Isosceles triangle theorem

Hello friends. In this blog we will learn about the isosceles triangles and the theorems related to such triangles. The isosceles triangles are used in the geometry of grade IX standard of the geometry mathematics. As per the definition of the isosceles triangles, the isosceles triangles are those triangles whose two sides are congruent to each other; this means that a triangle which has its two sides equal is an isosceles triangle. Now if the two opposite sides of any triangles are equal then the both opposite angles will also be equal in the magnitude. The figure of the isosceles triangles can be given as:

Now, let us talk about the isosceles triangle theorems. The theorem for isosceles triangles states that the angles apposite the both equal sides are also equal. The isosceles triangle theorems with their relative proof can now be given as:
Theorem: The first theorem for isosceles triangle states that if the two sides of a triangle are congruent (Congruent stands for similarity in the shape and size of the sides in triangle) then the angles opposite them will also be congruent. To prove the theorem, let us draw a triangle XYZ in which two of the sides (side XY and side YZ) of triangle are congruent to each other. For the purpose of proving let us draw a bisector line YM, which intersects the base line XZ at a mid point M. Now the two triangles XYM and ZYM in the original triangle XYZ are congruent to each other. They are congruent because the line YM is common between them and also congruent sides for both triangles and that’s why the other lines XY and YZ both are also congruent.  The angles between them (angle XYM and angle ZYM) are also congruent to each other and from figure angle YXZ and YZX are both corresponding angles of the triangles XYM and ZYM. So, they are also congruent. Hence it is proved that the angles opposite the both equal sides are also equal. (To get help on ICSE Board Syllabus click here)

The same theorem in some other words can also be explained as, that if two opposite angles of a triangle are equal then the opposite sides will also be congruent and In the next session we will discuss about time and angle measures. 

Rectangular coordinate system in Grade IX

In this section, we will discuss about the topic Rectangular Coordinate System, which you need to study in grade IX.
Mainly there are two types of coordinate System.
1-cartesian coordinate System
2-polar coordinate System

Cartesian coordinate system:
As you can see in above figure, every point on x Axis and y axis is shown uniquely in the plane. Every point either on x axis or on y axis are at equal distance from each other. If you are given 3 Cartesian coordinate in the plane you can determine the exact position of point with the help these coordinates. Using coordinate system, geometrical things like curve can be described by Cartesian equation.
For example, a circle of radius 16 can be described in x and y coordinate can be written as x²+y²=4.
There are 3 types of dimensions
One dimension System-
In one dimension only x coordinate is there rest coordinate are 0
For example-(5, 0, 0) here dimension of x is 5 and dimension of y & z is 0
Two dimension System-
For two dimensions there are two coordinate one is in x direction and other is in y direction and z coordinate is 0.
For example – (5, 4, 0) here dimension of x is 5 ,dimension of y is 4 & dimension of z is 0.
Three dimension System-
For three-dimension all the 3 coordinate are present
(4, 5, 6)
Here dimension of x is 4, dimension of y is 5 and dimension of z is 6.
Now we will move to next coordinate system that is polar coordinate system
In this system the given points are in two dimensions and each point in the plane is determined by a distance and direction is decided by angle. The point which is fixed is called is called pole. And the ray from the pole in the fix direction is called polar axis. The distance from the pole is called radius and angle is called polar angle. The radial coordinate is given by r and the angular coordinate is given by θ or t. Angle in polar System are generally expressed as degree or radian (1 radian=360degree).
Conversion of polar coordinate to Cartesian coordinate:
The two coordinate r and θ can be converted to x and y coordinates by the following formulas X=rcosθ and y=rsinθ. (To get help on cbse sample papers click here)
In the next topic we are going to discuss Rate, distance, time, angle measures, and arc lengths and In the next session we will discuss about Isosceles triangle theorem. 

Math Blog on Planar cross-sections, perpendicular lines and planes

A plane section consists of two-dimensional flat surface. The two dimensional analogues are the points with zero-dimensions, points with one-dimension and also with three-dimensional plane. From the theory of Euclidean geometry planes are seen as the parts of that space with higher dimensional space.
If we take a two-dimensional Euclidean space then we can say that it is a whole space and most of tasks in geometry and trigonometry are performed by using it. Planar cross sections are the important concepts in math. When a plane surface can be crossed by the intersection of a solid plane then it is known as the cross section. Some properties of planar cross section are given below,
Two planes either intersect in a line or they are parallel.
Two lines are parallel to a plane surface, they are contained in the same plane or meet at a single point.
Two lines are said to be parallel to each other if these lines are perpendicular to each other.
Perpendicular lines: If two lines intersect then four angles are formed at the intersecting points of the two lines. The condition of perpendicular line is that two lines are said to be perpendicular if four angles which are formed are equal and also if the slope of one is the negative reciprocal of the other then two lines are perpendicular. Let us suppose that if the slope of one line is n, then slope of the other is equal to the negative reciprocal that is -1/n. The perpendicular makes a right angle or 90° and it shows a relationship between two lines. On other hand parallel Lines are those lines which never intersect each other and so that parallel lines have the same slope. Perpendicular lines are those lines that intersect only at right angles. Because of this reason product of slopes of parallel lines is equal to – 1. (To get help on CBSE Sample papers click here) and In the next session we will discuss about Rectangular coordinate system in Grade IX. 

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