; Either leg of the given right triangle is the mean proportional between the hypotenuse of the given right triangle and the segment of the hypotenuse adjacent to that leg.


(PS CLICK ON THE PICTURE, IT MAKES IT EASIER TO SEE!)

Hope you enjoy this, I tried. Failed, but ... "It is hard to fail, but it is worse never to have tried to succeed" Words of wisdom. :)

-Maggie Ridenour

QUIZ CANCELED!!

Sorry for the late notice, everybody.  I'm canceling the quiz.  We really need to have a test on Friday, so there's no room in the schedule to postpone it.  I hope you're not too disappointed.

Please spread the word.

-Mr. Wilhelm

p.s. -- The assignment from 8.5 is still due tomorrow.

yet another random blogviewer!

-Game (Tyler)

8.5: Three Theorems Involving Proportions

Hi everyone! Today in class, we learned three new theorems.

First is the side-splitter thereom:

            side-splitter: a line in a triangle that is parallel to one side and intersects the two other sides

                                 

            Proportion: 

Next is the theorem with parallel lines:

If three or more parallel lines are intersected by two transversals, the parallel lines divide the transversals proportionally.

Proportion:

 The last thereom is the angle bisector thereom:

If a ray bisects an angle of a triangle, it divides the opposite side into segments that are proportional to the adjacent sides.

Other:

corollary-  a thereom that is a direct result of a previous one

That's all! Hope you liked the blog!

~Natalie

Sections 8.1 and 8.2

Hey guys!  Sorry this didn't go up yesterday, hope you didn't have to much trouble on the homework...

Also, heres a funny video about ratios if you watch it all the way through. Ratios: Bad Date

After your done reading,  Click here for a new video from the makers of one dozen monkeys!

8.1: Ratios and Proportions

First of all, let's start with some basic definitions that boil these concepts down.

Ratio: Comparison of two numbers (Using fraction or colon).

Example: 5:6 or 5/6

Proportion: Setting two ratios equal to each other.

Example:
  

<----That form will be the basis for all the properties discussed below.

Oh, and also here are two necessary definitions.

		Means: Second and third terms (b and c)
		Extremes: First and fourth terms (a and d)

  

Now that we have the basics out of the way, let's talk about some of the properties of proportions.

               
 Theorem 59: In a proportion, the product of the means is equal to the products of the extremes.

   ad=bc

This theorem uses basic cross-multiplying to simplify proportions.

Other theorems summed up to the following simple equations:


  






And also, an unexpected, yet true statement,





One last portion of this chapter included a new concept: arithmetic and geometric means.

The arithmetic mean of two numbers is found using the following equation:




While on the other hand, the geometric mean of two numbers is found using this equation:





That concludes Section 8.1.


8.2: Introduction to Similarity

In simple terms, similar figures are figures that proportionally have the same shape, but are not necessarily the same size.

In order for two polygons to be considered "similar", the must have one of two basic qualities:

-All corresponding angles congruent
-All corresponding side lengths are proportional













Some other terms that were also mentioned in the book were:

Dilation: An enlarged figure that is similar to the original figure.

Reduction: A figure reduced in size that is similar to the original figure.

This problem shows an example of a common problem which ties in similar figures, ratios, and proportions.

Another theorem that was also briefly mentioned in the book is as follows:

Theorem 61: The ratio of the perimeters of two similar polygons equals the ratio of any pair of corresponding sides.

This means that if, for example, the perimeter of an original square is 20 units, and the perimeter of the dilation of the square is 40 units, than the ratio of the perimeters as well as the ratios of corresponding sides should be equal.


That is about it for today's information! Thanks, and hoped it helped!



Juliaaaaaaa

Sections 8.3 and 8.4

So today we learned how to prove that triangles are similar.
The 3 ways are:
AA~
SSS~
SAS~

When proving triangles congruent we had also learned ASA, AAS, and HL. But these are uneccessary for proving triangles similar because in each of these you already have some form of the first 3.



