Thursday, September 29, 2011

Kim's Blog Post

Today we had a quiz on surface area.

Question 1

Draw the net of the exposed areas of this shape. Assume that this shape is resting on a table.

This is the net.

There are only three rectangles because the base is not exposed. Since it's resting on a table, you can't touch it.

Question 2

Calculate the surface area of the exposed areas.

S.A. = 3(lw) + 2(s²)
= 3(4)(2) + 2(2²)
= 3(8) + 2(4)
= 24 + 8
= 32m²

Question 3

If a can of paint covers 0.5m², how many cans would it take to cover 32m²?

1 can = 0.5m²
x cans = 32m²
32 = 0.5x

32/0.5 = 0.5x/0.5
64 = x

It will take 64 cans to cover this shape.
________________________________________________________________

After the quiz we learned how to find the surface area of a shape after you put two shapes together.

We have these two shapes and we have to put them together.

When they're connected they look like this.

The red part is the part that is covered. (On Shape 1, the red part is the same as the back.)

Shape 1

Shape 2

So to find the surface area of the the connected shape, it's easier to do two separate formulas then subtract the area that's touching.

Shape 1
S.A. = 2(s²) + 4(lw)
= 2(2²) + 4(3)(2)
= 2(4) + 4(6)
= 8 + 24
= 32u²

Shape 2
S.A. = 2(lw) + 2(lw) + 2(lw)
= 2(1)(4) + 2(2)(4) + 2(1)(2)
= 2(4) + 2(8) + 2(2)
= 8 + 16 + 4
= 28u²

32 + 28 = 60u²

Area covered = 2(s²)
= 2(2²)
= 2(4)
= 8m²

60 - 8 = 52u²

The surface area of the shape is 52u².

Here's another way to find out the surface area.

S.A. = 2(s²) + 4(lw) + 2(lw) + P(h) - 2(s²)
= 2(2²) + 4(2)(3) + 2(1)(2) + 6(4) - 2(2²)
= 2(4) + 24 + 4 + 24 - 2(4)
= 8 + 24 + 4 + 24 - 8
= 52u²

Homework:

- Do all the homework that Mr. Backe assigned before if you haven't finished it.
- Be ready for the test next week.
- Come to Mr. Backe if you need any help with anything.

The person that I choose to do the next blog post is Chelsea.

Wednesday, September 28, 2011

Anabelle's Blog Post: Surface Area

Recap/Review:

The day before yesterday we discussed whether or not a certain rectangular prism with a piece missing would have the same surface area as one that is whole. We came to the conclusion that it would.

When looking at them in one view/one face at a time it will be the same.

Today's notes:

We were asked if this figure was any different. Once we really thought about it, we found that it was. 2 small square units would be missed if we were to simply use the method for the last two figures.

With that in mind we easily solved the surface area for the figure:

SA= 2(lw)+P(h)+(lw)
= 2(2)(3)+10(2)+2(1)(1)
= 2(6)+20+2
= 12+22
= 34u^2

The next figure we inspected looked like this:

Once again, we figured out that the surface area for this one would also have a different surface area in comparison to one that is whole.

SA= 2(Top/bottom)+2(side)+2(front/back- missing units)+(inside)
= 2(lw)+2(lw)+2(s^2-s^2)+4(lw)
= 2(2)(3)+2(2)(3)+2(3^2-1^2)+4(1)(2)
= 12+12+2(9-1)+8
= 12+12+16+8
= 48u^2

Afterwards we discussed finding the surface area of a cylinder.

We were asked to make the net of a cylinder.

The circumference of one of the circles will give you the width/length of the rectangle that makes up the cylinder.

You can find the surface area of a cylinder using these formulas
SA= 2πr^2+2πrh
= 2πr^2+πdh
= 2πr^2+Ch

We then were given a cylinder to solve the surface area for. To make things different we were also asked to solve in pi. This means that we do not multiply anything by pi, but to simply leave it in to solve for later.

SA= 2πr^2+2πrh
= 2π5^2+2π(5)(12)
= 2π(25)+2π(60)
= π(50)+π(120)
= (170)π u^2 (solve by pi)

= 534.07 u^2 (solve to two decimals)

Homework:
• "Apply"
• number 22, 23: textbook
• 1.3 in the homework book and extra practice sheets
• Journal
Oh, and ....

That's right, Kim. I went there.

Saturday, September 24, 2011

Olivia's Blog Post

Today we went over our homework from yesterday.

Mr. Backe had given us an isosceles triangular based prism to find the surface area for. We were supposed to use the Pythagorean Theorem to find the length of the hypotenuse.

