Thursday, June 6

Draw isosceles Triangle


The triangle is a closed geometric shape that contains three sides. There are several types of triangles in the geometry world. We can divide the triangles according to their angles and sides. We can define the isosceles triangle by using its sides.
      The triangle that has two equal or congruent sides in its measure is called as isosceles triangle. We can say that the equilateral triangle is also as an isosceles triangle,since the equilateral triangle has three equal sides . while we draw a isoscles triangle we have to see that two sides triangle must be equal.


draw Isosceles Triangle:



  • when we draw a isoceles triangle, angles are congruent i.e., an angles of each sides are equal.
  • when we draw a isoceles triangle diagonals of an isosceles triangle must be congruent.
  • The measurement of the adjacent angles is giving the 180 degree. x = y = 180o


Formulas for Isosceles triangle:


  • Hight,  h = √(b2 - `(1/4)` a2)
  • Perimeter of Isosceles Triangle = A + B + C
  • Area of isosceles triangle A = (b* h)/2

Example problems:



1) Find the area of isosceles triangle with the base and height are 5 cm and 7 cm.
solution:
We can find the area of given problem using the following formula:
Area A= (b*h)/2
Substitute the values of b and h into the above formula. Then we get,
         A= (5*7)/2
 Here multiplying to the values of 5 and 6 then dividing by 2.
  Then we get the final solution.   
 Answer A=17.5 cm2
2) Find the perimeter of Isosceles triangle that has side, Side 1 =10 cm, Side 2 = 10, Side 3 = 7 cm.
Solution:
   Given, Side 1 =10 cm, Side 2 = 10, Side 3 = 7 cm.
  Perimeter of Isosceles triangle   = Side 1+ Side 2+ Side 3
                                                         = 10 + 10 + 7
  Perimeter of Isosceles triangle  = 27 cm
3) Find the height of the isosceles triangle of the base a = 7 cm and equal sides b = 11 cm.
Solution:
The height of the isosceles triangle is given by the formula:
             h = √(b2 - `(1/4)` a2)
substitute the a = 7cm and b = 11 cm. in the formula,
             h = √(112 - (1/4)*72 )
             h = √(121 - `(49/4)` )
             h = √(121 - 12.25 )
             h = √(108.75) = 10.428 ≈ 10.4 cm





Height of the isosceles triangle = 10.4 cm

Straight Triangle

In geometry, a triangle is the fundamental shapes. The polygon with 3 corners and 3 sides are the line segments. The corner otherwise called as the vertices. The sides otherwise called as the edges. A triangle with corners A, B and C is represented by ∆ABC.Here we will see about the triangle types and formulas.


Classification of Straight Triangles


The straight triangle has three types. The names are as follows,
  • Equilateral Triangle
  • Isosceles Triangle
  • Scalene triangle
Equilateral triangle
     Here all three sides are the equal length. An equilateral triangle is also a normal polygon with every angles calculate 60 degree. The equilateral triangle is as follows,
              In this triangle sides are equal length. 
Isosceles triangle
     These triangle two sides only the equal length. The figure is as follows,
                 In this traingle has only the two equal sides.
Scalene triangle
Here all sides are the different type. The three angels are also various in measures. This triangle figure is as follows,

           All sides are different and angles are different.

Classified Based by the Internal Angels
  • A triangle does not have an angle that calculates 90 degree is known as the oblique triangles.
  • It has all interior angles calculating less than 90 is known as the acute triangle or acute-angled triangle.
  • A straight triangle has one angle that calculates more than 90 degree is known as obtuse triangle or obtuse-angled triangle.
Formula
Triangle area is A=(1/2).b.h
Base is denoted by b and height is denoted by h

Formula for angles
     Sin(q)=opposite/Hypotenuse
     Cos(q)=Adjacent/Hypotenuse
     Tan(q)=Opposite/Adjacent

Properties
  • Angle sum property
  • Exterior angle property
  • Triangle Inequality property
  • Pythagoras theorem
Algebra is widely used in day to day activities watch out for my forthcoming posts on Solving Inequalities Word Problems and 8th samacheer kalvi books. I am sure they will be helpful.