With the 2 triangles to the left we cannot prove them similar, because we do not have AA~, SSS~, or SAS~.



With these two triangles we can prove them similar using AA~.

The same principles apply to SSA~ and SSS~. Although you need to always remember that in order to prove similar the angles you use must be congruent, but the sides must be proportional.

I won the second game. Curse you blog post.
Big Money!.... that was for you Joey
Ah isn't math fun?
-Andrew

Sections 7.3 and 7.4

Hey!

So today we did section 7.3 and 7.4, which are all about formulas involving polygons.
Lets start with 7.3

So Each polygon has a unique name (until you get up really high)
Sides:                    Polygon                                                         
3                           Triangle      
4                           Quadrilateral   
5                           Pentagon    
6                           Hexagon   
7                           Heptagon
8                           Octagon     
9                           Nonagon          
10                         Decagon                       
12                         Dodecagon     
15                         Pentadecagon     
n                           n-gon

Each polygon has a certain number of diagonals. These are formed when you connect different vertexes to eachother (except ones that form the outer edge of the shape)

 

  

The pink lines are the diagonals of this hexagon

The number of diagonals in a polygon can be found using

                      n(n-3)

# of diagonals=     2 

(where n is the number of sides in the figure.)

You can also find the measure of all of the angles in a polygon using this formula:

 sum of the angles= (n-2)180

This works because you can divide the shape into triangles with each triangle equaling 180 degrees. 

See the triangles?

<---------------

The sum of the exterior angles of any polygon will ALWAYS be 360 degrees.  

This is because if you start at one angle and slowly rotate around to the next angle, and the next and the next, you will have made a whole circle when you get back to the first, or 360 degrees!

We also learned a formula that isn't in the book, which is to find the measure of one angle of a shape with a certain number of sides.

                                                                180n-360

measure of one angle in an n sided polygon=       n

Now for section 7.4

Section 7.4 focused on regular polygons. Here are examples of a few:

  

 Regular Polygon- a polygon that is both equilateral and equiangular

They also showed some equiangular polygons, which are like regular ones but they dont necessarily have all sides congruent. (like a rectangle.)

A formula to find the measure of any exterior angle of an equiangular polygon:

                                             360

measure of the exterior angle=  n

And that was really all that we learned!

Random Math things:

NONAGON SONG!  (click on it!)

ONE DOZEN MONKEYS 

 

 Hope you liked the blog!!!!!!!!!!! 

:) Katie

Chapter 7: 7.1 and 7.2

12.1.10

• 7.1 Triangle Applications Theorems
• 7.2 Two Proof-Oriented Triangle Theorems

7.1

3 new theorems

1. The sum of the measures of the 3 angles of a triangle is 180⁰

 http://www.wyzant.com/Help/Math/Geometry/Triangles/Angles_of_Triangles.aspx

This can be proven using the parallel postulate, saying that line M is parallel to NS, and measure 180⁰, being a straight line and all.  Like Mr. Wilhelm demonstrated for us in class, when the three angles are lined up along that straight line they complete the 180⁰.

http://www.tutorvista.com/topic/parallel-lines-triangle-angle-sum-theorem

2.      The measure of any exterior angle is equal to the sum of the measures of remote interior angles.

http://schoolwires.henry.k12.ga.us/4376205615266500/blank/browse.asp?A=383&BMDRN=2000&BCOB=0&C=66757

An exterior angle is formed by extending on of the sides to form an angle that is exterior to the polygon and is supplementary to its corresponding interior angle.  It is, by definition, adjacent and supplementary to an interior angle of the polygon. 

Since all the angles of a triangle add up to 180⁰, then any interior angle is supplementary to the two other angles combined.  And if the exterior angle is supplementary to the interior angle, then by a loose use of the transitive property the exterior angle must be equal to the two remote interior angles. 

