Sunday, March 13, 2011

Chapter 11: Area, A.K.A The one before THE Curve

Area is measured in square units and is one of the most important features of Geometry.

common-geometry-formulas.jpg


Those are all the Formulas for the area of shapes, except for the kite which will be mentioned later on.

However, there is more to area than just formulas. For example the trapezoid comes with a few nifty little theorems as well.


Trapezoid: Every trapezoid has a median. The median of a trapezoid is the segment connecting the midpoints of the non parallel sides. The length of a median is found multiplying the sum of the parallel sides by 1/2.

This equation is represented as

m= (a+b)/2

trapezoid_median.gif

also, the area of a trapezoid can be found by multiplying the median by the trapezoids height.

trap3.GIF.gif




Lastly, the median of a trapezoid bisects the diagonal of the trapezoid, providing a basis for some interesting questions from Mr. Wilhelm.








The area of a kite is

A=1/2d^1,d^2


formula-area-kite.jpg


The diagonals of a kite form a right angle, and also sides d and a are congruent as are sides c and b. Lastly a kite has one pair of congruent angles, these angles are the ones that d^2 runs through.

kite3.gif





In a rhombus, the diagonals form right angles, and bisect the other diagonals. Also all sides are congruent.


rhombusstuff.gif




Next are the areas of regular polygons. For an equilateral triangle the formula is


A= ¼(s²⋅√3)


apothem.gif




The area of any regular polygon is ½ apothem ⋅height


A=½ap


KEY SOLUTION TO HEXAGONAL FIGURES AND EQUILATERAL TRIANGLES!!! ITS CALLED 30-60-90 TRIANGLES!!!


properties- hexagon_0.JPG.jpg


Picture TIME


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I don't know the exact equation but a segment is a sector minus its inner triangle.



As far as ratios go if its area you square the equation and for volume you cube the equation.


Oh and a median of a triangle divides the triangle into two triangles with equal areas.


Lastly Hero's and Brahmagupta's Formulas


first, Hero's.

heron1.gif

next, brahmaguptas


ts.png





Now For a Trip Down Memory Lane


First, Proofs.


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Now that I think about it all we really did 1st tri were proofs soo...

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HEY its our book.









I'd like to give a shout-out to Mr. Wilhelm, the greatest teacher ever, and a kudos to my fellow bloggers, wheel spinners, book lickers, trigons, game losers, and applauders. Thank you all for making this honors geometry class

a little more enjoyable and interesting than most other classes,


Oran Lieberman


Chapter 12 Final Review

In chapter 12, we learned about surface area and volume.


Section 12.1: Surface Areas of Prisms


In every prism there are two bases and n number of lateral faces, where n is the number of the sides of the face. The lateral faces are always parallelograms. The lateral area of a prism is the surface area of all of the lateral faces and total surface area of a prism is the sum of the lateral areas and the surface area of the two bases.



The formula: S=2B+L, where L is the lateral area and B is the base of the prism. Also, the perimeter * height equals the lateral area, so a more useful formula is S=2B+ph, where p is the perimeter of the base and h is the height of the prism.



Section 12.2: Surface Areas of Pyramids




In every pyramid, there is one base and n lateral faces, when n is the number of sides on the base. In a regular pyramid, the base is a regular polygon and all of the lateral edges are congruent.




The formula: S=B+L, when B is the area of the base and L is the lateral area of the pyramid. One half of l, the slant height, and p, the perimiter of the base, equals the lateral area, so a more useful formula is S=B+1/2 pl, where p is the perimiter of the base and h is the height of the pyramid.




Section 12.3: Surface Area of Circular Solids





A cylinder is an infinitely sided prism, so the formula for its surface area is S=2B+ph. Because the cylinder's base is a circle, the area of the base is πr2(pi * radius squared, superscript doesn't work), and the perimiter is 2πr, so the formula is S=πr2+2πrh, where h is the height and r is the radius of the cylinder.





