Sunday, March 13, 2011

Chapter 6 Review

So i wrote this entire blog. and then it got deleted. SO sorry for this super bad blog because my computer wont let me upload certain pictures or use color on the blog :/

Plane: extends infinitely in all directions, 2-D.

Four ways to determine a plane:
three non-collinear points,


line and non-collinear point,


2 intersecting lines,

or
two parallel lines.




Point of intersection of a line and a plane is called foot of line.
- for line to be perpendicular to plane, line must be perpendicular to every line in the plane that goes through the foot.



If line intersects plane not containing it, then intersection is exactly one point. If two planes intersect, their intersection is exactly one line.





GIVEN: m||n
S intersects m and n
conclusion: line AB|| line CD


Sorry again for the borginess :/ i was really mad it all got deleted and wouldnt upload, ugh :(
To make up: heres a picture of me on Bo in Florida :D

-Maggie

Ch. 9 final review. Go time.

Hello.
First off, some background music. (Press control N, then click on it to play it in the background)

Alright, chapter 9 was all about triangles with one exception:
Brahmagupta's formula.
This states that any cyclic quadrilateral (and quadrilateral that can be inscribed in a circle) can have it's area calculated by taking half the perimeter (s) and finding the value of the radicand (s)(s-a)(s-b(s-c)(s-d).

This is a play off of Heron's formula, in which the area of ANY triangle can be found by taking the value of
these theorems require you to know all of the side lengths. (SSS)


 Then we had that wonderful little add on of the law of sines and cosines.

Law of sines:       a       +       b      +        c       = Area
                       sinA           sinB             sinC
This requires knowing a side and it's optional angle. (SSA, AAS)

Law of cosines: this took a really long time to derive in class I remember.


­a­2= b2+c2-2bcCosA
Fig. 1 - A triangle.

Law of cosines works with SAS.

Okay, now for the easy part.


angle bisector theorem:
A picture is worth a thousand words, is it not?

With any right triangle, you can drop an altitude that goes through a vertex and all the triangles now formed are similar. You can then use geometric means to figure out sides.
Por ejemplo, if AB=6, and AD=3, we can set up a proportion using AB as the geometric mean.

AB= x 
AD  AB

So we have 3x=36
x=12.


The pictures make it easier to read.

Side splitter. Here we go.

Just proportions. The triangle formed by the sidesplitter (sounds like a sick wrestling move, doesn't it?) is similar to the original. The ratio of areas of these triangles is 1:9. So, VX=8.
 Also, these triangles are similar because they have the mutual angle and the parallel lines do the rest.

I think it safe to assume you all know about the basics of circles, the pythagorean theorem, about the multiplication and addition of radicals, and distance formula, but here you go.


Oh wait! 30-60-90 right triangles always have side ratios that a=s b=s(root)3 and c=2s.
45-45-90 right triangles have congruent a and b=s, and c=s(root)2.



Oh, wait! Trigonometry!

so, by taking the tangent, ground distance, and angle of elevation, you can find how far away the lighthouse is from this man's head! Important to all you light house philosophers.
This also works for the angle of depression, the only difference is you use the angle of depression instead of the angle of elevation.


This blog post is quite the keyboard full.



 that's a lot said with a little.


Sweet Caroline, I think that's everything.

slant height, a'ight?

Wow. This is only a small fraction of what we've learned with our beloved teacher Mr. Wilhelm educating us and challenging us to prepare us for the future. Glad to be out of my two hardest classes, but I'm going to miss Mr. Wilhelm. He has been a great inspiration to us all.











We love you Mr. Wilhelm.

I'm off.
-Shane
P.S. Don't let anyone but yourself decide who you will become.

Friday, March 11, 2011

Chapter 8: Similar Polygons

To the left, you see a glorious motion picture of a HYPERCUBE. This demonstrates the fourth dimension and is not at all related to this chapter.
When you manage to tear your eyes from it, we will continue with our review. You could also continue looking at it and flunk the final. Your choice.
.
In the eighth chapter of the textbook Geometry for Enjoyment and Challenge, the author(s) demonstrated the use of similar polygons in geometry.
RATIO:
a quotient of two numbers a:b or a/b

PROPORTION:
equation stating that ratios are equal

GEOMETRIC MEAN:
a proportion in which the means are equivalent a / x=x / d
.
ARITHMETIC MEAN: the average of two numbers x=(a+d) / 2 I don't think we used this outside of the first section.
.
.
SIMILAR FIGURES: same shape but not necessarily the same size.
These are turtles. I've always wanted a turtle. I got an cat instead. He's fat.
.
In SIMILAR POLYGONS:The corresponding sides are proportional
The ratio of the perimeters are equal
Corresponding angles are congruent

△ABC∼△XYZ
c/a=z/x c/b=z/y a/b=x/y etc.
.
You can proove that triangles are similar if:
2 angles are congruent AA∼ or AA∼
3 sides' ratios are equal SSS∼
2 sides' ratios and the included angles are congruent SAS∼
.
CSSTP: Corresponding Sides of Similar Triangles are Proportional
CASTC: Corresponding Angles of Similar Triangles are Congruent
I don't know if you can use that abbreviation but there you go.
.
.
SIDE SPLITTER THEOREM
A line divides two sides of a triangle prop
ortionally if it is parallel to the the third.