3.      The segment joining the midpoints of two sides of a triangle is parallel to the third side and its length is one-half the length of that third side. 

http://www.tutorvista.com/topic/proof-of-midpoint-formula

7.2

2 new theorems

1.      If two angles of a triangle are congruent to two angles of a second triangle, then the third angles are congruent. (No-Choice Theorem)

Proof: since the sum of all the angels must equal 180, the sums must be set equal to each other.  If we then apply the Subtraction Property, we see that the last two angles must be equal, and therefore congruent

DISCLAIMER!!!: the two triangles need not be congruent for us to apply the No- Choice Theorem.

Also, can be written ass AA→ AAA

2.      Proving AAS.  If two angles and one side, not between those two angles, are all congruent between triangles, then the triangles are congruent. 

This can be proven using the No- Choice Theorem to say first that each pair of angles is congruent, then that the third angle is congruent, then that the sides are congruent.  Technically its AAA or even ASA, but it’s one less step to say AAS.

http://hotmath.com/hotmath_help/topics/AAS-postulate.html

And thats what we learned today! homework was:

7.1: 9,12,15-19

7.2: 18

blessings, Em J

6.1-6.3: Planes. 11-30-10, 25 days till Christmas excluding today

Once I finished my biology notes, I moved on to this task.

Here is what we learned...

1) not all planes fly

a) a geometric plane is a 2D surface that extends infinitely in all directions. It is defined by any 3 or more points where at least 1 is non-colinear.

b) When planes intersect, the intersection is a line. If a line intersects a plane not containing it, the inter section is exactly one point.

2) Any two lines prove a plain. Parallel, intersecting... BUT WAIT! NOT SKEW! You could probably come up with over 9000 planes to two lines as long as they are not skew.

Let this remind you that skew does not work.

a) When a line intersects a plane, the point of intersection is a "foot." For a line to be perpendicular to a plane, it must be perpendicular to every line passing through the foot.
3) A plane is parallel to a line if they never intersect. A plane is parallel to another plane if they do not intersect, as well.

(They aren't touching)
a) in this picture, with closer review, you'll notice that the parallel planes are being cut by a transversal plane. WHAT DOES THIS MEAN? Not a double rainbow, but all of the lines of intersection are parallel. Makes enough sense, right?

I hope you find this new blog much more helpful.
-Shane McPartlin
Remember: everyday is only as good as you make it.

Chapter 3 Review

 I had to do this blog on a chapter 3 review.
Chapter 3 was all about congruent figures. 

We learned how to prove triangles congruent by three postulates. Those were the SSS, SAS, and ASA postulate. 

SSS-
Given :  AB ≅ DE
BC ≅ EF
AC≅DF

Prove: ΔABC ≅ ΔDEF  

 Sense we know that if 3 corresponding sides are congruent we know that the triangles are congruent by SSS.











SAS- Given :  AB ≅ DE
BC ≅ EF
B ≅ E
Prove: ΔABC ≅ ΔDEF

 We can prove two triangles congruent if we have one angle in between two congruent sides. So by SAS we can prove these two triangles congruent. 










ASA-Given :  AB ≅ DE
B ≅ E
A ≅ D
Prove: ΔABC ≅ ΔDEF

 We can prove two triangles congruent if we have one congruent side in between two congruent angles.  So by ASA we can prove triangles congruent.











My definitions are in simple terms.  The exact definitions for each postulate is as follows. 
SSS postulate: If there exists a correspondence between the vertices of two triangles such that three sides of one triangle are congruent to the corresponding sides of the other triangle, the two triangles are congruent.
SAS postulate: If there exists a correspondence between the vertices of two triangles such that two sides and the included angle of one triangle are congruent to the corresponding parts of the other triangle, the two triangles are congruent. 
ASA : If there exists a correspondence between the vertices of two triangles such that two angles and the included side of one triangle are congruent to the corresponding parts of the other triangle, the two triangles are congruent. 