Here is a net of the triangular prism:

Here is the 3D drawing of the triangular prism:

To find the surface area of this triangular prism, you must first find the length of the hypotenuse for the triangle

Pythagorean theorem:
a² + b² = c²
3² + 4² = c²
9 + 16 = c²
√25u² = √c²
5u = c

 Sorry if the letters are small and fuzzy. :S

Now you can find the surface area for the triangular prism
Use the formula,
S.A. = A1 + A2 + A3 + 2(A4)

Now plug in the numbers and solve the surface area:
* pythw = Pythagoras(width)

S.A. = A1 + A2 + A3 + 2(A4)
= s² + lw + pythw + 2(bh/2)
= 3² + (3)(4) + (5)(3) + 2[(4)(3)/2]
= 9 + 12 + 15 + 12
= 48u²

Next, we went over question 4. a) from page 32.
Here is a net of the shape:

To find the area for this shape, Mr. Backe made us find the faces that were similar to the Back and Base, and put them together as 2(A1), 2(A2), etc.

 Sorry, this one is really complicated, I didn't have enough room, though I thought I did.

The formula you use to find the surface area of this shape is:

S.A. = 2(A1) + 2(A2) + 2(A3) + 2(A4)
= 2(lw) + 2(lw) + 2(lw) + 2(s²)
= 2(4)(5) + 2(2)(5) + 2(1)(2) + 2(2²)
= 2(20) + 2(10) + 2(2) + 2(4)
= 40 + 20 + 4 + 8
= 72u²

If you got 76u², that was wrong, because the two shapes have a different net:

As you can see, the nets are different, causing the surface area to be different.

Next, we talked about this shape from question 5. a) on page 32 (it was turned vertical):

Mr. Backe asked us if the view of that shape was the same as the view as this shape (turned vertical):

At first, the majority of the class thought that the view was different, but then we realized that the view was the same

Mr. Backe then asked us to draw a net for the views of both of the shapes:

:

Both shapes have the same view, so therefore, the surface area is still the same

To solve:
S.A. = 2(A1) + 2(A2) + 2(A3)
= 2(s²) + 2(lw) + 2(lw)
= 2(2²) + 2(2)(3) + 2(3)(2)
= 2(4) + 2(6) + 2(6)
= 8 + 12 + 12
= 32u²

Homework
- journal (also previous day's journal if you didn't do it)
- Textbook page 32 - 33 # 4-9
- homework book chapter 1.3 for extra practice if you haven't already finished it

*Hint: use graph paper to draw out the nets of the shapes to help you

Have a great long weekend everybody! :)

Thursday, September 22, 2011

Lizelle's Blog Post.

Today, we had a little minor quiz.
Formula for a parallel: A=BH
Formula for a trapezoid: (B1+b2) / 2 H
Formula for a rectangle: LW
Formula for a square: s2
Formula for a circle: A= πr2
Perimetre for a triangle: S+S+S
Perimeter for a qualadrateral : S+S+S+S
Perimetre for an n-gon: Add all sides.
Circumfrance = πD
Diametre = R2 or C/π

So then we worked on this:

We discussed what this net makes, and it made a rectangular prism.

FORMULA:

SA= 2(lw) +Ph
= 2(3.2) + (3.3+2+2)4
= 2(6) + 4(10)
= 12 + 40
= 52 units squared.

Also, we discussed about our homework last night.

Many people got this wrong, because we didn't use the pytha. theory!

We figured out that the homework was in the shape of a triangular based prism!
Pythagorean:
A2+b2=C2
2^2 + 3^2 = C2
4 + 9 = C2
13 = C2
square root 13.

Alright, so let's use this information and plug it into our math part!

Area blue+Area Red+Area Green+2 area black= S2+lw+pythagorasw+2)bh/2+2squared+(3)(2)+3.87+2(3)(2)2
But since the 2's at the end get cancelled out, the formula should look like this:
Area blue+Area Red+Area Green+2 area black= S2+lw+pythagorasw+2)bh/2+2squared+(3)(2)+3.87+2
(3)(2)

For todays homework: It's

To be honest, I am having trouble solving this. If any of you guys see this, may you help me. It would mean a lot!

Oh yes, and guys don't forget to do

Textbook: 1.3 4, 5, 6,7
and to write in your journal, okay. :)
bye guys!

Wednesday, September 21, 2011

Roemer's Blog Post

Today, we went over the Show You Know's.

I saw how the Gr.9's from last year chose the next scribe makers so..
Jieram Ramos
, I CHOOSE YOU.

Jieram's Post

Quiz
1) Area of a Square
Formula: Area of Square = S²
S = Side

2) Area of a Rectangle
Formula: Area of Rectangle = lw lh

3) Area of a Circle
Formula: Ao = πr²

4) Area of Triangle
Formula: bh/2

5) Circumference
Formula: πd or 2πr

6) d= c/π or 2r

7) r = d/2

Area of Parallelogram
Formula: bh

Trapezoid
Formula: Area of Trapezoid = (b1 + b2/2) h
Isosceles TrapezoidTrapezoid

Net
3D

S.A = 6s²
= 6(2²)
= 6(4)
= 24u²

S.A = 2s² + large perimeter h
= 2 (2²) + 8 (2)
= 2(4) + 16
= 8+16
= 24u²

P= Perimeter
PA= s+s+s
Psquare= s+s+s+s
P3Dsquare= s+s+s+s

Net

3D

S.A = 4lw+25²

= 4(4)(3)+2(3²)
= 48 + 2 (9)
= 48+18
= 66u²