Class of May 8, 2009 :D

iushaiuhsaiuhYesterday we started class by learning new subject instead of correcting homework.

We learned the formulas for fiding lateral and surface areas of right cylinders, just as we had done for right prisms on thursday.
SOOO:

The formula for the Lateral Area is L=(p)(h), L being lateral area (!), p being perimeter of the base (or the crcumference of the base/circle) and h being the height of the CYLINDER.

The formula for Total Surface Area is T=L+2B, T being total surface area, L being the lateral area and B being the area of the base. Just a reminder: the formula for the base of the cylnder (or for the area of a circle) is A=πr².

After, we corrected our homeworks. Something that gave me a hard time was finding the perimeter of the solid, which was required to do like... 99% of the hw... So, teh perimeter of the solid is the same as the perimeter of ONE BASE.
>>>Exercise 25 gave a hard time to something like.. ahn.. all of us (except one or two). The thing is you cannot divide the shape into two and find the total surface areas of both and add - cause then, you'll add the area of the part they connect, which is not filled. Sooo, you have to find the area of everything: divide the solid into polygons and fint the areas of these.

And that was the class... In the end:
HOMEWORK!! \o/
Page 596 #s 26-30 all + 32-34 all
It's a good idea for those who didn't do the last hw (pg 596 #s15-25 all + 32 + 33) to do it, because Ms. Saatkamp said she was not going to explain the exercises for us if we don't do it... So, PLEASE, DO IT!)

;*

Today Class of 9A

Hey guys! So today after we correct the homework which was page 588 and the problems were 23, 24, and 25. We went on to our lessons called The Surface of Right Prisms and The Surface Area of Right Cylinders.
Also we got a packet that contains figures and we cut only one figure today though. But for all of you classmates out there DON'T FORGET THEM SINCE WE'LL NEED THEM FOR LATER CLASSES!
Now back to the lessons that we learn today, this connects to the figures that we were cutting. The figure below is the one that we cut today (It’s not exactly right or even congruent).

Now look at the figure, can you see that there is a big rectangle and the two bases? Also can you see that the green part equals a side of the base? And that if we get the length the rectangle that means we also got the perimeter of the bases? Also the red part is the height of the prisms. So in the end the big rectangle can tell you: perimeter of one base, the height of the prism, and area of the lateral area. Easy huh?


But we need some hand-dandy formulas!!!

So let’s say we want the Lateral Area then we'll use this formula:





And P= perimeter of one base

H= Height of the Prism

Later we need to use another formula to find the Total Surface Area which is:



Where
L= Lateral Area
B= Area of Base


Then we go on the next lesson which is Surface Area of Right Cylinders....

So pretty much the formulas are the same expect there are some minor changes where you should check twice

With the Lateral Area is:




Where

P= Perimeter/ Circumference which is if you have forgotten,

H= Height of Prism


Afterwards with Total Surface Area



Where

L= Lateral Area

B= area of bases (Base of Circle...)

And that is pretty much what we did in class today! So I hope you have a great weekend but don't forget these things

1. Bring the Packet to class because you never know when you may need them
2. Project Deadlines have been changed from Friday to Monday so you have at least one more weekend to work on it!!
3. Also with the projects, if you don't have class with Ms. Saatkamp on Monday, then find a way to give the project to her or else...
4. We have a make-up test this Monday, so if you sign up for it get ready!!!!

HAVE FUNNN! =3

07/05/09

Hey!
Today in class we went over the Home work, (it was page 588 #23, 24, 25) although they were only a few problems they were complicated. To simplfy things it is much easier if you seperate them into different parts, number 23 for example can be the two bases (the bases being the L shapes, since they are parallel and congruent) two equal rectangles, the back rectangle and the two smaller rectangles. And after you have added up all the seperate areas you will have 120 units squared. The second problem is very similar, find the area of the pentagon using the idea of apothem times perimeter divided by two, and then multiply it by two since they are two hexagons. Then find the area of the rectangle and multiply by six (the number of rectangles). The last one was the most complicated, you have to pay attention to realize that the top edge is actually smaller than the middle edge.
Now, in class we began with some definitions, Ms. Saatkamp stressed the fact that many times the problem will actually give us a description of the solid rather than a diagram so it is important to understand the definitions. There are basically 3 names when naming prisms, such as Right Hexagonal Prism. The first name may be either right or oblique, the second being the shape of the bases such as hexagonal or triangular, the third name is prism, just to give it an ending :P
Then Ms. Saatkamp gave us some shapes to cut up, which was pretty fun. One of the shapes we cut up was an open Right Pentagonal Prism. Which by looking at opened looks like 5 rectangles and 2 pentagons. I'm sorry i'm having a lot of trouble getting pictures on here. Anyways, by looking at it you can see that the perimeter of a penatgon is = to the side the big rectangle(if you join all 5 rectangles it forms a big one).
To find the total surface area you use the following formula:
T = L + 2B
T being total area. L being Lateral area (which you can find by multiplying perimeter and height) 2B being twice the area of the base.
Homework is pg 596 #15-25 +32 and 33
bjos

Class Summary-9A- 5/7/09

Today Mrs. Saatkamp began the class by showing us our test grades and by explaining to us that we would have to sign up for recuperation in advance. After this, Saatkamp helped us define what a sphere is.