Next we learned about circles.
Theorem 19: all radii of a circle are congruent. This is helpful in proving triangles congruent. 
We also learned about CPCTC. Which means that the corresponding parts of congruent triangles are congruent.  Which means that if two triangles are congruent that every corresponding side and angle are congruent. 
Here is a proof that involves both CPCTC and radii.  (O MEANS CENTER OF CIRCLE)

Given: OP
Prove: AB ≅ CD

 In this we know that there are 4 radii, and that they are all congruent.  We also know that                      CPD ≅APB. So by SAS we can prove the triangles congruent. And sense we learned CPCTC we know that sense the two triangles are congruent, that every corresponding side and angle is congruent too.  So we can prove that AB≅CD.






Next we learned about medians and altitudes.  
Altitude: of a triangle is a line segment drawn from any vertex of the triangle to the opposite side.  The altitude is perpendicular to one of the sides.  Every triangle has three altitudes.  Also if used in a proof: altitude -implies- perpendicular-implies- right angle.
Example:










Medians: of a triangle is a line segment drawn from any vertex of the triangle to the midpoint of the opposite side. Every triangle has three medians.  Also if used in a proof: median -implies- segment divided into two congruent segments.
Example:

In isosceles triangles the altitude is the median, and vice versa.  

Next we learned about auxiliary lines. An auxiliary line is a line segment drawn that do not appear in the original figure.  Such that they connect two points already in the diagram, that help to prove what needs to be proven. 

Auxiliary lines can help prove proofs.  Draw them when needed.










Proof examples:

Given: CD and BE are altitudes of ΔABC
AD≅AE
Prove: DB ≅ EC

In this proof we use the postulate of ASA to prove triangles congruent.  Then we use CPCTC to prove segments congruent.  But then we go BEYOND CPCTC to prove more.  This proof involved altitudes, ASA, CPCTC.








Given: CFD ≅ EFD
FD is an altitude
Prove: FD is a median

In this proof we have to prove that an altitude is a median (which means it's isosceles).  We prove the triangles congruent, then use CPCTC to prove it is a median. 









Next we learned about overlapping triangles.  They are still proven the same way, but sometimes it's hard to recognize where the triangles are.  You may want to redraw them, or make the lines more apparent. 

 This is an example of an overlapping triangle. 
ΔCBF and ΔFDE are overlapping.  I'm not going to prove this one because you solve it the same basic way. By using SSS, SAS, or ASA. 







In the next section we just learned about types of triangles.
Scalene triangle- a triangle i which no two sides are congruent.










Isosceles triangle-a triangle in which at least two sides are congruent.










Equilateral triangle-a triangle in which all sides are congruent. (all equilateral triangles are equiangular)













Equiangular triangle-a triangle in which all angles are congruent. (all equiangular triangles are equilateral)







Acute triangle-a triangle in which all angles are acute.(angles<90)








Right triangle-a triangle in which one of the angles is a right angle.












Obtuse triangle-a triangle in which one of the angles is an obtuse angle.(90<obtuse<180)

We also learned about Angle-Side theorems. 

-If two sides of a triangle are not congruent, then the angles opposite them are not congruent, and the larger angle is the opposite the longer side.
-If two sides of a triangle are not congruent, then the angles opposite them are not congruent, and the longer side is opposite the larger angle. 
Also shorter side opposite shorter angle, and shorter angle opposite shorter side.
SOOO basically this-

There was also another theorem about angles that we learned before this section that was-if two sides of a triangle are congruent, the angles opposite the sides are congruent. (congruent angles -> opposite sides congruent too)












Proof example:
Given: AB ≅AC  
Prove: 1≅2

HL postulate- if there exists a correspondence between the verticies of two right triangles such that the hypotenuse and a leg of one triangle are congruent to he corresponding parts of the other triangle, the two right triangles are congruent. 

so even thou it does not fall into SSS, SAS, or ASA, these triangles are congruent by HL.  There is only one line that fits the distance created by the hypotenuse and one of the legs if it is a right triangle. 







Proof:
Given: GJ is the altitude of HK
HG≅KG
Prove: ΔHGJ≅ΔKGJ

And we are finally done....YAY!!!!!!!!! (sorry the pictures are soo small)


-Jennifer Kendall