5 Properties of Parallelogram

Let us study about the 5 properties of parallelogram. Parallelogram is also a type of quadrilateral. Parallelogram is said to have two pairs of parallel sides.
Each pair has equal length. Also their diagonals are seemed to bisect each other in the center point.
Parallelograms have 5 important properties used in their construction point and evaluation point. All those 5 properties of parallelogram are discussed in detail with their examples below.




5 properties of parallelogram:

The 5 important properties of parallelogram are as follows:

a)      Parallelogram opposite sides are said to be parallel.

b)      Opposite sides are seemed to be congruent in parallelogram.

c)      Also parallelogram is said to have congruent opposite angles.

d)     Parallelograms two consecutive angles are supplementary angles.

e)      Parallelogram diagonals seemed to be bisecting each other right at their center point.

Parallelogram opposite sides are said to be parallel:

two pairs of parallel sides

This is the 1st property of parallelogram.
The above figure ABCD is seemed to be having two pairs of parallel sides as (AD, BC) and (DC, AB). Therefore it is a parallelogram.



Opposite sides are seemed to be congruent in parallelogram:

two pairs of congruent sides

This is the 2nd property of parallelogram.
The above figure MNOP is seemed to have the two pairs of congruent opposite sides as (MP, NO) and (PO, MN).



Parallelogram is said to have congruent opposite angles:

two pairs of congruent opposite angles

This is the 3rd property of parallelogram.
The above figure KLMN is seemed to have the two opposite pairs of congruent angles as (angle K, angle M) and (angle L, angle N).


Parallelograms two consecutive angles are supplementary angles:

two consecutive angles are supplementary

This is the 4th property of parallelogram.
The above figure ABCD is showing that all their two consecutive angles are supplementary angles as (A, B) or (A, C) or (C, D) or (B, D).




Parallelogram diagonals bisect each other right at their center point:

diagonals bisecting each other

This is the 5th property of parallelogram.
From the above figure ABCD we prove that the two diagonals of the parallelogram one which extends form A and other which extends from B will meet C and D respectively.
Also they bisect each other exactly at the center position of that parallelogram.

Mathematical Induction

The mathematical induction is one of the numerical proof naturally it is used to find the given statement is accurate for all normal numbers. It can be a deducted by proving that the 1st statement may be the countless succession of the statement is accurate and  we have to prove that if any other statement in the countless succession is real. In this article we shall discuss about the examples involved in induction proof.

Mathematical induction proof:

Principle of Mathematical Induction proof examples:

illustrate the  true for n = 1
suppose true for n = k
explain true for n = k + 1
Conclusion: Statement is true for all n >= 1


Examples in mathematical induction:

Examples in mathematical induction proof 1 :

Prove for n >= 1

1 * 1! + 2 * 2! + 3 * 3! + ….. + n * n! = (n+1)! – 1

This can be denoted by Sigma Notation

`sum_(k=1)^n` kk! = (n+ 1)! - 1

Proof:

Base case: n = 1

The left hand side is 1* 1!. The right hand side is 2! - 1. They are equal.


Examples in mathematical induction proof 2 :

Give the proof  when  amount of the n natural numbers by this formula

1+ 2 + 3 + … + n = n (n+ 1) / 2

Proof:

First, we have to denote the formula is real for n=k ,and it will be

1 + 2 + 3 + … + k = k (k + 1) / 2       (1)

This is the induction statement.  In order to prove  the method  is correct  means then

n = k + 1.

1+ 2 + 3 +… + (k + 1) = (k+1) (k + 2) / 2             (2)

Now add the ( k + 1) on both sides of the above equation in (2)

1 + 2 + 3 + … + k + (k + 1) = k (k + 1) / 2 + (k+ 1)

= k (k+ 1) + 2 (k+ 1) / 2

.

By adding  the  above  fractions we get ,

= (k + 1) (k+ 2) / 2

We can take  (k + 1) as a ordinary factor.

This is line (2), which is the first thing we need to explain

Next, we must explain that the formula is true for n = 1.  We have:

1 = ½· 1· 2

Thus the formula is true for n = 1. And  this fulfilled the both situation of the mathematical induction.

Percent in Math

The term percent in math is used to represent a value relative to hundred.  The same comparison can be expressed as a fraction if we compare a value with respect to another value. If the second value is 100, then we call it as percentage. For example:3/6 , Here we consider 3 relative to 6. If we consider relative to 100, it becomes 50 /100, we call this as 50 percent and is represented as 50 % .

