Learn Different Types of Triangles

A triangle is one of the basic shapes of geometry: a polygon with three corners or vertices and three sides or edges which are line segments. A triangle with vertices A, B, and C is denoted ABC.

            

In Euclidean geometry any three non-collinear points determine a unique triangle and a unique plane (i.e. a two-dimensional Euclidean space). (Source : Wikipedia)

Understanding Types of Manipulation is always challenging for me but thanks to all math help websites to help me out. 

List of learndifferent types of triangles

Different types of triangles classified by sides:

1. Equilateral triangle
2. Isosceles triangle
3. Scalene triangle.

Different types of triangles classified by angles:

1. Acute
2. Right
3. Obtuse

Brief description for learn different types of triangle:

Learn Equilateral triangles:

                 

Equilateral triangle is the one of the different types of triangles. In equilateral triangles all sides are equal length. The equilateral triangles are regular polygons. The equilateral triangles each angle is measuring 60°.

Learn Isosceles triangles:

                       

Isosceles triangle is the one of the different types of triangles. These isosceles triangles two angle are equal measure. The Isosceles triangles in any of the two sides are equal length. These isosceles triangles the base angles are the angles opposite the two equal sides.

Learn scalene triangles:

               

Scalene triangles are the one of the different types of triangles. The Scalene triangles don’t usual triangles.  Some triangles are drawn at random would be scalene. The scalene triangles interior angles are always different measure. Scalene triangles all three angles are different measure and all the sides are different lengths.

My forthcoming post is on 10th model question paper 2012 samacheer kalvi will give you more understanding about Algebra

Learn acute triangles:

               

Acute triangles are one of the different types of triangle. It’s classified by its angles. Acute triangles each angles are less than 90 degree.

Learn Right triangles:

                   

Right triangles are the one of the different types of triangles. The right triangles are also called as right angle triangles or rectangle triangles.  This right triangles one angle must be 90 degree. The right angle triangles opposite sides are the hypotenuse; hypotenuse is the longest side of the right triangle. Remaining two sides are called as legs. In right triangles the lengths of the sides are connected by the Pythagorean Theorem.

Learn obtuse triangles:

                    

Obtuse triangle is one of the different types of triangles. It’s classified by its angles. Triangles any one angle is measures more than 90° is an obtuse triangle or obtuse-angled triangles.

Solve for Cube Root

In mathematics, a cube root of a number, denoted (^3sqrt (x) or x^1/3, is a number such that a³ = x. All real numbers have exactly one real cube root and a pair of complex conjugate roots, and all nonzero complex numbers have three distinct complex cube roots. For example, the real cube root of 8 is 2, because 2³ = 8.                                                                                                       (Source: Wikipedia)

Example to solve for cube root:

Let us see some of the example for solve for cube root for better understanding,
Example 1: solve for cube root `(^3sqrt (512))`
Solution:
Given: `(^3sqrt (512))`
Here, 512 is obtained by multiplying 8 three times. That is,
512 = 8 * 8 * 8`
(^3sqrt (512))` ` =` `(^3sqrt (8))`
= 8 `
Answer: solve for cube root `(^3sqrt (512)) = 8`

Example 2: solve for cube root` (^3sqrt (10^2 * 10^7))`
Solution:
Given: `(^3sqrt (10^2 * 10^7))`
(10^2 * 10^7) = (10 * 10) * (10 * 10 * 10 * 10 * 10 * 10 * 10)`
= (10 * 10 * 10) * (10 * 10 * 10) * (10 * 10 * 10) `
= 10^3 * 10^3 * 10^3 `
(^3sqrt (10^2 * 10^7)) = (^3sqrt (10 ^3* 10^3 * 10^3)) `
= 10 * 10 * 10 `
= 1000`
 Answer to solve for cube root:  `(^3sqrt (10 * 10 * 10)) = 1000`

Example 3: solve for cube root` (^3sqrt (17^2 * 17^4))`
Solution:
Given: `(^3sqrt (17^2 * 17^4))`
(17^2 * 17^4) = (17 * 17) * (17 * 17 * 17 * 17)`
= (17 * 17 * 17) * (17 * 17 * 17)`
= 17^3 * 17^3 * 17^3 `
(^3sqrt (17^2 * 17^4)) = (^3sqrt (17^3 * 17^3 * 17^3)) `
= 17 * 17 * 17 `
= 4913`
Answer to solve for cube root:  `(^3sqrt (17 * 17 * 17)) = 4913`

Example 4: solve for cube root  ` (^3sqrt (1/729))`
Solution:
Given: `(^3sqrt (1/729))`
1/729 = 1/8 * 1/8 * 1/8`
(^3sqrt (1/729)) = (^3sqrt (1/8^3))`
= 1/8 `
Answer: solve for cube root  `(^3sqrt (729)) = 1/8`

Example 5: solve for cube root  `(^3sqrt (1/8))`
Solution:
Given:  `(^3sqrt (1/8))`
1/8 = 1/2 * 1/2 * 1/2`
(^3sqrt (1/8)) = (^3sqrt (1/2^3))`
= 1/2 `
 Answer: solve for cube root  `(^3sqrt (1/8)) = 1/2`

Example 6: solve for cube root `(^3sqrt (343))`
Solution:
Given: `(^3sqrt (343))`
Here, 512 is obtained by multiplying 8 three times. That is,
343 = 7 * 7 * 7`
(^3sqrt (343))` ` =` `(^3sqrt (7^3))`
= 7 `
Answer: solve cube root `(^3sqrt (343)) = 7`

I am planning to write more post on Trig Reference Angles and tamil nadu samacheer kalvi 10th books. Keep checking my blog.

