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gf11221
06-11-2011, 04:08 AM
GUYS, I REALLY NEED TO FINISH THIS FREAKIN' PROJECT FOR OUR CLASS DISCUSSION TOMORROW!

HERE IS THE ACTIVITY's INSTRUCTION THAT OUR PROFESSOR WROTE ON THE BOARD:

A:
-USING A POSTCARD/PAPER, CUT A SQUARE OF SIDE 10 CM
-CUT OFF THE SMALL SQUARE OF SIZE 2CM BY 2CM FROM ONE OF THE CORNERS OF THE POSTCARD/PAPER
-LABEL THE SIDE OF THE POSTCARD/PAPER WITH 10CM AND (10-2)CM
-CUT THE REMAINING POSTCARD/PAPER
-REARRANGE THE REMAINING PARTS TO FORM RECTANGLE
-FIND THE AREA OF THE A.) ORIGINAL PAPER B.) THE SMALL SQUARE C.) NEXX(<<idk what does this mean) SQUARE
-AFTER CUTTING OUT THE SMALL AQUARE, WHAT IS THE AREA OF THE REMAINING PART OF THE PAPER?
-MAKE A GENERALIZATION OF WHAT YOU'VE DISCOVERED(<<I'll do this but i need your help making the equation)

B:
-REPEAT THE PROCESS IN A USING X METERS AND Y METERS AS THE LENGTHS, INSTEAD OF 10CM AND 2CM RESPECTIVELY/PROSPECTIVELY(<<Can someone explain this to me and i need a help too)
-RELATE THE PROCESS OF FINDING FACTORS OF THE DIFFERENCE OF TWO SQUARES USING INTERGERS AND THE FACTORS OF THE DIFFERENCE OF TWO SQUARES USING VARIABLES(<<I need your help doing this part too)

soroban
06-11-2011, 12:31 PM
Hello, gf11221!

I'll get you started . . .


A

We have a paper square, 10cm x 10cm.
Cut off a small square of size 2cm x 2cm from one of the corners.
Cut the remaining paper into some parts.
Rearrange the parts to form a rectangle.

(a) Find the area of the original square.
(b) Find the area of the small square.
(c) NEXX SQUARE (??)

After cutting out the small square, what is the area of the remaining paper?
Make a generalization of what you've discovered.
\text{The original square has area: }\:10^2 \,=\,100\text{ cm}^2

\text{The small square has area: }\:2^2 \,=\,4\text{ cm}^2

\text{The remaining paper has area: }\:10^2- 2^2 \:=\:100 - 4 \:=\:96\text{ cm}^2




: 2 : - - 8 - - :
*---*-------------* -
2 |///| | :
- *---* | :
: | : | :
: | : B | 10
8 | A : | :
: | : | :
: | : | :
- *---+-------------* -
: - - - 10 - - - :
After removing the small square,
cut the remaining paper into two parts, A and B.


Move part A directly below part B.


: - - 8 - - :
- *-------------* -
: | | :
: | | :
: | | :
10 | | :
: | | 12
: | | :
: | | :
- *-------------* :
2 | A | :
- *-------------* -
: - - 8 - - :
And we have an 8 x 12 rectangle.

\text{This is possible because: }\:10^2 - 2^2 \:=\:(10-2)(10+2) \:=\:8\cdot12


\text{In general, we have an }x\text{-by-}x\text{ cm square and remove a }y\text{-by-}y\text{ cm square.}

\text{The remaining area is: }\:x^2-y^2 \:=\:(x-y)(x+y)\text{ cm}^2.

Therefore, the remaining area can always be converted to a rectangle.

gf11221
06-11-2011, 07:48 PM
Thanks! YOU ARE AWESOME AND YOUR EXPLANATION AND EXAMPLES IS MUCH EASIER TO UNDERSTAND THAN MY TEACHER! :lol:

Edit:
So this is what i did:
10^(2)-2^(2)=(10-2)(10+2)

Squaring a number is the same as multiplying the number by itself (10*10). In this case, 10 squared is 100.
100-2^(2)=(10-2)(10+2)

Squaring a number is the same as multiplying the number by itself (2*2). In this case, 2 squared is 4.
100-4=(10-2)(10+2)

Subtract 4 from 100 to get 96.
96=(10-2)(10+2)

Subtract 2 from 10 to get 8.
96=(8)(10+2)

Add 2 to 10 to get 12.
96=(8)(12)

Multiply 8 by 12.
96=(8*12)

Multiply 8 by 12 to get 96.
96=(96)

Remove the parenthesis around the expression 96.
96=96

Since 96=96, the equation will always be true.
Always True

is it right?

JeffM
06-12-2011, 11:29 AM
Yes it is right.

x[sup:10sqfc78]2[/sup:10sqfc78] - y[sup:10sqfc78]2[/sup:10sqfc78] = (x + y)(x - y).

However, what both your teacher and Soroban were trying to do was to imprint this basic relationship in your mind through a SPECIFIC PHYSICAL EXAMPLE.

Here is a GENERAL LOGICAL PROOF:

x[sup:10sqfc78]2[/sup:10sqfc78] - y[sup:10sqfc78]2[/sup:10sqfc78] = (x[sup:10sqfc78]2[/sup:10sqfc78] - y[sup:10sqfc78]2[/sup:10sqfc78]) + (xy - xy) = (x[sup:10sqfc78]2[/sup:10sqfc78] + xy) - (xy + y[sup:10sqfc78]2[/sup:10sqfc78]) = x(x + y) - y(x + y) = (x + y)(x - y).

.

gf11221
06-13-2011, 12:57 AM
Already did that on the white board yesterday, thanks for your help guys!