If you need to convert a number into percentage, then multiply the number by 100 and put % symbol.

Now, Let us discuss about some example on percentages

Example problems to find percentage in math:

Percentage is mostly used in banking sector to find the interest.

The formula to find interest is,

Interest =Principle amount x time x rate of interest.



Example 1:

Suppose we need to find the interest will $2000 earn in one year at an annual interest rate of 5%

Solution:

5% = 5 /100

(5/100 ) x 2000 =100

So the answer is $100

Example 2:

Find the interest if amount if $1000 and month is 6 and rate of interest is 6%

Solution:

Interest =Amount x time x rate of interest.

Note : 6 month  = ½ of years.

Therefore,

6%  =>6/100  or 3/50

Substitute these values into interest formula. We get,

Interest = $1000 x ½ x 6/100

=$1000 x1/2  x 3/50

=$500  x 3/50

=$10 x 3

=$30

Example 3:

There are 45 boys  out of 90 students in the class room. Find the percentage of boys in the class room.

Solution:

Total students =90

Out of these 90 students 45 students are boys ,it is represented in terms of fraction.

= 45/90            [fraction form]

Convert the fraction into percentage,

So multiply the fraction number by 100,we get,

(45/90)  * 100

=50%

Example 4:

40 percent of X = 200 then find x?

Solution:

40 % * X =200

40 x */100  =200

40x  =200 *100

40x  = 20000

X =20000/40

X= 2000/4

X=500

Therefore, X value = 500

Verification:

40 percent of 500 =200

(40 x 500)/100  =200

20000/100   =200

200 =200

Hence Verified.


Practice problems to find percentage in math :

1)10 percent of p =100 find P?

Answer: 10

2) Find the interest will $5000 earn in one year at an annual interest rate of10%

Answer : 500

Combinatorial Learning Formula

Combination are selections ie. it involves only the selection of the required number of things out of the total number of things. Thus in combination order does not matter. For example, consider a set of three elements a,b,c and combination
made out of the set with


i) One at a time: a, b, c
ii) Two at a time: a,b, b,c, c,a
iii) Three at a time: a,b,c

The number of combinations of n things taken r, (r < n) is denoted by nCr or ([n],[r])

Learning the evaluation of combinations and the combinatorial formula

Learning to derive the formula for nCr:( Combinatorial formula)
No of combinations from ‘n’ things taken ‘r’ at a time = nCr
No of permutations from ‘n’ things taken ‘r’ at a time = nPr
Number of ways ‘r’ things that can be arranged among themselves = r!
Each combination having r things gives rise to r! permutations

Therefore  nPr = (nCr) r!


=> ((n - r )!) / (n!) = ( nCr) r!


nCr = r!(n - r )!

Learning observations related to combinatorial formula:

(i)   nCo = (n!)/(0! (n - 0 )!) = (n!)/(n!) = 1


(ii) nCn = (n!)/(n! (n -n)!) = (n!)/(n!0!) = 1


(iii) nCr = nCn-r


(iv) If nCx = nCy then x = y or x+y = n


(v) nCr =(^nPr)/(r!)

Learning examples on combinatorial formula

Learning to solve examples using the combinatorial formula :

Ex 1:Evaluate 8C3

Solution:

8C3  =  (8!)/(3! (8 - 3)!) = (8!)/(3! . 5!) = (8 . 7 . 6 . 5!)/((3 . 2 . 1) . 5!)  =  (8 . 7 . 6)/(3 . 2 . 1) = 8 . 7 = 48

OR another method

8C3 = (8 . 7 . 6)/(3 . 2 . 1) = 8 . 7 = 48

Ex 2:

Evaluate: 9C7

Solution:

9C7 = 9C2 = (9 . 8)/(2 . 1) = 9 . 4 = 36

Ex 3:    A committee of seven students is formed selecting from 6 boys and 5 girls such that majority are from boys. How many different committees can be formed?

