Length of Lines

Introduction:

The Length of lines is the main topic in Geometry. The line length between two points, it is used in all geometric concepts. Let P1 (x1, y1) and P2 (x2, y2) be two distinct points in the Cartesian plane and denote the line length between P1 and P2 by d (P1, P2) or by P1P2. The length formula is used to find the length between two points.  With the help of geometric properties and the length formula we have to identify the given points make which shape.


Brief Description of Line Length Formula

Consider the line segment P1 P2. Here two cases arise.

Case (I):

The segment P1 P2 is parallel to the x-axis Then y1 = y2. Draw P1L and P2M, perpendicular to the x-axis. Then d (P1, P2) is equal to the length between A and B. But A is (x1, 0) and B is (x2, 0). So the length AB =|x1 -X2|. Then d (P1, P2) =|x1 -X2|.

Case (II):

The segment P1 P2 is parallel to the y-axis. Then x1 = x2 .Draw P1A and P2B, perpendicular to the y-axis. Then d (P1, P2) is equal to the line length between A and B. But A is (0, y1) and B is (0, y2). So the length AB =| y1- y2|.then d (P1, P2) = | y1- y2|

`sqrt(((x2-x1)^2 + (y2-y1)^2))`

This is called the line length formula which gives the line length d between the two given points (x1, y1) and (x2, y2). We observe that d (P1, P2) = d (P2, P1). The formula has been derived for two points which are not on a horizontal line or vertical line.

Examples Problems

Given points are P (5,4) and Q (8, 6). Find the length of PQ using line length formula.

Solution:

Let d be the line length between P and Q. Here we have to find the length between two points using line length formula.

Then d (P, Q) =`sqrt(((x2-x1)^2 + (y2-y1)^2))`

= `sqrt(((8-5)^2 + (6-4)^2))`

Simplifying this we get the length is v13.

Example 2:

Given points are W(2, -3), X(6, 5), Y(-2, 1) and Z(-6, -7), Show that this points make a square.

Solution: One way of showing that WXYZ is a rhombus is to show that all its sides are of equal length. One way is showing that a rhombus is not a square is to show that the diagonals are of unequal length. Here we find

WX= `sqrt(((6-2)^2 + (3+1)^2))`

Simplifying this we get v80

XY = `sqrt(((-2-6)^2 + (1-5)^2))`

Simplifying this we get v80

WY= `sqrt(((-2-2)^2 + (1+3)^2))`

Simplifying this we get v32

XZ = `sqrt(((-6-6)^2 + (-7-5)^2))`

Simplifying this we get v288

YZ = `sqrt(((-6+2)^2 + (-7-1)^2))`

Simplifying this we get v80

WZ= `sqrt(((-6-2)^2 + (-7+3)^2))`

Simplifying this we get v80

Therefore WX=XY=YZ=ZW

WY is not equal to XZ

Therefore WXYZ is a rhombus but not a square.

My forthcoming post is on solving linear equations by graphing, a list of prime numbers will give you more understanding about Algebra.

Area of a Cone Shape

Introduction :
                    Cone is the one type of three dimensional shapes. It has one base and one vertex. The base of the cone is circle in shape. It surface is tappers smoothly towards the top. The distance between center of the base and any point on the boundary of base is called radius. In this article we shall see how to calculate the volume and surface area of the cone.


Area of a Cone Shape – Formula:

Formula for volume of the cone (v) =`1/3` p r^2 h cubic units

                                                    v - Volume of cone

                                                    r – Radius

                                                    h – Height 

Surface area of cone (A) = lateral surface area of cone + area of base 

                                              = p r s    + p r^2 square unit 

                                                s – Slant height

Area of a Cone Shape – Example Problems:

1. The cone has the radius = 7 cm, height = 14 and slant height = 15.6 cm. Find the volume of the cone.

Solution:

 Given:

                        Radius (r) = 7 cm

                        Height (h) = 14 cm

                      Slant height (s) = 15.6 cm

Formula:

Total surface area:

Total Surface area of cone (A) = lateral surface area of cone + area of base

Lateral surface area (L.S.A) = p r s    square unit.

Substitute the r and s value in formula and simplify,

                                                           = p x 7 x 15.6

                                                            = 3.14 x 7 x 15.6

                                                            = 342.88 cm3

       Lateral surface area (L.S.A) = 204.1 cm^2

                          Area of base = p r^2 square unit

                                                    = 3.14 x 72

                           Area of base = 153.86 cm^2

Total Surface area of cone (A) = lateral surface area of cone + area of base

                                                    = 204.1 +153.86

                                                    = 357.96 cm^2

   2. The cone has the radius = 8 cm, height = 15 and slant height = 17 cm. Find the volume of the cone.

Solution:

 Given:

                        Radius (r) = 8 cm

                        Height (h) = 15 cm

                     Slant height (s) = 17 cm

Formula:

Total surface area:

Total Surface area of cone (A) = lateral surface area of cone + area of base

Lateral surface area (L.S.A) = p r s    square unit.