Practice problem for solve for cube root:

Practice problem for solve for cube root 1: `(^3sqrt (125))`
Answer: `(^3sqrt (125)) = 5`

Practice problem for solve for cube root 2: `(^3sqrt (216))`
Answer:` ( ^3sqrt (216)) = 6`

Practice problem for solve for cube root 3: `(^3sqrt (1331))`
Answer: `(^3sqrt (1331)) = 11`

Practice problem for solve for cube root 4:` (^3sqrt (1728))`
Answer: `(^3sqrt (1728)) = 12`

Practice problem for solve for cube root 5: `(^3sqrt (3375))`
Answer: `(^3sqrt (3375)) = 15`

Practice problem for solve for cube root 6: `(^3sqrt (27))`

Answer: `(^3sqrt (27)) = 9`

Graph Co-ordinates System

In geometry, a coordinate system is a system which uses a set of numbers, or coordinates, to uniquely determine the position of a point or other geometric element. The order of the co-ordinates is the significant and they are sometimes identified by their position in an ordered tuple and sometimes by a letter, as in 'the x-co-ordinate'. In elementary mathematics the co-ordinate are taken to be real numbers.    (Source : Wikipedia)

I like to share this The Rectangular Coordinate System with you all through my article. 

Examples for graph co-ordinates system:

Example 1:

Draw the graph of the line joining the co-ordinate points (2, 3) and (−4, 1).

Solution:

 Draw the x-axis and y-axis on a graph paper sheet and take 1 cm = 1 unit on both the axes. Let A and B be the co-ordinate points (2, 3) and (−4, 1). We mark these points on the graph paper sheet. We join the points A and B by a line  and extend it along the two directions. The required graphing is now obtained

Example 2:

Draw the graph of the line joining the co-ordinate points (-3, 3) and (4, 3).

Solution:

 Draw the x-axis and y-axis on a graph paper sheet and take 1 cm = 1 unit on both the axes. Let A and B be the co-ordinate points (-3, 3) and (4, 3). We mark these points on the graph paper sheet. We join the points A and B by a line  and extend it along the two directions. The required graph is obtained

Example 3:
Draw the graph of the line joining the co-ordinate points (-2,0) (0,3)

Solution:

 Draw the x-axis and y-axis on a graph paper sheet and take 1 cm = 1 unit on both the axes. Let A and B be the co-ordinate points (-2, 0) and (0, 3). We mark these points on the graph paper sheet. We join the points A and B by a line  and extend it along the two directions. The required graphing is now obtained

I am planning to write more post on 10th std samacheer kalvi question bank. Keep checking my blog.

Practice problem for graph co-ordinates system:

1. Draw the graph of the line joining the co-ordinate points (9, 3) and (21, 3).

2. Draw the graph of the line joining the co-ordinate points (5, 2) and (−4, 7).

Calculate the Area of Ellipse

In geometry, an ellipse is a plane curve that is formed as a result of conic section (ie) the intersection of a cone by a plane produces a closed curve called ellipse the figure is shown below. 

Ellipse has one major axis and one minor axis. Circle is a special case of ellipse where the major axis and minor axis are same. Here in this topic we are going to see more problems regarding area of ellipse.

Formula for area of the ellipse:

Area of Ellipse = `pi` * length of major axis * length of minor axis
in other words
Area of the ellipse = `pi`  * w * h

Area of the Ellipse - Example problems:

Example: 1
Calculate the area of the ellipse where the major radius is 5 cm and minor radius is 3 cm.
Solution:
w = 5  h= 3
Area of the ellipse = `pi` * 5 * 3
                           = 3.14 * 5 * 3
                             = 47.1 cm²
Example: 2
Calculate the area of the ellipse where the major radius is 6 cm and minor radius is 3 cm.
Solution:
w = 6  h= 3
Area of the ellipse = `pi` * 6 * 3
                             = 3.14 * 6 * 3
                             = 56.52 cm²
Example: 3
Calculate the area of the ellipse where the major radius is 4 cm and minor radius is 2 cm.
Solution:
w = 4  h= 2
Area of the ellipse = `pi` * 4 * 2
                            = 3.14* 4* 2
                            = 25.12 cm²
Example: 4
Calculate the area of the ellipse where the major radius is 7 cm and minor radius is 3 cm.
Solution:
w = 7  h= 3
Area of the ellipse = `pi` * 7 * 3
                           = 3.14 * 7 * 3
                             = 65.94 cm²
Example: 5
Calculate the area of the ellipse where the major radius is 8 cm and minor radius is 3 cm.
Solution:
w = 8  h= 3
Area of the ellipse = `pi` * 8 * 3
                            = 3.14 * 8 * 3
                            = 75.36 cm²

I am planning to write more post on The Exponential Function and  samacheer kalvi tamil book. Keep checking my blog.

Example: 6
Calculate the area of the ellipse where the major radius is 4 cm and minor radius is 3 cm.
Solution:
w = 4  h= 3
Area of the ellipse = `pi` * 4 * 3
                            = 3.14 * 4* 3
                            = 37.68 cm²

Area of a Standard Triangle

Area of a shape is to find out the amount of space it consumes two dimensionally. In math area of different shapes are studied and specific formulas have been derived to calculate problems on area of some frequently used shapes like triangle, square, cylinder etc. Area is measured in square units. Here we are going to study about the area of standard triangles. A standard triangle is one which as sides in such a way that, the sum of two angles gives the third angle. Here we are going to learn how to calculate the area of  a standard triangle.

Standard triangle:

A triangle which has the angles in such a way that, the sum of two smaller angles give the third angle. It can also be defined as the triangles having sides in such a way that the sum of the square of two smaller sides is equal to the square of the bigger side (i.e. satisfying Pythagoras theorem). The formula for finding the area of  standard triangles is given below,

                                  Area of a standard triangle = `1/2` b * h
                                  where,
                                   b = base width of the triangle,
                                   h = height of the triangle.