Examples

1)Find the area of straight triangle with the base 5 cm and the height 7cm.
Solution
     Given b=5 cm and h=7 cm.
     Formula A=(1/2).b.h
                    =(1/2).5 cm.7cm
                    =(1/2).35cm2
                    =17.5cm2

2)what is the obtuse triangle with the base 4 inches and the height 6 inches.
Solution
     Given b=4 in and height h=6 in
     Formula A=(1/2).b.h
                    =(1/2).(4 in).(6 in)
                    =(1/2).24 in2
                    =12in2

Solve Regular Polygons

If all sides of the polygon is same length in all of its sides and all of its angles are equal, then the polygon is known as regular polygon.  The regular polygon cannot be a concave polygon. Most regular polygons are convex or star. We will see about Solve regular polygons in this article.


Regular polygon examples:
Few examples of Regular polygons are,
                      Square                                 Equilateral triangle                  Pentagon                                    
            squareEquilateral triangle     pentagon

                        Hexagon
Hexagon
                    

To solve regular polygons formulas:


     The following formulas are used to solve regular polygons,
  • Interior angle of each side = `(180(n-2))/n ` degrees 
  • Exterior angle of each side = `360/n ` degrees 
  • Diagonal=` (n(n-3))/2` 
  • Area = n x area of triangle 
      = ½ * (apothem * perimeter)  (Or)
  • Area  = ½ *( n* s*r) ( or )
    A= `(s2 n)/ (4tan (pi/n))`
    And there is lot of formulas for area.
 Where, n = number of sides, s = side length, r = radius or apothem

Solve regular polygons Example problems:


Example 1:
The regular hexagon has the apothem 7 cm and side is 5 cm. Calculate the area.
Solution:
Now, let us solve regular polygons using the first formula (above mentioned)
By the formula,
Area = (½) * (apothem) *(perimeter)

Perimeter of hexagon = Length of the side * Number of side
                                    = 5 * 6
                                    = 30cm
   Area of the hexagon = (1/2) * 7 * 30
                                     = 105cm2
Example 2:
The octagon has the apothem of 9 cm and the side length is 6 cm. Find its area.
Solution:
Given, n=8, s=6,  r = 9
    Let us solve regular polygons using the second formula of area (above mentioned)
           By the formula
                                    Area = (½) * n * s * r
                                            = (1/2) * 8 * 6 *9
                                            = 216cm2
Example 3:
A regular pentagon has the side of 5 inches. Determine its area.
Solution:
Given,
N= 5(pentagon), s= 5 inches
We can use the third formula now,
                                      Area= `(s^2 n)/ (4tan (pi/n))`
                                       = ` (5^2 * 5) / ( 4tan(pi/5))`

                                      = `125/ 2.906`
                                     =  43.01inches2   


I am planning to write more post on Sine Cosine and Tangent Chart, tamilnadu state board books. Keep checking my blog.

Example 4:
 An equilateral triangle has the side of 5inches length. Find its area and perimeter.
Solution:
Length = a= 5 inches
By formula,
     Area of equilateral triangle =   `sqrt(3)/4` a2  
                                                     = `sqrt(3)/4` * 25
                                                     = 10.825 sq.inches
    Perimeter of the triangle= a+b+c
                                               = 5+5+5

                                               = 15 inches.

Wednesday, June 5

Surface Normals

Let us see about surface normals. A surface normal also called the simple normal, to a flat surface is a vector that is vertical to that surface.A line normal to a flat, the normal factor of the force, the normal vector, etc. The concept of routine generalizes to orthogonality.A normal is used for computer graphics.

