## Wednesday, September 28, 2011

### John Chua's Blog Post

Today in math class we learned how to find the surface area of 3D shapes with missing pieces and find an area of a cylinder by solving it in pi.

first shape look like this:
top bottom rest of shape extra piece
S.A.= 2 (lw) + Ph(perimeter-height)+ 2(lw)
=2(2)(3)+10(2)+2(1)(1)
=12+20+2
=34u^2

(information)
to find the perimeter of the rest of the piece is 3+2+3+2.
the extra piece that is not shown. It has two pieces that has a dimension of 1-1.

2nd shape:

large-small
top and bottom sides front back inside
S.A.= 2(lw) + 2(lw) +2(s2+s2) +4 (lw)
= 2(2)(3) + 2(2)(3) +2 (3^2-1^2)+4 (1)(2)
= 12+12+2(9-1) +8
= 48^2

3rd shape is a cylinder:

here is the net of the image:

Circumference= (pi.D)---->(2 pi r)

diameter of the circle is 10 and the height of the cylinder is 12.

S.A.=2 pi r^2 + 2 pi r h ( to find the radius you divide the diameter by 2)
=2 pi 5^2 + 2 pi (5)(12)
=2 pi 25 + 2 pi 60
=50 pi + 120 pi
= 170u^2 <------ solve in pi (accurate)
= 534.07 u^2 <------- solve to the hundredths (approximation)

HOMEWORK!! : Journal, all apply 1.3(22 or 23), all practice sheet , all workbook, Study for test( Tuesday or Wednesday)

Aj will be doing the next blog.

## Sunday, September 25, 2011

### Ishaka's Blog Post

During our math class on Friday we didn't do much because we had an evacuation drill, but we answered question 4 a) from page 32 with the enitre class.

Question a) was asking us to find the surface area of a 3D shape.

(Question 4 a) as a net)

STEP 1

We had to split the 2 sides into to parts because we didn't have a formula that would help us with that kind of shape.

STEP 2

After spliting the two sides into two parts, we started to label A1, A2, A3, and A4, for the parts that were the same to help us with the formula.

(There are 3 A2's because both the tops equal the base, and there are also 3 A1's because both the fronts equal the back.)

STEP 3

Now that we had all the information that we needed to find the surface area, we just filled in the information into our formula.

S.A = 2(A1) + 2(A2) + 2(A3) + 2(A4)
2(lw) + 2(lw) + 2(s^2) + 2(lw)
2(4.5) + 2(2.5) + 2(2^2) + 2(2.1)
40 + 20 + 8 + 4 = 72units^2

HOMEWORK: Page 32 #'s 4 - 9, and also write in your math journal.

John will be doing the next blog post.

## Thursday, September 22, 2011

### Surface Area

Area of rectangle: lh or lw

Area of circle: πr²
Area of triangle: bh/2
Area of square:
Circumference: πd or 2πr
Diameter: 2r or c/π
Perimeter of triangle: s + s + s
Perimeter of quadrilateral: s + s + s + s
Perimeter of any n-gon: Add all side lengths
_________________________________________

Parallelogram:

Formula:

a = bh

Trapezoid/Isosceles Trapezoid

Formula:
a = ( b1 + b2/2 )h

Surface Area of nets

Square:

First way:
SA = 6s²
= 6(2²)
= 6(4)
= 24u²

Second way:
SA = 2(s²) + (Pob)h
= 2(2²) + 8(2)
= 2(4) + 16
= 8 + 16
= 24u²

Rectangle:

First way:
SA = 2(s²) + 4(lw)
SA = 2(2²) + 4(3x2)
SA = 2(4) + 4(6)
SA = 8 + 24
SA = 32u²

Second way:
SA = 2(lw) + (Pob)h
SA = 2(3x2) + 10(2)
SA = 2(6) + 20
SA = 12 + 20
SA = 32u²

Homework:
Identify 2 ways to find the surface area of this net.

## Wednesday, September 21, 2011

### Jocelle's Post

1.)Formula for the area of a square
A= s² (side x side)
2.)Formula for area of a rectangle
A= lw or lh (length x width)(length x height)
3.)Formula for area of a triangle
A= bh/2
4.)Formula for area of a circle
A= πr²
5.)Formula for a circumference
C= πd
6.)Diameter is equal to?
D= 2r or D= C/π
R= d/2
8.)Perimeter of a triangle formula
P= s+s+s
9.)Perimeter for a rectangle formula
P= s+s+s

Brackets
Exponents
Divide
Multiply
Subtact

What's the 3d shape of the net?

Surface Area of a cube:
SA= 6s²
SA= 6(2²)
SA= 6 (4)
SA= 24 u²

What's the 3d shape of the net?

Rectangle:
SA= lw
SA= 4lw + 2s²
SA= 4(4)(3) + 2 (3²)
SA= 4(12) + 2(9)
SA= 48 + 18
SA= 66u²

Homework:1.3 Homework Book