A cone is an infinitely sided pyramid, so the formula for its surface area is S=B+1/2 pl. Because the cone's base is a circle, the area of the base is πr2 (again, radius squared), and the perimiter is 2πr, so the formula is S=πr2+πrl, where l is the slant height and r is the radius of the cylinder.




A sphere's Surface area is S=4πr2, and a hemisphere's formula for surface area is 3πr2, not 2πr2, because you are adding another circle when you cut the sphere in half.





We also learned about a frustum during this section.


A frustum is a pyramid with a pyramid similar to the original one sliced off, leaving the bases parallel. In a conical frustum, the Surface area is S=πR2-πr2+πRL-πrl, where R is the radius of the big cone, r is the radius of the small cone, L is the slant height of the big cone, and l is the slant height of the small cone.





Section 12.4: Volumes of Prisms and Cylinders







Volume is the measure of the space enclosed by a solid. in prisms, the formula for volume is V=Bh, where B is the area of the base and h is the height of the prism. A cylinder's base is a circle, so B=πr2, and the formula is V=πr2h, where h is the height of the cylinder and r is the radius of the cylinder.




Section 12.5: Volumes of Pyramids and Cones












The formula for volume in a pyramid is 1/3 Bh, where B is the area of the base and h is the height of the pyramid. A cone's base is a circle, so its area is πr2, and the formula for volume is V=1/3 πr2h. Also, the volume for a frustum is 1/3 πR2H-1/3 πr2h, where R and H are the radius and height of the large cone and r and h are the radius and height of the small cone.









Section 12.6: Volumes of spheres









The volume of a sphere is 4/3 πr3 (pi * radius cubed), and the volume of a hemisphere is half that, or 2/3 πr3.









I hope that you found this helpful and good luck on the final!



-Jacob

Chapter 7 Final Review

Chapter 7: Polygons

7.1: Triangle Application Theorems
Theorem 50     The sum of the measures of the three angles of a triangle is 180o

Proof:  According to the Parallel Postulate, there exists exactly one line through point A parallel to line BC, so the figure below can be drawn

Because of the straight angle, we know that < 1+ < 2+ < 3 = 180o.  Since < 1 is congruent to < B (Parallel lines à alternate interior angles congruent), and < 3 is congruent to < C, we may substitute to obtain < B + < 2 + < C = 180o Hence, m< A + m< B + m< C = 180o
Definition        An exterior angle of a polygon is an angle that is adjacent to and supplementary to and interior angle of the polygon. 
Theorem 51     The measure of an exterior angle of a triangle is equal to the sum of the measures of the remote interior angles.

Theorem 52     A 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 the third side.

7.2: Two Proof- Oriented Triangle Theorems
Theorem 53     If two angle of one triangle are congruent to two angles of another triangle, then the third angles are congruent. (No- Choice Theorem)
Since the sum of the angles in each triangle is 180o, the sums may be set equal.  If we then apply the Subtraction Property, we see that the third angles must be congruent.
Disclaimer: The two Triangles need not be congruent for us to apply the No- Choice Theorem.
Theorem 54     If there exists a correspondence between the vertices of two triangles such that two angles and a non-included side of one arc ore congruent to the corresponding parts of the other, then the triangles are congruent.  (AAS)
                        Really, one uses the No-Choice Theorem to make an AAS into an ASA.

7.3: Formulas Involving Polygons
Theorem 55     The sum Si of the measures of the angles of a polygon with n sides is given by the formula Si + (n – 2)180.
On Occasion, we may refer to the angles of a polygon as the interior angles of the polygon.
Theorem 56     If one exterior angle is taken at each vertex, the sum Se of the measures of the exterior angles of a polygon is given by the formula Se = 360

Theorem 57     The number d of diagonals that can be drawn in a polygon of n sides is given by the formula:

7.4: Regular Polygons
Definition        A regular polygon is a polygon that is both equilateral and equiangular. 

Theorem 58     The measure E of each exterior angle of an equilateral polygon of n sides is given by the formula:
And that was chapter 7! Good luck everyone and study hard! special shoutout thanks to Dobes for the ride home :)

Blessings, Em J

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.