Example problems on standard triangles:

Here are few examples illustrating the calculation of area for a standard triangle,

Example 1:
Find the area of the standard triangle shown in the figure.

Solution:
From the figure it is clear that,
b = 5cm
h = 12 cm.
Area of the triangle = `1/2` (b)* (h)
                                   = 1/2 (5) (12)
                                   = 60/2
                                   = 30 cm²

Algebra is widely used in day to day activities watch out for my forthcoming posts on Solving Nonlinear Differential Equations and samacheer kalvi books for 10th. I am sure they will be helpful.

Example 2:
Find the area of the standard triangle shown in the figure.

Solution:
From the figure it is clear that,
b = 3cm
h = 4cm.
Area of the triangle =` 1/2` (b)* (h)
                                   = 1/2 (3) (4)
                                   = 12/2
                                   = 6 cm²

Types of Angles Geometry

In geometry, an angle is the figure formed by two rays sharing a common endpoint, called the vertex of the angle. The magnitude of the angle is the "amount of rotation" that separates the two rays, and can be measured by considering the length of circular arc swept out when one ray is rotated about the vertex to coincide with the other.(source: WIKIPEDIA). In this article we see the types of angles in geometry. There are nine types of angles in geometry.

Types of geometric angles with diagrams

Types:
Acute angle:
An angle whose measure is smaller than 90 degrees, then the angle is acute angle. The following is an acute angle.
                                                                
Right angle:
An angle whose measure is exactly 90 degrees, then the angle is right angle. The following is a right angle.
                                                                
Obtuse angle:
An angle whose measure is above 90 degree and also less than 180 degree. Thus, it is between 90 degrees and 180 degrees, and then the angle is obtuse angle. The following is an obtuse angle.
                                                             
Straight angle:
An angle whose measure is 180 degrees. Thus, a straight angle looks like a straight line. The following is a straight angle.
                                                              
Reflex angle:
An angle whose measure is above 180 degrees but less than 360 degrees, then the angle is reflex angle. The following is a reflex angle.
                                                              
Adjacent angles:
Angle with a common vertex and one common side. <1 adjacent="" and="" angles.="" are="" nbsp="" p="">                                                              
Complementary angles:
Two angles whose measures add to 90 degrees. Angle 1 and angle 2 are complementary angles, as together they form a right angle.
Note that angle 1 and angle 2 do not have to be adjacent to be complementary as long as they add up to 90 degrees.
                                                               
Supplementary angles:
Two angles whose measures add to 180 degrees. The following are supplementary angles.
                                                                

Types of geometric angles with examples:

Vertical angles:
Angles that have a common vertex and whose sides are formed by the same lines. The following (angle 1 and angle 2) are vertical angles.
                                                              
When two parallel lines are crossed by another line is called Transversal, 8 angles are formed. Take a look at the following figure.
                                                               
Angles 3,4,5,8 are interior angles
Angles 1,2,6,7 are exterior angles

Alternate interior angles:
In transversal pairs of opposite sides are interior angles.
Such as, angle 3 and angle 5 are alternate interior angles. Angle 4 and angle 8 are also alternate interior angles.

Alternate exterior angles:
In transversal pairs of opposite sides are exterior angles.
Angle 2 and angle 7 are alternate exterior angles.

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Corresponding angles:
Pairs of angles that are in similar positions.
Angle 3 and angle 2 are corresponding angles.
Angle 5 and angle 7 are corresponding angles

Practice for geometry angles:
1) Angle 180° is -------- a) Straight angle b) right angle Answer: Straight angle
2) Angle less than 90° is ------------ a) obtuse angle b) acute angle Answer: Acute angle

Volume of Triangular Prism

Geometry deals with shapes, structures, lines, planes and angle’s. Geometry learning is also known as architectural learning. Basic shapes of geometry are square, triangle, rectangle, parallelogram, trapezoid etc. Triangle is one of the basic shapes in geometry. The total internal angle of the triangle is 180 degree.

Prism:

In mathematics, prism is one interesting shapes in geometry. It is a polyhedron of n-sided prism which has a base of n-sided polygonal, copy of a translated which has joining n faces with the sides.

Definition of volume:

In mathematics, Volume is one interesting measure of geometry. It is an amount of three-dimensional space of an object occupies.

Formula for finding the Volume of triangular prism:

The formula used for finding the volume of triangular prism, is given as,

Volume of triangular prism = Base area x Length of prism

Base area of prism =` (bh)/2`

Where, b = base of the triangle

            h = height of the prism.

           First the base area of the triangular prism is to be calculated. Using that value the volume of the prism can be calculated.

Volume of triangular prism Problems:

Example 1:

Find the volume of the triangular prism, whose base is 7 cm, height is 9 cm and length is 12 cm.


Solution:
The formula used for finding the volume of triangular prism, is given as,
Volume of triangular prism = Base area x Length of prism
Given: b = 7 cm, height = 9cm and length = 12cm
Base area =` (bh)/2`
                = `(7*9)/2`
                = `63/2`
                = 31.5 cm²
Volume     = 31.5*12
                = 378 cm³.
The answer is 378 cm³.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Area of an Octagon and ix std cbse question papers. I am sure they will be helpful.

Example 2:
Find the volume of the triangular prism, whose base is 5.5 cm, height of the prism is 9 cm and length is 7 cm.



Solution:
The formula used for finding the volume of triangular prism, is given as,
Volume of triangular prism = Base area x Length of prism
Given: b= 5.5 cm, height = 9 cm and length = 7 cm.
Base area = `(bh)/2`
                = `(5.5*9)/2`
                = `49.5/2`
                = 24.75 cm².
Volume     = 24.75*7
                = 173.25 cm³.
The answer is 173.25 cm³.         