Substitute the r and s value in formula and simplify,

                                                     = p x 8 x 17

                                                     = 3.14 x 8 x 17

                                                     = 427.04

Lateral surface area (L.S.A) = 427.04 cm^2

                          Area of base = p r^2 square units

                                                  = 3.14 x 8^2

                         Area of base = 200.96 cm^2

Total Surface area of cone (A) = lateral surface area of cone + area of base 

                                                         = 427.04 + 200.96

                                                         = 628 cm^2
Understanding how to solve proportions with variables is always challenging for me but thanks to all math help websites to help me out.

Word Problems based on Ratio and Proportion

Introduction :

A ratio is the comparison of two quantities by division. It is a relation that one quantity bears to another with respect to magnitude. If A and B are two numbers, than the ratio of A to B is A/B and is denoted by A:B. The ratio does not have any unit.

The equality of two ratios is called Proportion. If (A/B) = (C/D) , then A,B,C,D are said to be in proportion and can be written as

A : B :: C:D

The Ratio and Proportion world problems are a kind of problems in which , the relation between different quantity has to be determined and then using the definition of ratio and proportion, the unknown has to be calculated.
Here are some of the ratio and proportion word problems:

Ratio and Proportion: Word Problems.

Problem: Find the value of k that must be added to 7, 16, 43, 79 so that they are in proportion. (Answer: 5)

Problem: Find the fourth proportional to the numbers 60, 48, 30. (Answer: 24)

Problem: The Income of Alex and Bob are in the ratio of 3:2 and their expenditure in the ratio of 5:3. Find the income of Alex if each saves dollars 1000. (Answer:  $6000)

Problem: A mixture contains alcohol and water in the ratio of 12:5. On adding 14 litres of water, the ratio of alcohol to water becomes 1:1. Find the quantity of alcohol in the mixture. (Answer: 24 litres)

Ratio and Proportion: Multiple Choice Word Problems:

Problem: If the ratio of ages of Alex and Bob is 6:5 at present and fifteen years from now, the ratio will get changed to 9:8, then find Alex's age.

(A). 24 years

(B) 30 years

(C) 18 years

(D) 33 years

(Answer: (B) 30 years)

Problem: If dollars 58 is divided among 150 children such that each girl and each boy gets 25 dollars and 50 dollars respectively. Then how many girls are?

(A) 52

(B) 54

(C) 68

(D) 62

(Answer: (C) 68)

Problem: The number that must be added to each of the numbers 8, 21, 13 and 31 to make the ratio of first two numbers equal to the ratio of last two numbers is

(A) 5

(B) 7

(C) 9

(D) None of these.

(Answer: 5)

Introduction to improper fraction to decimal

Introduction :

A fraction is a part or parts of a whole. If the numerator value is larger than its denominator value, the fraction is called an Improper fraction. For example 4/3, 7/2 is all improper fractions. An improper fraction value is always larger than the value1. The special kind of fraction is known as decimal fractions. That having the denominator to the powers of 10. 

I like to share this solve a math problem with you all through my article.

Convert the Improper Fraction to Decimal

Step 1: First we can convert the divide the numerator value by denominator value. It involves two cases. They are described in step 2

Step 2: On the first step numerator is surely greater than the denominator. But the remainder of this step value is lesser than the denominator. This is the type of proper fraction.

Step 3: So we put a point on the quotient value then we can add zero to the remainder value.

Step 4: Now the numerator value is greater than the denominator. Again we can do the same procedure. Now we can get the decimal form of the given improper fraction.

These are the main steps to followed by converting the improper fraction to decimal value.

Example Problem-improper Fraction to Decimal:

Consider the improper fraction 8/3.

This is an improper fraction because its numerator is greater than the denominator.
Two times 3 is 6 so we can get the remainder 2.
Then 2 is not divided by 3.
So we can add zero to 2 and put a point on the quotient value
That is 20
Now 20 can be divided by 3.
6 x3 are 18.
Then the remainder is 2.
Again 2 is not divided by 3
So we can use the same procedure.
Then the decimal value is 2.66…..
This value is going on.
So we can convert this decimal as 2.67
This is the procedure to convert the improper fraction to decimals.

Consider another improper fraction 15/4

This is an improper fraction because its numerator is greater than the denominator.
Three times 4 is 12 so we can get the remainder 3.
Then 3 is not divided by 4.
So we can add zero to 3 and put a point on the quotient value
That is 30
Now 30 can be divided by 4.
7 x4 are 28.
Then the remainder is 2.
Again 2 is not divided by 4
So we can use the same procedure.
Now the value is 20
20 is divided by 4
5 times 4 is 20
So the remainder is zero
Therefore the quotient value is 3.75
This is our required decimal

Introduction to rectangular prism volume

Introduction to rectangular prism volume.

                          Rectangular prism volume article deals with the volume of the rectangular prism and the model problems related to volume of the rectangular prism.