Sphere: a set of points in space that are all equidistant from a given center.

After this Saatkamp explained to us what Platonic Solids are:

Platonic Solids:

Solids formed by the same polygon and that polygon needs to be regular.

THERE ARE ONLY FIVE PLATONIC SOLIDS, THEY ARE:

Tetrahedron: has 4 faces and all faces are equilateral triangles.

Hexahedron, more commonly known as "cube": has 6 faces and all of the faces are squares.

Octahedron: has 8 faces and all of the faces are equilateral triangles.

Dodecahedron: has 12 faces and all faces are regular pentagons.

Icosahedron: has 20 faces and all faces are equilateral triangles.

After all of this, Saatkamp introduced us to the idea of surface area which is, for example: A company has make a soda container for 200 ml of soda, several people design containers, but, the company wants to find out which container uses the least amount of material. TO FIND THIS OUT SURFACE AREA IS NEEDED.

That was the class. Our homework is the continuation of the in-class examples:

pg 588/# 23, 24, and 25

Wednesday's class, 06/05/09

In today's class we were continuing what we were learning last class about solids. 

Cone: a solid with one circular base. In a cone, the vertex is in the middle of the base.

 

Cylinder: a solid with two circular bases. There are oblique cylinders and right cylinders.

Sphere: a solid formed by a set of points in space, equidistant to a given center. This definition is very similar to the definition of a circle, but the only different is that a circle in on a plane while a sphere is in space. 

Then we learned about platonic solids, which are solids that were discovered by Plato, that are only formed by the same regular polygons. There are only 5, it is impossible to have any other solids formed by the same regular polygons. 

Tetrahedron: - it has 4 faces

                     - all faces are equilateral triangles

Hexahedron: - it has 6 faces

                     - all faces are squares

                    (- this is most commonly known as a cube)

Octahedron: - it has 8 faces

                    - all faces are equilateral triangles

Dodecahedron: - it has 12 faces

                        - all faces are regular pentagons

Icosahedron: - it has 20 faces

                     - all faces are equilateral triangles  

Afterwards, we did some practice problems to find the area of a solid, which is when you find the area of the shapes that form the sides, and then add them all. 

Our homework was page 588, #23, 24 and 25. This is a continuation of the exercises we were doing in class. 

(: 

05/05/2009

Todays calss we started reviewing the quiz and the we started with learning about solids,

there are 4 types of solids; the Polyhedral/Polyhedrons, Cone, Cylinder and Sphere.

The Polyhedral/ Polyherdons are divided in two sections the Prisms and the Pyramids and those in other two regular and irregular.
Prism:
It's a solid with:

• 2 congruent and parallel bases
• the lateral faces are rectangles or parallelograms
• a prism is regular when the bases are right on top of each other and the lateral faces are rectangles, their bases are also regular polygons.
• a prism is oblique when the bases aren't right on top of each other and the lateral faces are parallelograms.

Pyramid:
It's a solid with:

• only 1 base
• the lateral faces are triangles
• a pyramid is right when the vertex and the center of the base are in top of each other.
• in an olique pyramid don't

Cone:
It's a solid with one cirular base and a vertex.

Cylinder:
It's a solid with two circular bases.

Class of Tueday, 28 of April

Today we started by learning how to solve for the area of regular polygons.




First, we solved for the area on the pentagon, on the sheet of paper that she gave us. Then, we solved for the area of the octagon, when she told us that the raduis of the circumscribed circle was 10. To solve:







y= i/2

a= apothem


x= side/2



(circumscribed circle means a cirle on the outside of the octagon that touches all the vertices of the octagon)



That means that the radius is one of the sides of a triangle made from the diagonals.


First, you have to calculate the sum of interior angles to find one interior angle, in order to find angle y:


si= 180(8-2) = 1080


i= 1080/8 = 135


y= 135/2 = 67.5

Now, we calculate the apothem:

sin 67.5= a/10

a= 9.238


Then we calculate x:


cos 67.5= x/10


x= 3.826 (remember always to use at least 3 decimal places)

Then calculate the base (side of the octagon):


b= 2x= 7.654


Then calculate the perimeter:


p= 8b = 61.229


then calculate the area:


A= (pXa)/2


A= (61.229X9.238)




A= 282.8u²


So here are the Formulas for the Area of Regular Polygon:





p= perimeter
a= apothem

(Area of Circles):