Application Integration Standards

Calculus has two parts. There are integrtaion and differentiation. The definition of differentiation is "rate of change of input". The Process of differentiation is called integration. For example, f(x) is a function, The integration of given function  can be  expressed as `int` f(x) dx .Here, f(x) is a continuous function. Integral calculus has two types. One is definite and indefinte integral. In this article, we shall discuss about application of standard integration.

Basic standard integration formulas:

1. `int` x ⁿ dx = `(x^n+1) / (n+1)`
2.`int`cos x .dx = sin x + c
3. `int`sin x.dx  = - cos x + c
4.`int`sec²x dx = tan x + c
5. `int`cosec x . cot x .dx = - cosec x + c
6. `int` sec x . tan x. dx = sec x + c
7. `int`cosec² x . dx = -cot x + c
8.`int`a f(x) dx = a `int`f(x) dx
9. `int`[f(x) ± F(x)] dx =`int` f(x) dx ± `int`F(x) dx
10. `int` `(1/x)` dx = ln x + c

These above formulas helps to solve  the standard integration. Area and Volume of the sphere, cylinder, average value these are the main application of integrals.

Application integration standards - problems:

Application integration standards - problem 1:
Determine the area of shaded portion of given graph.

                        
     Solution:
Given  y = x²
The boundary limits are 2 and 5
So we can integrate the given equation is  `int_a^b ` x² dx.
we know the lower limit is 2 and upper limit is 5.
So,   `int_2^5 ` x² dx. =  `[(x^3)/3]_2^5`
= `[(5^3)/3]`.- `[(2^3)/3]`.
= `125/3`  -  `8/3` .
= `(64 - 1)/3` .
= `117/3`     .
= 39 .square units.

Answer:    Area of shaded portion is 39 square units.    

If you have problem on this topics please browse expert math related websites for more help on cbse website.

Application integration standards - problem 2:

Determine the total area of graph x³ + x² - x and the x axis. the points x = -1   and x = 2

           
    Solution:
Given  x³ + x² - x
the limits are   x = -1  and x = 2
In the above graph x axis -1< x< 0  and 0 < x < 2.
So, the total area = `int_(-1)^0` [x³ + x² - x] dx + ` int_0^2` [x³ + x² - x] dx.
`int_(-1)^0` [x³ + x² - x] dx  = ` [(x^4/4) + (x^3)/3 - ((x^2)/2)]_(-1)^0`
= `[0] - [(-1)^4/4 + (-1)^3/3 - ((-1)^2/2)] `
= ` - [1/4 - 1/3 - 1/2] `
= ` - [(3-4-6)/12] `
= ` - [(-7)/12] `
=   ` [(7/12)] ` .

int_0^2` [x<sup>3</sup> + x<sup>2</sup> - x] dx.      = `[(x^4/4) + (x^3)/3 - ((x^2)/2)]_0^2` ..
                                            = `[(2^4/4) + (2^3/3) - (2^2/2)] - [0]`    .
                                            = `[(16/4) + (8/3) - (4/2)]`    .
                                            = `[(4) + (8/3) - (4/2)]`    .
                                            =  `[(24 + 16 - 12)/6]`  =   `[(28/6)]` .
           `int_(-1)^0` [x<sup>3</sup> + x<sup>2</sup> - x] dx + ` int_0^2` [x<sup>3</sup> + x<sup>2</sup> - x] dx. = ` [(7/12)+(28/6)] ` .
 =  `[(7+56)/12]` .
 =  `63/12` .
     Answer:  Total area is `63/12` .square units.

Decimals and Fraction Solving

We are going to see about the topic of decimals and fraction solving with some related problems.  Decimal have point in between the numbers.  For example, 12.34 is a decimal number.  Fraction is formed by division method.  For example, `45/2` is a fraction.  Both the decimal and fraction are involved in the method of addition, subtraction, multiplication and division.

Decimal and fraction methods

Adding decimal:  Two decimal numbers are added.
For example,
                  1 2 . 3 4
                +   1 . 2 2
                -------------------
                  1 3 . 5 6
Subtracting decimal:  Two decimal numbers are subtracted.
For example,
                 1 2 . 3 4
                -   1 . 2 2
                ----------------
                  1 1 . 1 2

Multiplying decimal:  More than one decimal numbers are multiplied.
For example,
                1 . 2 × 1.2
               -------------------
                  1 . 4 4
Dividing decimal:  Divide any two decimal numbers.
For example,
                2)2.4   (1.2
                    1
                ----------
                       4
                       4
                ------------
                       0
In fraction,
Adding fraction:  More than one fraction numbers are added.  It has two method same denominator and different denominator.
For example,
               ` 5/12` + `3/12` = `8/12`
Subtracting fraction:  Subtract the fraction number from another fraction number.  It has two method same denominator and different denominator.
                `5/12` - `3/12 ` = `2/12`
Multiplying fraction:  More than one fraction numbers are multiplied.
                `5/12` * `3/12` = `5/48`
Dividing fraction:  Divide the fraction numbers.
                `5/12` % `5/4` = `1/3`

Example problems – Decimal and fraction solving

Example problem 1 – Decimal and fraction solving
Solving the fraction in addition and subtraction method:  `2/5` and `4/15` .
Solution:
The given fraction is `2/5` and `4/15`
First we are going to add the given fractions
`2/5` + `4/15`
In this given fraction, the denominators are not equal so we need to take the L.C.M
L.C.M of `2/5 ` and `4/15`
                = `2/5` + `4/15`
                = `(6 + 4)/15`
                = `10/15`
We can simplify the answer
                =` 10/15`
                = `2/3`
Now we are going to subtract the given fractions
`2/5 ` - `4/15`
In this given fraction, the denominators are not equal so we need to take the L.C.M
L.C.M of `2/5` and `4/15`
                = `2/5` - `4/15`
                =` (6 - 4)/15`
                = `2/15`
Answer for add and subtract is `2/3` and `2/15` .