Sol:

Number of students in the committee = 7
Number of boys = 6
Number of girls = 5
The selection can be done as follows


Boy (6)    Girl (5)
6              1
5              2
4              3


ie. (6B and 1G) or (5B and 2G) or (4B and 3G)

The possible ways are ([6],[6]) ([5],[1]) or ([6],[5]) ([5],[2 ]) or ([6],[4]) ([5],[3])

The total number of different committees formed

=  6C6  x  5c1 + 6C5  x  5C2  +  6C4  x  5C3

= 1 x 5 + 6 x 10 + 15 x 10 = 215

What is Continuous Integration

In what is continuous integration, Let f (x) be continuous and single valued for a <= x <= b. Divide the interval (a, b) of the x axis into n parts at the points (a -= X_0), X_1, X_2… ( X_n -= b) as in below figure.

Define Delta x_i = x_i - x_(i-1) and let xi_i be a value of x in the interval x_(i-1) < xi_i <= x_i. Form the sum of the areas of each approximately rectangular area to get

sum_(i=1)^(n) f(xi_i) Delta x_i

continuous integration

If we take the limit as n ->oo and Delta x_i -> 0, the approximation of the area under the curve as a rectangle becomes more accurate and we obtain the definition of the definite integral of f (x) between the limits a, b; thus

^lim_((n -> oo)_(Delta x_i -> 0)) sum_(i=1)^n f(xi_i) Delta x_i -= int_a^b f(x)dx

The points a, b are referred to as the lower and upper limits of integration, respectively. The definite integral

int_a^b f(x) dx

equals the area under the curve f(x) in the interval (a, b) for what is continuous integration.

Fundamental Theorem - what is continuous integration:

In what is continuous integration, the connection between differential and integral calculus is through the fundamental theorem of calculus. In particular, if f(x) is continuous in the interval a <= x <=b and G(x) is a function such that (dG)/(dx) = f(x) for all values of x in (a, b), then

int_a^b f(x)dx = G(b) - G(a)

The definite integral acts like an antiderivative of f(x), or the inverse operation to differentiation. The concept of inverse differentiation can be made more explicit by introducing the notation

[G(x)]_(a)^(b) = G(b) - G(a)

and using (dG)/(dx) = f(x) in the definite integral above equation to give

int_a^b ((dG)/(dx))dx = [G(x)]_a^b

Example - what is continuous integration:

In what is continuous integration, let f(y) = ln y, u(alpha) = a, and v(alpha) = alpha. In this case, the function f does not depend on alpha. Consequently, we substitute u, v, and f into find.

(d)/(d alpha) int_a^(alpha) In y dy = f(alpha)(d a)/(d alpha) = In alpha

because (da)/(d alpha) = 0 and f(alpha) = ln alpha. This result is verified by first evaluating the integral

int_a^b In y dy = [y In y –y ]_a^alpha = alpha In alpha - alpha - (a In a - a)

Taking the derivative of above equation with respect to alpha gives

(d)/(d alpha)[alpha In alpha - alpha - (a In a –a)] =(d)/(d alpha) [alpha In alpha - alpha]

Answer is     = In alpha

Arithmetic Adding Fractions

A fraction is a part of a whole. A fraction can be altered to a decimal by dividing the upper number, or numerator, by the lower number, or denominator. Fractions are to indicate ratios, and to represent division which is one of the basic arithmetic operations. Thus the fraction 3/4 is also used to signify the ratio 3:4 (three to four) and the division 3 ÷ 4 as well.(three divided by four). Here in this topic we see about adding fractions.

Arithmetic adding Fractions with same denominator - Example problems:

Arithmetic adding Fractions - Example 1:

Add: 2/7   +    3/7

Solution

Here the denominator is equal

So we keep the denominator same as 7 and add the numerator alone

So the solution is   (2+3)/7 =  5/7

Arithmetic adding Fractions - Example 2:

Add:  5/8 +  3/8

Solution

Here the denominator is equal

So we keep the denominator same as 8 and add the numerator alone

So the solution is ,

(5+3) / 8 = 8 / 8  = 1.

Arithmetic adding Fractions with different denominator - Example problems:

Arithmetic adding Fractions - Example 1:

10/8   +    12/9


Answer:

Here the denominators are different that are 8 and 9. So take LCD of 8 and 9 first.

LCD of 8 and 9 is 72.

Therefore,
10 x 9        12 x 8          90      96       186
------------  +   ------------  =  -------- +  ------ = ----------
8 x 9          9 x 8          72       72        72


Simplifying this we get the answer as 31/12 .
Arithmetic adding Fractions - Example 2:

16/3     +   10/10

Answer:

Here the denominators are different that are 3 and 10. So take LCD of 3 and 10 first.