Calculating a surface normals


For a convex polygon such as a triangle, a surface usual will be calculated as the vector cross product of two (non-parallel) edges of the polygon.
For a plane given by the equation ax + by + cz + d = 0.
For a plane is represent the following equation r = a + αb + βc.
where a is a vector to get onto the level surface and b and c are non-parallel vectors lying on the plane, the normal to the plane defined is given by b × c .
For a hyperplane in n+1 dimensions equation given from following equation r = a0 + α1a1 + α2a2 + ... + αnan,
where a0 is a vector to get onto the hyperplane and ai for i = 1, ... , n are non-parallel vectors two-faced on the hyperplane, the normal to the hyperplane can be approximated by (AAT + bbT) − 1b where A = [a1, a2, ... , an] and b is a random vector in the space not in the linear span of ai.

Cross product partial derivation


I am planning to write more post on Area of a Semicircle Formula, icse syllabus. Keep checking my blog.


In surface normal,a surface S is given completely   as the set of points (x,y,z) it satisfying F(x,y,z) = 0, so the normal point (x,y,z) on the plane.
For a surface S given clearly as a function f(x,y) of the independent variables x,y (e.g., f(x,y) = a00 + a01y + a10x + a11xy).
The normals can be identified in at least two equivalent ways.
The first getting its contained from F(x,y,z) = zf(x,y) = 0, from the normal follows readily.

Angle Sum Theorem

In a triangle the sum of all the three interior angles will be equal to 180o.  If one side of a triangle is produced, the exterior angle so formed is equal to the sum of the interior opposite angles. Here we are going to see about the angle sum theorem.

Understanding Angle Bisector Theorem is always challenging for me but thanks to all math help websites to help me out.

Angle sum theorem:

Angle sum theorem of a triangle is equal to two right angles, i.e., 180 degrees
Given:
ABC is a triangle
To Prove
Angle A + Angle B + Angle ACB = 180o
Produce BC to D. Through C draw CE || BA.

Proof of angle sum theorem of a Triangle:

Statement
Reason
1. Angle A = Angle ACE Alternate angles angles BA is parallel to CE
2. Angle B = Angle ECD Corresponding angles BA is parallel to CE
3. Angle A + angle B = Angle ACE + Angle ECD statements (1) and (2)
4. Angle A + angle B  = Angle ACD statement (3)
5. Angle A + Angle B + Angle ACB = Angle ACD + Angle   ACB adding Angle ACB to both sides
6. But Angle ACD + Angle ACB = 180o linear pair
7. Angle A + Angle B + Angle ACB = 180 ° statements (5) and (6)


I am planning to write more post on Circle Circumference, cbse vi question papers. Keep checking my blog.

Corollary of Theorem on Angle theorem of a Triangle

If one side of a triangle is produced, the exterior angle so formed is equal to the sum of the interior opposite angles.
Given:
In Triangle ABC, BC is produced to D.
To Prove Corollary of Sum of Angles of Triangle:
Angle ACD = Angle A + Angle B



Proof:


Statement Reason
1. Angle ACB + Angle ACD = 180o. linear pair
2. Angle A + Angle B + Angle ACB = 180o sum of the angles of a triangle = 180
3.  Angle ACB + Angle ACD = Angle A + Angle B + Angle ACB statements (1) and (2)
4. Angle ACD = Angle A + Angle B Reason statement (3); Angle ACB is common

Tuesday, June 4

Hard Fraction Problems

Hard fraction problems:

This articles discusses hard fraction problems solving. A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development were the common or "vulgar" fractions which are still used today (`1/2` , `5/8` , `3/4` etc.) and which consist of a numerator and a denominator. (Source – Wikipedia)
Hard fraction problems solve for proper fraction, improper fraction and complex fraction problems.