Example problem 2 – Decimal and fraction solving
Solving the decimal in addition and subtraction 15.36 and 5.78
Solution:
First, we are going to add the given decimal numbers
Arrange the numbers one by one and then add the decimal number starts from right side,
                  1 5 . 3 6
                +   5 . 7 8
                  ---------------
                  2 1 . 1 4
Now, we are going to subtract the given decimal numbers
Arrange the numbers one by one and then subtract the decimal number starts from right side,
                  1 5 . 3 6
                -   5 . 7 8
                  ---------------
                     9 . 5 8
Answer for add and subtract is 21.14 and 9.58.

Practiced problems – Decimal and fraction solving

Practiced problem 1 – Decimal and fraction solving
Solving the fraction in multiplication method `32/5` and `55/64`
Answer:  `11/2`
Practiced problem 2 – Decimal and fraction solving
Solving the decimal in multiplication method 32.1 and 11.2
Answer:  359.52

Solving Expanded Binomials

In binomial is the distinct probability of the number of success in a series of n experiments, all of that defer success by probability p. Such experimentation is also identified a Bernoulli experiment. Actually, when n = 1, the binomial is a Bernoulli distribution. The binomial is the base for the accepted binomial analysis of statistical importance.

solving expanded binomials:

It is often used to form number of success in an example of size n since a population of size N. as the example are not self-sufficient the resultant distribution is hyper arithmetical distribution, not a binomial one. But, for N a large amount of n, the binomial distribution is a good estimate, and widely used.

Definition of binomial: (source: Wikipedia)

The binomial a2 − b2 can be factored as the product of two other binomials:

a2 − b2 = (a + b)(a − b).

an+1-bn+1 = (a-b)`sum_(k=0)^n a^kb^(n-k)`

This is a special case of the more general formula:

The product of a pair of linear binomials (ax + b) and (cx + d) is:

(ax + b)(cx + d) = acx2 + adx + bcx + bd.

A binomial increase to the nth power stand for as (a + b)n  know how to be expanded through way of the binomial theorem consistently. Taking a simple instance, the ideal square binomial (p + q)2 know how to be establish by squaring the first term, calculation twice the product of the first also second terms and finally calculation the square of the second term, to give p2 + 2pq + q2.

A simple however interesting application of the refer to binomial formula is the (m,n)-formula for make:

for m < n, let a = n2 − m2, b = 2mn, c = n2 + m2, then a2 + b2 = c2.

Example for solving expand binomials:

Example 1:

Expanded (4+5y)2 to solving binomial the equation.

Solution:

Step 1: given the binomials are (4+5y)2

Step 2: `sum_(k=0)^2 ((2),(k))4^(2-k)(5y)^k`

Step 3: `((2),(0)) 4^2(5y)^0 + ((2),(1)) 4^1(5y)^1 +((2),(2)) 4^0(5y)^2`

Step 4:  16 + 20y + 25y2.

Example 2:

Expanded (6+3y)3 to solving the binomial equation.

Solution:

Step 1: given the binomials are (6+3y)3

Step 2: `sum_(k=0)^3 ((3),(k))6^(3-k)(3y)^k`

Step 3: `((3),(0)) 6^3(3y)^0 + ((3),(1)) 6^2(3y)^1 +((3),(2)) 6^1(3y)^2 + ((3),(3)) (3y)^3`

Step 4:  216 + 108y + 54y2+27y3.

My forthcoming post is on taylor series cos x and cbse sample papers for class 12th will give you more understanding about Algebra.

Example 3:

Expanded (2+3y)2 to solving binomial the equation.

Solution:

Step 1: given the binomials are (2+3y)2

Step 2: `sum_(k=0)^2 ((2),(k))2^(2-k)(3y)^k`

Step 3: `((2),(0)) 2^2(3y)^0 + ((2),(1)) 2^1(3y)^1 +((2),(2)) 2^0(3y)^2`

Step 4:  4 + 6y + 9y2.

Subtracting Fractions with Common Denominators

In this article we are going to discuss about Subtracting fractions with common denominator or  like denominators.

A fraction involves two numbers. The top number is said to be numerator and the  bottom number is said to be denominator.

Fraction    =  (Numerator of a fraction) / (Denominator of a fraction)

To subtract the fractions with the common denominator, first subtract the numerators and then put that difference over that common denominator.

How to Subtract fractions with like denominators

Below are the examples on Subtracting fractions with common denominators:

Example problem 1:

Simplify the given fraction, 5/3 – 7/3

Solution:

= 5/3  – 7/3

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (5-7)/3

=(-2)/3

So the result is -2/3

Example problem 2:

Simplify the given fraction, 10/7 – 9/7

Solution:

= 10/7 – 9/7

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (10-9)/7

=(1)/7

So the result is 1/7

Example problem 3:

Simplify the given fraction, 13/9  – 10/9

Solution:

= 13/9 – 10/9

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (13-10)/9

=(3)/9

Equivalent fraction is 1/3

So the result is 1/3 .

Example problem 4:

Simplify the given fraction, 7/9 – 3/9

Solution:

= 7/9 – 3/9

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (7- 3)/9

= (4)/9

Equivalent fraction is 4/9

So the result is 4/9 .

Example problem 5:

Simplify the given fraction, 8/7 – 3/7

Solution:

= 8/7 – 3/7

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (8-3)/7

= (5)/7

Equivalent fraction is 5/7

So the result is 5/7 .