LCD of 3 and 10 is 30.

Therefore,
16 x 10       10 x 3       160       30       190
------------  +   ------------  =  -------- +  ------ = ----------
3 x 10          10 x 3       30       30        30


Simplifying this we get the answer as 190/30 .

Learning Tangent Line

If a line and a curve intersect at a point only then the line said to be tangent to the curve at that point. And the point is called the point of contact.

Tangents at (x, y) to y = f(x)

Lt y=f(x) be a given curve and P(x, y) and Q(x + δx, y + δy) be two neighbouring points on it

Equation of the line PQ is

Y – y = y + δy – y/x + δx – x(X – x)

Or       Y – y = δy/δx(X – x)                                             …(1)

Equation of line PQ

Equation of the line PQ



The line (1) will be a tagent to the give curve at P if Q → P which in turn means that δx → 0 and we know that

Lim δy/δx = dy/dx

δx → 0

Therefore the equation of the tangent is

Y – y = (dy/dx) (X – x)

Geometrical Intrepetation - slope and tangent

Geometrical meaning of dy/dx

From equation (1) we observe that dy/dx represents the slope of the tangent to the given curve y = f(x) at any point (x, y)

...    dy/dx = tan Ψ

Where Ψ is the angle which the tangent to the curve makes with +ive direction of x-axis. In case we are to find the tangent at any point (x1, y1) then (dy/dx)(x1, y1) i.e. the value of dy/dx at (x1, y1) will represent the slope of the tangent and hence its equation in this case will be

y – y1 = (dy/dx)(x1, y1) (x – x1)

Algorithm to draw tangent of a circle

Algorithm to obtain the equations of tangents draw from a given point to a given circle.

Step i. Obtain the point, say (x1, y1)

Step ii. Write a line passing through (x1, y1) having slow m i.e., y – y1 = m (x – x1)

Step iii. Equate the length of the perpendicular from the centre of the circle to the line in step ll to the radius of  the circle

Step iv. Obtain the value of m from the equation in step iii

Step v. Substitute m in the equation in step ll

Normal to the tangent:

The normal to a curve at a given point is a straight line passing through the given point, perpendicular to the tangent at this point. From the definition of a normal it is clear that the slope of the normal m′ and that of the tangent m are connected by the equation m′ = –`1/m .`

i.e., m′ = –1/f ′(x1)

= − 1/  dx/dy    (x1, y1)

Hence the equation of a normal to a curve y = f(x) at a point P(x1,y1) is of the form y – y1= –1/f ′(x1)( x – x1).

The equation of the normal at (x1,y1) is

(i)                x = x1 if the tangent is horizontal

(ii)               y = y1 if the tangent is vertical and

(iii)              y – y1 = –1 / m (x – x1) otherwise.

Example problems:

Example 1 :  Find the equations of the tangents through (7, 1) to the circle x2 + y2 = 25

Solution :  The equation of any line through (7, 1) is

y – 1 = m (x – 7)

mx – y – 7m + 1 = 0             …(i)

The coordinates of the centre and radius of the given circle are (0, 0) and 5 respectively

The line (i) will touch the given circle if,

Length of the perpendicular from the centre = radius

| (1 - 7m)/√(m2 + (-1)2 ) | = 5

│(1 – 7m)/√(m2 + 1)│ = 5

(1 – 7m)2/(m2 + 1) = 25

24m2 – 14m – 24 = 0

12m2 – 7m – 12 = 0

(4m + 3) (3m – 4) = 0

m = –3/4, 4/3

Substituting the values of m in (i), we obtain

3/4x – y + 21/4 + 1 = 0 and 4/3x – y – 28/3 + 1 = 0

3x + 4y – 25 = 0 and 4x – 3y – 25 = 0 which are the required equations.


Example 2: Find the equation of the tangent and normal to the curve y = x3  at the point (1,1).

Solution:

We have y = x3 ; slope y′= 3x2.

At the point (1,1), x = 1 and m = 3(1)2 = 3.

Therefore equation of the tangent is y − y1 = m(x − x1)

y – 1 = 3(x – 1) or y = 3x – 2

The equation of the normal is y − y1 = −`1/m `  (x − x1)

y – 1 =–`1/3` (x – 1) or y = –`1/3` x +4/3 .

normal tangent line graph