How to tackle problem based on Rate?

Hello friends’ Previously we have discussed about longhand subtraction and today we are going to discuss a very interesting topic Rate and Work which is included in grade IX of Maharashtra Board Syllabus and In my opinion it can play a vital role in your Homework help, friends rate and work problems involve the calculations of time in which a work can be done, by a single person or machine or together by a group of two or more people or machines, friends’ rate represents the capacity of a person or machine of doing work with respect to time, friends’ the method of solving work and rate problems is not obvious, there is a trick to solve the rate and work problems, and the trick is we have to think of the problem of how much work  each person or machine does in each unit of time, rate and work problems can better be understood by taking some examples:
Example.  Suppose a painter can paint a entire house in 12 hours, and the second painter takes 8 hours to paint it, how long will it take the two painters to paint the entire house together,
Friends’ here we can proceed in the following way, as first painter can paint the house in 12 hours,
So here first of all we will find his rate of doing  work which can be find out by dividing 1 by 12,
Here it means that the first painter can do 1/12^th part of the job per hour similarly for the second painter, we can find his rate of doing work by dividing 1/8, here it means that second painter can do 1/8^th part of the job per hour, now to find out how much can they do if they work together, to find this we have to add their individual rate of doing work
So adding their rates we obtain   1/12 + 1/8 = 5/24 which means that if they work together than they can do 5/24 of the job together per hour.
Now let’s say‘t’ is the time taken to complete the job if they work together,
So work completed per hour will be equal to 1/t which is also equal to 5/24 as we have calculated above, now equating to expressions we obtain   1/t  =  5/24, or
                                                                                         t  =  24/5  =4.8 hours
Now we can say that they will take 4.8 hours to complete the job if they work together,
Example: one pipe can fill a pool 1.25 times faster than a second pipe, when both the pipes are opened they fill the pool in five hours. How long will it take to fill the pool if only the slower pipe is used?
Now again we proceed in the same we lets say faster pipe can fill the pool in t hours, that is why slower pipe will fill the pool in 1.25t hours,
Now converting in their rate form faster pipe can do 1/t job per hour and slower pipe can do 1/1.25t job per hour and together they can do 1/t + 1/1.25t  job per hour which should be equal to 1/5
Now equating the two expressions we obtain     1/t + 1/1.25t = 1/5 or
                                                                                   5  + 5/1.25   =  f  or  f =  5+ 4 =  9
Hence faster pipe will take 9 hours to fill the pool and the slower pipe will take
1.25f = 1.25×9 = 11.25 hours to fill the job.

This is all about the problem based on Rate and if anyone want to know about Isosceles triangle theorem then they can refer to internet and text books for understanding it more precisely.Read more maths topics of different grades such as symmetry in the next session here.

How to Tackle Measurements problems of Geometric Figures

Hello Friends, in today's session  we all are going to discuss about one of the most interesting and a bit complex topic of mathematics, geometry. In  grade IX of every education board ,we  get  familiar  with  algebraic expressions  of  mathematics , but  not  so  familiar  with geometry  .  First  of  all,  we  discus  what is  geometry ? Basically  geometry  is  a  mathematical  branch   and  in  this   branch   we  study  about  the designing of different shapes, calculation of their area like area of circle, depth, volumes, height, etc.  In this  session  we  learn  how to measure  geometrical  figure , like  square,  rectangle  etc.  and  introduction  about  composite  figures  .  First  of  all  we  learn   Measurements of geometric figures  here  measurement means finding  distance , area  and  volume  of  geometric figures  and  so  we  divide  measurement  of  geometrical  figures  in  three  units ,For more measurement related problems click here,

• measuring   distance 
• measuring   area  of   geometrical  figures
• measuring  volume of  geometrical  figures