I am planning to write more post on unseen passages for class 7 and unseen passages for class 8. Keep checking my blog.

Example problem 6:

Simplify the given fraction, 8/13 – 5/13

Solution:

= 8/13 – 5/13

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (8-5)/13

= (3)/13

Equivalent fraction is 3/13

So the result is 3/13 .


Practice problems

Here are some practice problems on Subtracting fractions with like denominator

Problem1:  9/13 –5/13

Answer: 4 /13

Problem2:  7/10  – 5/10

Answer: 1/5

Problem3:  11/12 – 5/12

Answer: 1/2

Problem4:  6/7 – 5/7

Answer: 1/7

Problem5:  2/5 – 1/5

Answer:1/5

Figure Measurements

Measurements are used to measure the length of a cloth for stitching, the area of a wall for white washing, the perimeter of a land for fencing and the volume of a container for filling. The measurements consist of lengths, angles, areas, perimeters and volumes of plane and solid figures. If a student wants to know about the measurements of figure, they can be referring the following examples.

Figure Measurements – Examples 1:

These are the examples for measurements of a figure.

figure measurements

Find the area of the triangle whose height is 10cm and base is 6cm.

Solution:

Given base = 6cm

height = 10cm

Area of triangle = `1 / 2 (base xx height)`

= `1 /2 (6 xx 10)`

= `1 / 2 (60)`

= 30

Therefore, the area of the triangle is 30cm2

Find the area of the triangle whose height is 12cm and base is 8cm.

Solution:

Given base = 8cm

height = 12cm

Area of triangle = `1 / 2 (base xx height)`

= `1 /2 (8 xx 12)`

= `1 / 2 (96)`

= 48

Therefore, the area of the triangle is 48cm2

Figure Measurements – Examples 2:

These are the examples for measurements of a figure.

Measurements of Figure 1:

Three angles of a triangle are x + 34˚, x + 40˚ and x + 46˚. We have to find x for the triangle.

Solution:

x + 34 + x +40 + x + 46 = 180˚

The sum of the three angles of a triangle is equal to 180˚

3x + 120 = 180˚

3x = 180˚ - 120˚

= 90˚

x = `60/3`

= 20˚

Measurements of Figure 2:

The triangle has a three angles  x + 20˚, x + 10˚ and x + 30˚. We have to find x for the triangle.

Solution:

x + 20 + x +10 + x + 30 = 180˚

The sum of the three angles of a triangle is equal to 180˚

3x + 60 = 180˚

3x = 180˚ - 60˚

= 120˚

x = `120/3`

= 40˚

Algebra is widely used in day to day activities watch out for my forthcoming posts on Multiplying Mixed Number Fractions and Strategies for Addition. I am sure they will be helpful.

Measurements of Figure 3:

The measurements of the angles whose triangle are in the ratio 2:1:3. Calculate the angles of the given triangle values.

Solution:

Ratio of the angles of a triangle = 2:1:3

Total ratio = 2 + 1 + 3

= 6

Sum of the three angles of a triangle is 180˚. Therefore,

First angle = `2/6 xx 180`

= 60˚

Second angle = `1/6 xx 180`

= 30˚

Third angle = `3/6 xx 180`

= 90˚

Math Symbols Terms

There are many terms and symbols used in math. The symbols can be used to perform some operations such as addition, multiplication, subtraction, division; relationship symbols such as less than, greater than, equal, not equal, etc. The math terms can be used to describe the topics such as algebra, arithmetic, array, and axis, etc. Let us see about math symbols and terms in this article.

I like to share this Symbol for Correlation with you all through my article. 

Some of the Math Symbols

Addition:
It can be represented as +. It is used for addition operation and also for logical OR purpose. It can be read as plus or logical or.
Subtraction:
It can be represented as –. It is used for subtraction operation. It can also spelled out as minus.
Multiplication:
Multiplication can be represented as ×. It is used for multiply the given terms.  It can also spelled out as into.
Division:
It can be represented as ÷. It is used for dividing the given terms.  It can also spelled out as divide by.
Percentage:
It can be represented as %. It is used for calculating the ratio that compares to the number 100.  It can also spelled out as percent.
Summation:
Summation can be represented as `sum`   . It is used for calculating the sum of many or infinite values.  It can also spelled out as sum.
Dot Product:
It can be represented as (.). It is used for calculating the scalar (dot) product of two vectors.  It can also spelled out as dot.
Cross Product:
It can be represented as (×). It is used for calculating the vector (cross) product of two vectors. It can also spelled out as cross.



Algebra is widely used in day to day activities watch out for my forthcoming posts on Fibonacci Number Sequence and number to words converter. I am sure they will be helpful.

Some of the Math Terms

Algebra:
An algebraic equation represents the scale.
Algorithm:
A step-by-step problem solving technique is used in math computations.
Area:
It can be used for representing the two-dimensional shapes such as polygon, octagon, hexagon, triangle and circle.
Array:
A set of numbers can be occurred in a specific size of pattern. Matrix or array consists of columns and rows.
Axis:
Axis consists of horizontal and vertical axis having co-ordinates in a plane.
Attribute:
Describing the object with data consists of shape, color, or size.
Arc:
A circumference of the circle or the portion of a segment draws with a compass.

Maxima and Minima

Maxima and Minima are the largest value (maximum) or smallest value (minimum), that a function can take at a point either within a given boundary (local) or on the whole domain of the function in its entirety (global). In general, maxima and minima of a given set are the greatest and least values in that set. Together, Maxima and Minima are called extrema (singular: extremum). We need to learn minima to determine the nature of the curve or the function and various other applications like projectiles, astrophysics to microphysics, geometry etc.