Example for Hard fraction problems:


Example 1:
Subtract the fractions `4/5``3/4`

Solution:
The denominator (bottom number) is different so we have to take least common denominator (lcd).
LCD = 5 x 4 = 20
`(4 xx 4)/ (5 xx 4)` = `16/20` and `(3 xx 5)/ (4 xx 5)` = `15/20`
`16/20 ``15/20`
The denominators are equals
So subtracting the numerator directly = `(16-15)/20`
Simplify the above equation we get = `1/20`
Therefore the final answer is` 1/20`

Example 2:
Subtract the mixed fractions for given fractions,` 4 7/5``5 8/5`

Solution:
The given two mixed fractions are `4 7/5``5 7/5`
We need convert to mixed fraction to improper fraction `27/5``33/5`
The same denominators of the two fractions, so
                                         = `27/5``33/5`
Subtract the numerators the 27 and 33 = 27 - 33 = - 6.
The same denominator is 5.
                                         = `-6/5`
The subtract fraction solution is -`6/5` .

Example 3:
Multiply the mixed fractions for given two fraction,` 4 2/4` x `5 2/6`

Solution:
The given two mixed fractions are `4 2/4` x `5 2/6`
We need convert to mixed fraction to improper fraction `18/4` x `32/6`
Multiply the numerators the 18 and 32 = 18 x 32 = 576.
Multiply the denominators the 4 and 6 = 4 x 6 = 24
                                         =` 576/24`
The multiply fraction solution is 24

Example 4:
Convert `16/ (5/4)` to a simple fraction and reduce.

Solution:
The given complex fraction `16/ (5/4)`
Can be written as `16/1 -: 5/4`
First we have to take the reciprocal of the 2nd number, and then multiply with the second one
Reciprocal of `5/4 ` is `4/5`
`16/1` x `4/5`
Multiply the numerator and denominator
`(16 xx 4) / (1 xx 5)` = `64/5`
Therefore complex fraction solution is `64/5`

My forthcoming post is on cbse course for class 11, class 11 cbse sample papers will give you more understanding about Algebra

Practice for hard fraction problems:


Problem 1: Convert `5/ (5/4)` to a simple fraction and reduce.
Solution: 4.
Problem 2: Adding the mixed fractions for given two fraction,` 4 2/3` + `5 2/3`
Solution: `31/3`
Problem 3: Subtract the mixed fractions for given fractions,` 4 7/3``5 8/3`

Solution:`-4/3`

Written Translation

Introduction:
In word problem we can use the English words for translating words into mathematical expressions. In the translation we can use the math equation. This is used in the math problem to convert in the fairly simple problems. But we can’t do the figuring problem. This translating method is help to solve the word problems.
  1. Working clearly will help you think clearly, and
  2. Figuring out what you require will help you convert your final answer back into English.

Written translation - Translation methods:


  • Addition
  • Subtraction
  • Multiplication
  • Division
  • Power

Written translation - Addition
Words that mean adding
words that mean addingexample
the sum ofthe sum of a number and 6y + 6
more than6 more than a numbery + 6
plusa number plus twoy + 2
added to6 added to a numbery + 6
increased bya number increased by 3y + 3


Written translation - Subtraction

Words that mean subtracting
words that mean subtractingexample
less than6 less than a numbery - 6
less6 less a number6 - y
minus6 minus a number6 - y
subtract fromsubtract 6 from a numbery - 6
reduced bya number reduced by 6y - 6
decreased bya number decreased by 6y - 6
diminished by6 diminished by a number6 - y

Written translation - Multiplication


Words that mean multiplying
Words that mean multiplyingexample
multiplymultiply a number by 33y
multiplied bya number multiplied by 33y
times4 times a number4y
doubledouble a number2y
twicetwice a number2y
tripletriple a number3y


Written translation - Division

Words that mean dividing
Words that mean dividingexample
divided bya number divided by 6y/6
divide intodivide 6 into a numbery/6
the quotient ofthe quotient of a number and 3y/3
half ofhalf of a numbery/2
one third ofone third of a numbery/3

Power

Word that mean rising to the nth power
words that mean raising to the nth powerexample
a number raised to the nth powera number raised to the fifth powery6
raise a number to the nth powerraise a number to the fifth powery6
the nth power of a numberthe fifth power of a numbery6
squareda number squaredy2
cubeda number cubedy3