 Initially   we  discuss  standard  formula  to   measuring  distance  :
    if  there is  two  points P( x₁ , y₁ )  and   Q(x_{2 ,}y₂)  in  a  plane ,   then  formula  to   measuring distance  between  these  two  points   PQ  is  –
        sqrt ((x₂ – x₁ )²  +  ( y₂ –  y₁ )² )
        here   sqrt  means  square  root
Next  we   proceed  to  discuss  standard  formulas  to   measuring  area  of   geometrical  figures  :
1.   Area  of   triangle     =    1    *   b  *  h
                                             2
            here   b  refers    base  length   of  triangle   and   h  refers  to  height  length  of  triangle
2.   Area  of   square       =     a²
            here  a  refers  to  one  side  of  square
3.   Area  of  rectangle   =      a* b
                here  a  refers  to  length  of  rectangle and  b  refers  to  width  of  rectangle
4.   Area   of   parallelogram   =   b  * h
               here  b  refers  to  one  side  length  and  h  refers  to  width  between  two  sides
5.   Area  of   circle   =   pi * r²
               here  pi = 3.14  and   r  refers  to  radius  of  circle

6.   Area  of  eclipse  =  pi * r₁  * r₂
               here  pi = 3.14  and  r₁  and  r₂ are  two side  radius  of  eclipse

Next   we  are  going  to  discuss standard  formula  to  measuring  volume  of  geometrical  figures –
1.   Volume  of  cube  =  a³ 
            here  a  refers  to  each  side  of  cube
2.   Volume  of  Cuboid  = a * b * c
            here  a  shows  length , b  shows width  and  c  shows  height
3.   Volume  of  cylinder  =  pi *  r² * h
            here  pi = 3.14  , r  refers  to  radius  of  cylinder  and  h  refers  to  height  of  cylinder
4.    Volume  of  pyramid   =   1    *  b *  h
                                                3
           here  b  refers  to  width  of  pyramid  and  h  refers  to  height  of  pyramid
5.    Volume  of   sphere    =    4    *   pi  *  r² *  h
                                                3
          here  pi = 3.14  ,  r   refers  to  radius  of  sphere  and   h  refers  to  height  of  sphere
This  is  all  about  standard  measurement  formula  of  geometrical  figures .
Now ,  we  proceed  to  our  next  topic -  composite figures
We know that in  geometry ,  there  are  some  standard  figures   and  combination  of  these  standard  figures  make   composite figures  like  -

This  is  a  composite figure  because  it  includes  square  and  rectangle  in  one  figure .
So,  in  this  session  we  learn  how  to  measure  geometrical  figure  and  we  learn  basic  information  about  composite figures  and to know about Arithmetic and geometric series and Basic constructions grade IX refer Internet.
 

Geometrical Proofs in Grade IX

Hello friends today’s topic is Geometry. In Geometry we will learn what is Geometry.
We will also see some  Geometrical proofs. I think there are many students that always struggle in Geometrical proofs. So today I’ll try to explain it to you using examples and diagrams.
As a Grade IX student you should first know what is Geometry?
Geometry is every where around us whether it is man made thing, nature, arts, sports etc.
Geometry is basically about shapes and sizes based upon four imaginary ideas- point, line, plane and space.

Geometrical proofs illustrate step by step conclusion of a geometric statement using definitions, postulates, axioms and previously proved theorems.
There are two major types of  proofs:

Direct Proof is based on deductive reasoning in which the conclusion is made directly from the previous conclusions.

Indirect Proof includes the negative assumption of the statement our end result leads to contradiction which states that the inverse of assumed statement is true.

Geometrical proofs can be written either in two columns or in paragraph.
In two columns geometrical proofs we have two columns one consists of list of statements and other consists of the reasons which states why the statement is true.

Statements

Reason

Paragraph proof is only the two column proof written in paragraph.

Example 1:

Given: DEFG is a parallelogram.
Proof: angle G and E are congruent.

Solution:

Statements	Reason
1.DEFG is parallelogram	Given
2.DE parallel to GF	Definition of parallelogram
3.Angle GDF congruent to angle EFD	When transversal cuts parallel line, interior angles are congruent.
4.Angle GFD congruent to angle EDF	When transversal cuts parallel line, interior angles are congruent.
5.Segment DF is congruent to itself	Reflexive proporty
6.Triangle GDF congruent to triangle EFD	From postulate 3, 4, 5
7.Angle G and E are congruent	Corresponding parts of congruent triangles are congruent.

Example 2:
Proof that area of a circle is pi*r² .

Solution:
Here we prove that area of a Circle ( A ) is pi * r².
          