Learning analytical definition of minima:

A function f(x) is said to have a local minima point at the point x*, if there exists some ε > 0 such that f(x*) ≤ f(x) when |x − x*| < ε, in a given domain of x. The value of the function at this point is called minima of the function.

A function f(x) has a global (or absolute) minima point at x* if f(x*) ≤ f(x) for all x throughout the function domain.

We need to learn minima points of a curve by observing the involved function.

learning prologue regarding minima:

To learn Minima & Maxima, one needs to have a basic knowledge of calculus. The following points are some bare necessary (maybe not sufficient) definitions.

A function, y = f(x) is a mathematical relation such that each element of a given set ‘x’ (the domain of the function) is associated with an element of another set ‘y’ (the range of the function).

Closed interval of a domain is defined as an interval that includes its endpoints, as opposed to open interval which is an interval that does not include its endpoints.

A function, f(x) is said to be continuous at a given interval if it can assume all values within the interval i.e. the function is not broken anywhere inside the interval. Mathematically we determine this by ensuring the function has a finite value at the given point and taking the limit on both sides of the point and checking if they both exist and are equal (L.H.L. = R.H.L.).

Differentiability of a function is out of the scope of this article, but simply put, a function is said to be differentiable at a point if the curve at that point is smooth i.e. there is no drastic change of slope. Mathematically this is achieved by checking if both the left hand derivative and the right hand derivative of the function at the given point finitely exist and are equal (incidentally this common value is the value of the derivative of the function at the given point).

First Derivative is defined as the differentiation of a function, y = f(x), once, with respect to ‘x’. It is denoted by dy/dx or f’(x) and simply put, it gives the slope of the function at any given value of ‘x’ or the instantaneous rate of change of the function w.r.t. ‘x’ at any given value of ‘x’.

Second Derivative is defined as the differentiation of a function, y = f(x), twice, with respect to ‘x’. It is denoted by d2y/dx2 or f’’(x) and simply put, it gives the slope of the slope of the function at any given value of ‘x’ or the instantaneous rate of change of the slope of the function w.r.t. ‘x’ at any given value of ‘x’.

Critical points of f(x) are defined as the values of x* for which either f'(x*) = 0 or f’(x*) does not exist.

Tests for Minima:

Local Minima can be found by Fermat's theorem, which states that they must occur at critical points.

If f(x) has a minima on an open interval, then the minimum value occurs at a critical point of f(x).

If f(x) has a minima value on a closed interval, then the minimum value occurs either at a critical point or at an endpoint.

Critical points of f(x) are defined as the values of x* for which either f'(x*) = 0 or f’(x*) does not exist.

One can distinguish whether a critical point is a local maximum or local minimum by using the first derivative test or second derivative test.

learning minima -First Derivative Test

Suppose f(x) is continuous at a critical point x*.

If f’(x) <0 an="" and="" extending="" f="" from="" interval="" left="" on="" open="" x="">0 on an open interval extending right from x*, then f(x) has a relative minima at x*.

If f’(x) >0 on an open interval extending left from x* and f’(x) <0 a="" an="" at="" extending="" f="" from="" has="" interval="" maxima="" on="" open="" p="" relative="" right="" then="" x="">
If f’(x) has the same sign on both an open interval extending left from x* and an open interval extending right from x*, then f(x) does not have a relative extremum at x*.

An interesting point to NOTE:

Differentiability is not a criterion for the first derivative test. Suppose f(x) is continuous but not differentiable at x*, i.e. f’(x*) does not exist. Still the above holds true since the test is done in open intervals on the left and right sides of the point in consideration [see Figure below]. So the criteria is only that f(x) is continuous at x* and that f’(x) exists in the neighbourhood of x*.

In summary, relative minima occur where f’(x) changes sign.

learning minima -The Second Derivative Test:

Suppose that x* is a critical point at which f’(x*) = 0, that f’(x) exists in the neighbourhood of x*, and that f’’(x*) exists.

f(x) has a relative minima at x* if f’’(x*)>0.

f(x) has a relative maxima at x*if f’’(x*) <0 .="" p="">
f(x) does not have an extremum at x* if f’’(x) = 0.

NOTE:

Differentiability at the critical point is a criterion for the second derivative test as opposed to the first derivative test. Also, if f’’(x*) = 0, the test is not informative [see Figure below], it actually means there is no change of sign of f’(x) on going from the left to right of the given critical point (these points are called the points of inflection).

learning Absolute Minima and Maxima

For any function that is defined piecewise, one can learn minima (or maxima) by finding the minimum (or maximum) of each piece separately; and then seeing which one is smallest (or biggest).

Visualization to learn minima

max-min

Least Common Denominator Tutoring

Least common multiple of two rational facts a and b is the least positive rational number, that is an integer multiple of a as well as b. because it is a multiple, it can be divided by a and b without a remainder. Tutoring not only help students by giving answers, but also help students in their problem solving with step by step solutions.  In this article we shall discuss about least common denominator tutoring. The following are the examples and steps involved in least common denominator tutoring.

Least common denominator tutoring:

Least common denominator:

The least common denominator of two or more numbers is the least number which is a multiple of each of the given number.

Technique: Least common denominator tutoring

To find the l.c.d.

Step (1) Write the multiples of first number

(2) Write the multiples of second number

(3) Write the common multiples

(4) Write the least common denominator.

Example problems least common denominator tutoring:

Example 1: Find the L.C.D of `1/4` and `1/3`

Here least common denominator is 12

To find the l.c.d.

Step (1) Get l.c.m. for every denominators.

Step (2) Modify equivalent fraction with same l.c.m denominator

Step (3)  By taking l.c.m common in denominator and add all numerators.

= `(1*3)/12` + `(1*4)/12` = `(4+3)/12`

So the final result is `7/12` .