Definition of rectangular prism volume:

                      The amount of space occupied by the three dimensional rectangular prism. The volume of the rectangular prism is measured in cube units ( cm³ ,m³).volume can be calculated to three dimensional shapes only.

                                                                                              
                                                                                    

Formula Torectangular Prism Volume:

When the length, breadth and height of the rectangular prism are known, the formula to find the volume is

                               Volume of the rectangular prism = L*B*H cube. Units

                                     L is the length of the rectangular prism

                                     B is the breadth of the rectangular prism

                                     H is the height of the rectangular prism.

Model Problem to the Rectangular Prism Volume:

Problem: 1

                What is the volume of the rectangular prism when the length is 12cm, breadth is 8cm and the height is 10cm?

           Solution:

                      Length of the rectangular prism = 12cm

                      Breadth of the rectangular prism = 8cm

                      Height of the rectangular prism = 10cm

              Formula:

                               Rectangular prism’s volume = L*B*H cube. Units

                                                                             = 12*8* 10

                                                                             = 12*80

                                                                             = 960 cm³

                      Volume of the rectangular prism is 960 cm³

Problem: 2

                what is  the volume of the rectangular prism when the length is 9cm, breadth is 6cm and the height is 11cm

           Solution:

                      Length of the rectangular prism = 9cm

                      Breadth of the rectangular prism = 6cm

                      Height of the rectangular prism = 11cm

              Formula:

                          Rectangular prism’s volume    = L*B*H cube. Units

                                                                             = 9*6* 11

                                                                             = 54*11

                                                                             = 594 cm³

                      Volume of the rectangular prism is 594 cm³

Problem: 3

                Finding the volume of the rectangular prism when the length is 7cm, breadth 3cm and the height is 5cm

           Solution:

                      Length of the rectangular prism = 7cm

                      Breadth of the rectangular prism = 3cm

                      Height of the rectangular prism = 5cm

              Formula:

                               Rectangular prism’s volume = L*B*H cube. Units

                                                                             = 7*3* 5

                                                                             = 21*5

                                                                             = 105 cm³

                      Volume of the rectangular prism is 105 cm³

Algebra questions and answers

Algebra is a branch of math which mainly deals with the representation of numbers as letters or alphabets of English language.Algebra is studied by kids of grade 5 onwards.The first section is pre-algebra , then algebra 1 and algebra 2.

The algebra introduces the concept of the single variable. It is a part of a binary method (binary operators are addition, subtraction, multiplication, division) is replaced by box. The box is replaced by a letter such as x or y denote a variable, and a, b, or c denote constants. In this article we shall discuss about some algebra questions and answers.

For Ex : X + 3 = 7

Algebra deals with formulas which make simplification very easy and new techniques to solve expressions. For example the FOIL method is a new technique which can be used to simplify multiplication of expressions in an easy and accurate way. Same way the remainder theorem discussed here is a shortcut method of finding the remainder without actually dividing a polynomial.


Answers Related to some Questions of Algebra:

Qu 1:  In Algebra Foil Method to solve the following expression: (3+7x)(6+2x)

Sol:   From the foil method in algebra,

          Step 1:   (3) (6) + (3) (2x) + (7x) (6) + (7x) (2x)

          Step 2:   18 + 6x + 42x + 14 x^2

          Step 3:   18 + 48x +14x^2 (Simplify)

This is a very simple process.

Foil Method in algebra:

 The Foil method is used in algebra to multiply the two polynomials mainly binomials

  The general form is: (K+L)(M+N) = KM            + KN           + LM           + LN

                                                        |                 |                 |                |

                                                       (First)     (Outside)      (Inner)        (Last)

Answers Related to Remainder Theorem Questions in Algebra:

Remainder Theorem in Algebra

 Let f(x) be any polynomial greater than or equal to one and let b be any real number. If f(x) is divided by the linear polynomial x – b, then the remainder is f(b).

  Proof: Let f(x) be any polynomial with degree greater than or equal to one. Suppose that when f(x) is divided by x – b, the quotient is g(x) and the remainder is r(x), i.e.,

f(x) = (x – b) g(x) + r(x)

  Since the degree of x – b is 1 and the degree of r(x) is less than the degree of x – b, the degree of r(x) = 0. This means that r(x) is a constant, say r.

  So, for every value of x, r(x) = r.

Therefore, f(x) = (x – b) g(x) + r

 In particular, if x = b, this equation gives us

f (b) = (b – b) g(b) + r 

            = r,This proves the theorem.

Qu 2: Find the remainder when x^4 + x^3 – 2x^2 + x + 1 is divided by x – 1 in algebra.

Sol: Step 1:Here, f(x) = x^4 + x^3 – 2x^2 + x + 1, and the zero of x – 1 is 1.

 Step 2: Plug in x=1        So, f (1) = (1)4 + (1)3 – 2(1)2 + 1 + 1

            = 2

 By the Remainder Theorem, 2 is the remainder when x^4 + x^3 – 2x^2 + x + 1 is divided by x –1.