If parallel lines are intersected by two transversals, the parallel lines divide them proportionally.


ANGLE BISECTOR THEOREM
If a ray bisects a
 triangle's angle; it divides the opposite side into segments proportional to the adjacent sides.
.
.
AND that is about it... have fun studying and GOOD LUCK
~~~~~~OLIVIA~~~~~~

Thursday, March 10, 2011

Chapter 10 Review

This is a summary of chapter 10, enjoy!
They will be short and sweet...


Click this link for a breif lesson on being fooled.


10.1
In this chapter aka. "The Circle" were we learned about circles (which the title clearly tells you). A circle is the set of all points in a plane that are a givin distance from a given point on the plane. The given point is the center of the circle, and the given distance is the radius. We also learned what chords and diameters are.

The theoroms we learned were;

74-If a radius is perpindicular to a chord, then it bisects the chord.


75-If a radius of a circle bisects a chord that isnt a diameter, it is
perpindicular to that chord.

76-The perpendicular bisector of a chord passes through the center of the circle.






"Congruent Chords"

10.2
We learned 2 main theorems in this chapter...
77-If two chords of a circle are equadistant form the center, then they are
congruent.
78-If two chords of a circle are congruent, then they are equadistant from the
center of the circle.




10.3 is about "Arcs of circles"

An ark was a temporary boat used for river transport in eastern North America before canals and railroads made them obsolete. Arks were built primarily to carry cargo downriver on the spring freshet to carry lumber or logs and agricultural produce to a port city downriver...

But an arc consists of two points on a circle and points on the circle needed to connect the points by a single path.
There are two types or arcs;




Next is 10.4 "Secants and tangents"

A Secant is a line that intersects a circle at exactly two points(contains a chord)


A tangent is a linethat intersects a circle at exactly one point. This point is called the point of tangency or point of contact.


Another important thing is when you have your common external tangent, you want to add both circles radius and the difference for the height, then use the pythagorean theorem to solve the exterior tangent length.



Later we learn Theorom 85; If two tangent segments are drawn to a circle from an external ppoint,then those segments are congruent.

On to 10.5
Heres the big picture... click it to make it bigger!

Inscribed

Chord-Chord

Tangent-Chord

Secant-Secant angle

Secant-tangent angle

tangent-tangent angle


Fun right? good thing we still have more chapters to do...

Chapter 10.6: More Angle- Arc Theorems

89-If two inscribed or tangent-chord angles intercept the same arc, then they
are congruent.

90- If two inscribed or tangent- chord angles intercept congruent arcs, then
they are congruent.

91-An angle inscribed in a semicircle is a right angle.

92-The sum of the measures of a tangent-tangent angle and its minor arc is 180*
(degrees)

No pictures on this one, sorry ive been at this for over 2 hours and i want sleep...


10.8 (No 10.7 we didnt do it)
"Power Theorems"

Theorem 1-If two chords of a circle intersect inside the circle, then the product of the measures of the segments of one chord are equal to the product of the measures of the other chord.

Theorem 2- (during only 2 secant segments)If two secant segments are drawn from an external point to a circle, then the product of the measures of one secant segment and its external part is equal to the product of the measures of the other secant segment and its external part.

Theorem 3- (one tangent line and one secant line in the circle)If a tangent and a secant segment are drawn from an external point to a circle, then the square of the measure of the tangent segment is equal to the product of the measures of the entire secant segment and its external part.

That wraps up my chapter 10 review and I want to say thanks for the Great Honors Geometry class,
you booklickers...


-Tyler Rogers-

Dog on a Frog on a Blog.

Sunday, March 6, 2011

10.8 The Power Theorems


Today in class we learned the power theorems of circles. The theorems are.....


The first theorem is.... -If two chords of a circle intersect inside the circle, then the product of the measures of the segments of one chord are equal to the product of the measures of the other chord.
-so in this case, AP*PD=BP*PC The second theorem is... -If there are 2 secant segmants that go through the circle, then the product of the small part of one line and the entire line is equal to the product of the small part of the other line and its whole line.

- So in this case, PA*PD=PC*PB


A 3rd 2nd theorem for a special case of a circle in which there are only 2 secant segmants is... -If two secant segments are drawn from an external point to a circle, then the product of the measures of one secant segment and its external part is equal to the product of the measures of the other secant segment and its external part.


A 4th 3rd theorem for another special case of a circle in which there is one tangent line and one secant line is.... -If a tangent and a secant segment are drawn from an external point to a circle, then the square of the measure of the tangent segment is equal to the product of the measures of the entire secant segment and its external part


Well that's about all we learned in class.
-Michael Levitsky