            A = pi * r²

We will proof this by using some images as it is easy to understand.



Do you find any difference in both the images? No ! Look the area outside the circle but inside the square reduced as we move from 4 side square to 8 side octagon.
Now if we increase the number of sides, the space will become very small.
Keeping that statement in mind let’s look at the following image:

  

from the image above area of triangle AOB is :
            AOB = ½(base * height) = ½(s*r)
Where r = radius

As outer shape is octagon so we can create 8 triangles inside it. As the image below:

           

Now the area of octagon is:
                                    8* (1/2 (s*r ) )  =   1/2r * 8s
8s is the perimeter of octagon.

And as we have stated before that as we increase the number of sides to n resulting in n-gon will look like exactly same as circle. So that the perimeter of the octagon is almost same as the perimeter of circle.

Hence we can replace the 8s by  2 x pi x r.
Which is the perimeter of the circle. For calculating area of circle the number of side is very large.
By doing it (increasing the no of sides), the area of circle will be:
            1/2 r * 2 * pi * r = pi * r²

Remember this is what we are trying to prove.
So The Area of circle is = pi * r².

Gauss-Jordan elimination method in grade IX

Hello friends today we are going to learn the Gauss-Jordan Elimination method. Usually this method we study in grade IX. And also we try to Solve Linear Equations Using Gauss-Jordan Elimination Method. Gauss Jordan is a method to solve the system of linear equation. Gauss Jordan Elimination is the alternative of Gauss Elimination. In Gauss Jordan Elimination we transform the coefficient matrix into the matrix called augmented matrix, which is much easier to solve and gives the same solution as the original set of linear equation and solve the matrix by converting it into unit matrix.
Now question arises what is augmented matrix and unit matrix.

Augmented matrix is the matrix created by adding an additional column for the constants on the right of the equal sign. That is by appending columns of two matrices.
Example:
Suppose we have two matrices A and B, as:
A=1 4 5                       B =  9
     3 7 9           and             8
     6 2 8                              6
Augmented matrix ( A|B ) can be written as:
(A|B) = 1 4 5  | 9
             3 7 9 | 8
             6 2 8 | 6

Unit matrix is the matrix in which all the elements are 0 (zero) expects the diagonal elements. The diagonal elements are 1.
Example:
            1 0 0
            0 1 0
            0 0 1

We perform some elementary operations to solve the equation:

1. Row exchange.
2. Row replacement
3. Row scaling

In order to Solve Linear Equations Using Gauss-Jordan Elimination Method we:

1. represent the system as augmented matrix form.
2. perform elementary row operation on augmented matrix A|B to change it into diagonal form with each diagonal element having a non zero value. If any diagonal element has a zero, switch the row until a non zero value found. If after many efforts we are unable to find some then the system may have either infinite or no solution. Stop then.
3. Divide each diagonal and right hand side element of each row as each diagonal is equal to 1.
4. This form of row is called the reduce echelon form.

Example:
              2b + c=4
            a+b+2c=6
            2a+b+c=7
Step 1:
            Write the system in augmented matrix form.
                        0 2 1 | 4           (R1)
                        1 1 2 | 6           (R2)
                        2 1 1 |7            (R3)
Step 2:
            Perform elementary operations to convert augmented matrix into diagonal form.

                        1 1 2 | 6           (R2)
                        0 2 1 | 4           (R1)
                        2 1 1 | 7           (R3)


                        (R3)+(-2*R1)
                     
                        1   1   2 | 6
                        0   2   1 | 4
                        0 -1  -3 | -5

                        (R1)+(-1/2*R2) and (R3)+(1/2*R2)

                        1  0  3/2    | 4
                        0  2   1      | 4
                        0  0   -5/2  | -3

                        (R1)+(3/5*R3) and (R2)+(2/5*R3)

1    0      0   |  11/5
0    2      0   | -14/5
0    0  -5/2   |  -3

As in two rows we still have values which are not equal to 1. So we now try to convert in into row echelon form.

(R2)*(-1/2) and (R3)*(-2/5)

1   0   0  |  11/5
0   1   0  |   7/5
0   0   1  |   6/5


The right hand side of the augmented matrix is our result. So we have

a=11/5
b=7/5
c=6/5