useful tips to read maths book

HOW TO READ A MATHS BOOK
1. You can’t read a math book the way you read other books. It takes a special approach to read math, not just pass your eyes over it. You may need as long as half an hour to get through one page, but you will understand it when you’re done.

2.What not to Do
Don’t memorize too much
Don’t start with the homework problems. The assigned problems are not “the homework”. They’re the last part of the homework. The first part is reading and understanding today’s section of the textbook.
Don’t be afraid to go back. If you don’t understand something in today’s lesson, perhaps it’s because you didn’t really understand something in an earlier lesson, or you’ve forgotten it. Go back and re-learn that earlier material.
3.What to Do
First skim for an overview. Nobody understands something complex on first reading: about all you can hope to do is get a general sense of what it’s about and perhaps one or two of the main points of the argument.
Then reread with concentration. Read slower this time. Highlight important points for further study.
Go through each step of each example.
Fill in any gaps. Whatever you do, get those problems cleared up before the next class so that they don’t interfere with your understanding that lecture.
Think about what you’ve read.
Make it your own. Can you explain this to someone else? If you can do that, you probably understand it.
Do the homework problems. If you don’t understand how to do one, look back in the book for a similar problem. Don’t just push numbers at it; make sure you understand the example and see how to apply it.

class VII(sum of interior angles of a triangle)

Objective: To find the relationship between the interior angles of a triangle by paper cutting and pasting.

Material required:coloured paper, glue, a pair of scissors, and pencil.
Procedure:
step 1: On a coloured paper, draw any triangle and name its vertices as A, B and C.


Step 2: Label the interior angles as angle ABC, angle BCA and angleCAB.

Step 3: Cut triangle ABC.
Step 4:Cut the three angles.
Step 5: Draw a straight line on a sheet of paper.

Step 6: Place the three interior angles angleBAC , angleABC and angleACB on the straight line, adjacent to each other, without leaving any gap.



Result: angleABC + angleBCA + angleCAB = 180 degrees.

class VI activity- h.c.f,Integers

Objective: To find the h.c.f (highest common factors) of the given numbers experimentally through paper cutting and pasting.
Previous knowledge: 1. Factors of numbers
2. division of numbers

Material required: A rectangular strip of 36 cm in length and 2cm in width
A rectangular strip of 28 cm in length and 2cm in width
A pair of scissiors
fevicol
scale
procedure:
1.The first two strips represent numbers 12 and 20.Finding H.C.F means finding highest common factorof 12 and 24. that is, finding that biggest length of the strip which measures the length of 12 and20 exactly.





2. here from the first two strips it is clear that length can not be 12cm as 12 does not divide 20 and 12 exactly.
3. If we keep the first strip of 12 cm along 20 cm, we observe that length of the strip 8cm is left out in the second strip of 20cm.
20 / 12 leaves the remainder as 8.



4. We also observe that the length can't be 8 cm as it does not divide 12cm and 20 cm exact number of times.
5.now, on dividing, 12 with 8 (12/8) we see that the remainder is 4

6.the strip of 4cm divides cm exactly.
7.Thus we see that 4cm long strip divides 12 cm and 24cm exactly.

Result: the H.C.F of 20 cm and 12 cm is 4.
Assignment : Try the same activity for 36 and 28 cm.


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classVI(Integers)

Objective: Addition of two numbersexperimentally through paper cutting and pasting.
Previous knowledge: On addition of a positive integer to its negative integer gets nullified (or zero).
Procedure:
step 1: Firstly cut 25 unit squares which are kept unshaded and call them positive squares.
step 2: Then cut 25 unit squares which are kept shaded and call them negative squares.
step 3: case 1: Now, take (+8) + (+7)
we take 8 positive squares and 7 positive squares
count them (+8) + (+7) = 15
step 4: Case 2: now take,(-4) + (-5)
we take 4 negative squares and 5 negative squares
count them (-4) + (-5) = -9
Step 5: Case 3:Now take (+6) + (-5)
we take 6 positive squares and 5 negative squares
Count them (+6) + (-5) =1
Step 6: Case 4: Now take (-5) +(3)
we take 5 negative squares and 3 positive squares
Count them (-5) + (+3) = -2
Result:(i) If both the integers are positive then we add and the result is an positive integer.
(ii) If both the integers are negative then we add and the result is an negative integer.
(iii) If positive integer is greater than the negative integer then we subtract and the result is an positive integer.
(1v) If negative integer is greater than positive integer then we subract and the result is an negativ integer.
her sand
Assignment: -8 + 6 =
-4 - 5 =

class VIII activity

Objective: To visualize the factors of an algebraic expression.

Preparation:

1. prepare Algebra Tiles

2. Square tiles of dimension 10 X 10 each, representing x^2.•

3. Rectangular tiles of dimension 10 X 1 each , representing x .

4. Square tiles of dimension 1 X 1 each representing 1.·

Assumption:

We have assumed that

1. In the rectangular tiles each of dimension x sq units the top side represents (+ x) and the bottom side represents (- x).