Example 2: Find the L.C.D of `1/5` and `1/4`

Here least common denominator is 20

To find the l.c.d.

Step (1) Get l.c.m. for every denominators.

Step (2) Modify equivalent fraction with same l.c.m denominator

Step (3)  By taking l.c.m common in denominator and add all numerators.

=  `(1*4)/20 ` + `(1*5)/20`

So the final result is 9/20.

Example 3: Find the L.C.D of `1/25` and `1/15`

Here least common denominator is 75

To find the l.c.d.

Step (1) Write the multiples of 25: 25, 50,75

Step (2) Write the multiples of 15: 15, 30, 45, 60, 75

Step (3) common multiple is 75

Step (4) least common denominator is 75.

= `(1*3)/75` + `(1*5)/75` =`(5+3)/75`

So the final result is` 8/75` .

Practice problem for least common denominator tutoring:

Example 1: Find the L.C.D of `1/6` and `1/3`

Here least common denominator is 6

So the final result is` 3/6` .

Algebra is widely used in day to day activities watch out for my forthcoming posts on Least Common Multiples. I am sure they will be helpful.

Example 2: Find the L.C.D of `1/6 ` and `1/4`

Here least common denominator is 12

So the final result is `5/12` .

Example 3: Find the L.C.D of `1/4` and `1/2`

Here least common denominator is 4

So the final result is `3/4` .

A Proper Factor

A factor is a whole number which divides exactly by another whole number is called factor for  that number Any of the factors of a number, except the number itself. A factor is a portion of a number in general  integer or polynomial when multiplied by other factors  gives  entire quantity. The determination of factors is a factorization A proper factor of a positive integer is a factor of other than 1.Proper factor is a multiples for a given whole number which has multiples again it is aproper factor.Every whole number which has a factor of its own.proper factor is like a normal factor except 1
For example,
For 6, 2 and 3 are proper factors of , but 1 and 6 are not a proper factor.

Properties of proper factor:

The divisors for a any number other than 1 and  number itself are called  factors for that number.
A factor for N number is a number which divides  exactly N.

Example: the factors for 24 are 1,2,3,4,6,and 12

Generally for every number has itself and 1 as its factors.
When a number is greater than 1 and by itself and 1 as factors, then the number is prime.
A number or quantity that when multiplied with another number produces a given number or expression.

Example Problems for Proper factors:

Example 1 :
Find all the divisors and proper factors of  20

Solution :
The divisors of 20 are 1, 2, 4, 5, 10 and 20
The factors of 20 are 2, 4, 5 and 10

Example 2:
Find all the divisors and proper factors of 32.

Solution:
The divisors of 32 are 1, 2, 4, 8, 16 and 32
The proper factors of 32 are 2, 4, 8 and 16

Example 3:
Find Proper Factors of 30

Solution:
30=1*30
=2*15
=3*10
=4*15
=5*6
=6*5

Proper  Factors for 30 are 2,3,4,5,6

Example 4:
Find proper factors for 42

Solution:
42=1*42
=2*21
=3*14
=6*7
2,3,6

I am planning to write more post on Solve first Order Differential Equation and free 3rd grade math word problems. Keep checking my blog.

Example 5:
Find proper factors for 18

Solution:
18=1*18
=2*9
=3*6
=6*3
=9*2
Proper factors are 2,3,6,9

Equivalent Scale Factor

A scale area is a number which scales, or multiplies, some quantity. In the equation y=Cx, C is the scale factor for x. C is also the coefficient of x, and may be called the constant of proportionality of y to x. For example, doubling distances corresponds to a equivalent scale factor of 2 for distance, while cutting a cake in half results in pieces with a scale factor of ½.
SOURCE: WIKIPEDIA

Example problems of equivalent scale factor:

Equivalent scale factor problem 1:
Find the scale factor to the following figure:

Equivalent scale factor solution:
If we want to find the length of smaller rectangle then we can multiply the length of the one side of larger rectangle and the value of scale factor.
We can find the scale factor of the given rectangles by using the following formula,
Let D_l be the dimensions of larger rectangle and D_s be the dimensions of smaller rectangle and s be the scale factor.
Therefore the formula as,
                D_l*s=D_s
Substitute the values of dimensions into the above formula. Then we get
                  30*s=24
Divide by the value 30 on both sides,
                   `(30s)/30` `=` `24/30`
                       ` s ` `=` `24/30`
Divide by the value of 6 on both numerator and denominator. Then we get the value of scale factor.
                        s = `4/5 ` or 4:5
Therefore scale factor of smaller to larger rectangle= 4:5
Answer: 4:5
Equivalent scale factor problem 2:
Find the larger to smaller scale factor for the following figure:

Equivalent scale factor solution:
If we want to find the length of smaller rectangle then we can multiply the length of the one side of larger rectangle and the value of scale factor.
We can find the scale factor of the given rectangles by using the following formula,
Let D_l be the dimensions of larger rectangle and D_s be the dimensions of smaller rectangle and s be the scale factor.
Therefore the formula as,
                D_l*s=D_s
Substitute the values of dimensions into the above formula. Then we get
                  39*s=26
Divide by the value 48 on both sides,
`(39s)/(30)` `=` `26/39`
` s ``=` `26/39`
Divide by the value of 13 on both numerator and denominator. Then we get the value of scale factor.
                        s = `2/3` or 2:3
Therefore scale factor of larger to smaller scale rectangle= 3:2
Answer: 3:2

Algebra is widely used in day to day activities watch out for my forthcoming posts on Proof of Fundamental Theorem of Calculus and algebra 2 solver step by step. I am sure they will be helpful.

Practice problems of equivalent scale factor:

1. Find the scale factor to the following figure:

2. Find the larger to smaller scale factor for the following figure:

Answer:
1. 3:4
2: 6:5