2.Similarly in the square tiles each of dimension 1 sq unit the top side represents (+1) and the bottom side represents (-1).·

PROCEDURE

I) Representation of x^2 +5x + 6·

To represent this we need 1 square tile representing x^2 , 5 algebra tiles representing x and 6 algebra tiles representing 1.·

By splitting the middle term of the given polynomial we get the expression x^2 +3x +2x + 6.· Place a square tile of dimension 10X10 representing x^2 .·

Add 3 tiles of dimension 10 X 1 each to any side of the tile x^2.The area of new shape formed represents x^2 +3x.· Add 2 tiles of dimension 10 X 1 each to the side adjacent to the previous side.

The area of new shape formed represents x^2 +3x+2x.· Add 6 tiles of dimension 1 X 1 each to complete the rectangle.

The area of new shape formed represents x^2 +3x+2x+6.

II) Representation of x^2 -x - 6·

To represent this we need 1 square tile representing x^2 , 5 algebra tiles representing x and 6 algebra tiles representing 1.·

By splitting the middle term of the given polynomial we get the expression x^2 -3x +2x - 6.· Place a square tile of dimension 10X10 representing x^2 .·

Add 2 tiles of dimension 10 X 1 each to any side of the tile x^2.The area of new shape formed represents x^2 +2x.· Subtract 3 tiles of dimension 10 X 1 each to the side adjacent to the previous side.

The area of new shape formed represents x^2 +2x-3x.· Subtract 6 tiles of dimension 1 X 1 each to complete the rectangle.

The area of new shape formed represents x^2 +2x-3x-6

III) Representation of x^2 -5x + 6·

To represent this we need 1 square tile representing x2 , 5 algebra tiles representing x and 6 algebra tiles representing 1.·

By splitting the middle term of the given polynomial we get the expression x^2 -3x -2x + 6.·

Place a square tile of dimension 10X10 representing x^2 .· Subtract 3 tiles of dimension 10 X 1 each to any side of the tile x^2.

The area of new shape formed represents x^2 -3x.· Add 6 tiles of dimension 1 X 1 each to get 2 tiles of dimension 10 X 1 each to the side adjacent to the previous side.

The area of new shape formed represents x^2 -3x+6.· Subtract 2 tiles of dimension 10 X 1 each to complete the rectangle.

The area of new shape formed represents x^2 -3x+6-2x.

IV) Representation of x^2 +x -6·

To represent this we need 1 square tile representing x^2 , 5 algebra tiles representing x and 6 algebra tiles representing 1.·

By splitting the middle term of the given polynomial we get the expression x^2 +3x -2x - 6.·

Place a square tile of dimension 10X10 representing x^2 .·

Add 3 tiles of dimension 10 X 1 each to any side of the tile x2.The area of new shape formed represents x^2 +3x.· Subtract 2 tiles of dimension 10 X 1 each from the side adjacent to the previous side.

The area of new shape formed represents x^2 +3x-2x.· Subtract 6 tiles of dimension 1 X 1 each to complete the rectangle.

The area of new shape formed represents x^2 +3x-2x-6.

· Observation· In the representation of x^2 +5x + 6 , a rectangle is formed whose sides are (x+3) and (x+2) which are the factors of it.· In the representation of x^2 -x - 6 , a rectangle is formed whose sides are (x-3) and (x+2) which are the factors of it.· In the representation of x^2 -5x + 6 , a rectangle is formed whose sides are (x-3) and (x-2) which are the factors of it.· In the representation of x^2 +x - 6 , a rectangle is formed whose sides are (x+3) and (x-2) which are the factors of it.

Result:Thus we have observed that in all the four cases a rectangle is formed whose sides are the factors of the given polynomial.

class VI (integers)

Objective: Addition of two numbersexperimentally through paper cutting and pasting.

Previous knowledge: On addition of a positive integer to its negative integer gets nullified (or zero).

Procedure:

step 1: Firstly cut 25 unit squares which are kept unshaded and call them positive squares.

step 2: Then cut 25 unit squares which are kept shaded and call them negative squares.

step 3: case 1: Now, take (+8) + (+7)

we take 8 positive squares and 7 positive squares

count them (+8) + (+7) = 15

step 4: Case 2: now take,(-4) + (-5)

we take 4 negative squares and 5 negative squares

count them (-4) + (-5) = -9

Step 5: Case 3:Now take (+6) + (-5)

we take 6 positive squares and 5 negative squares

Count them (+6) + (-5) =1

Step 6: Case 4: Now take (-5) +(3)

we take 5 negative squares and 3 positive squares

Count them (-5) + (+3) = -2

Result:(i) If both the integers are positive then we add and the result is an positive integer.

(ii) If both the integers are negative then we add and the result is an negative integer.

(iii) If positive integer is greater than the negative integer then we subtract and the result is an positive integer.

(1v) If negative integer is greater than positive integer then we subract and the result is an negativ integer.

her sand

Assignment: -8 + 6 =

-4 - 5 =