solving online area of a circle

CIRCLE:

A line forming a closed loop, every point on which is a fixed distance from a center point.

A circle is a plane continuous figure connecting points which are equidistant from a given point, known as the center.

The word circle originates from the Latin word 'circus'. Chariot races, very popular in the Roman era were either circular or oblong and eventually the word was used to describe the shape as well.


Area of a circle - Important terms

In order to find the area of a circle, it is essential that we first understand a few important terms related to circles.

The outermost portion of the circle which separates the interior of the circle from the exterior is known as the circumference.

The center, as already mentioned, is the point inside the circle which is equidistant from all points lying on the circumference.

The distance between any point on the circumference from the center is known as the radius.

A line segment connecting any two points of lying on the circumference on the circle is called a chord.

The longest possible chord of a circle is known as a diameter. The diameter equals twice the radius.


Solving the area of a circle

Consider a circle with center P. Let us denote the radius of the circle by 'r' and the diameter by 'd'. Pi (denoted as π) is a numerical value, which is a crucial number used to calculate various attributes of circular figures. Its value up to 2 decimal points is 3.14. In terms of the radius.

we define the area (A) of the circle as:                  A = π r²

In terms of the diameter, which is twice the radius, this equation can be re-written as:

A = 0.25 π d²

The area in terms of the circumference (C) of the circle is given as:

A = C²/ 4π

Examples and exercises of area of circle.


Example 1: The radius of a circle is 3 inches. What is the area?

A = π r²

A  =  (  3.14  ) ( 3 ) 2

A = (3.14) (9)

A =  28.26 in 2

Answer  the following :

1) The diameter of a circle is 8 centimeters. What is the area?

2) The radius of a circle is 5 feet. What is the area?

Half Angle Properties

Introduction:

Let we will discuss about the half angle properties. The half angle is the angle that is half of original angle. Tha is, the product of two half angles at an edge is equal to the original(full) angle at that same edge. We will develop the properties formulas for half angle of both sine and cosine.


Half angle properties formula for sine:


We should start with formula for cosine of double angle. That is,
cos 2θ = 1 - 2sin^2 θ

Half angle properties formula - sine:

Let us consider,


Then 2θ = α and the formula becomes:


Solve for,


That is, we acquire sin(α/2) lying on the left of equation and everything else on right,



Solving of equation gives the following sine of half-angle identity:


The sign of sin α/2 represents on quadrant in which α/2 lies.
If α/2 is in the first or second quadrants then formula uses the positive case:


If α/2 is in the third or fourth quadrants, the formula uses the negative case:


Example:

Find the value of sin 45o using the sine half-angle properties.

Solution:

The sine half angle formula is ,



Therefore,

sin 45o = `sqrt((1 - cos 90 )/(2))`

= `sqrt((1 - 0 )/(2))`

= `sqrt((1)/(2))`

= 0.707


Half angle properties formula for cosine:


Half angle properties formula - cosine:

Using same process, with the similar replacement of θ = α / 2. We want to substitute into the identity,


We get,


Reverse process of the equation should be,


Addition of 1 with both sides of an equation. We get,


Making division process on both sides of equation by 2


Solving for cos(α/2), we get,


As before, the sign we need depends on quadrant.
If α/2 is in first or fourth quadrants that formula uses in positive case:


If α/2 is in the second or third quadrants, the formula uses the negative case:


Example:

Find the value of cos 115o using the cosine half-angle formula.

Solution:

We need to find cos 115o

That is, α = 230°, and so α/2 = 115°.

Therefore, cos 115o = `sqrt((1 + cos 230 )/(2))`

= `sqrt((1-0.643)/(2))`

= `sqrt((0.357)/(2))`

= 0.422

We have seen about the half angle properties.

Geometric Area

The geometric area refers to the size of the interior of a planar (flat) figure. Area (A) is a two-dimensional measure. The square units that  an area is measured with square inches, square feet and square centimeters. Hectares and acres are also considered in some special cases. In the International System of Units (SI), the standard unit of area is the meter squared (m ²).

Geometric area of Square

Formula to find the area of the square:

Area of the square is calculated by multiplying the base times itself.

Area of square = side x side square unit.

Area = a²

1. The side length of the square is 10 feet; find the area of the square.

Solution:

Area = a²

Given: Side length= 10feet.

                               = 10 x 10

                   Area    =100 feet²

2. The side length of the square is 4.5 m; find the area of the square.

Solution:

Area = a²

Given: Side length= 4.5m.

                              = 4.5 x 4.5

                      Area =20.25 m²

Geometric area of Rectangle

Area of the rectangle is calculated by multiplying the base times the height.

 Area of rectangle     =  (length x width) square unit.

Area = l x w square unit

3. The length and breadth of the rectangle are 10 meters and 5 meters respectively .find the area of that rectangle.

Solution:

Area = l x w square unit

           
Given: Length= 10 meters

                        Breadth=5 meters

                                    =10x5

                        Area    = 50 m²      

     

4. The length and breadth of the rectangle are 11 feet and 6 feet respectively .find the area of   that rectangle

Solution:

Area = l x w square unit.

            Given: Length= 11feet

                        Breadth=6feet

                                    =11x 6

                        Area    = 66 ft²

Geometric area of Triangle

Area of the triangle is calculated by multiplying the base times the one-half the height.

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

Area = `(1)/(2)bh`

5. What is the area of triangle with base 5 and height 10 feet?

Solution:

Area = `(1)/(2)bh`

Given: Base= 5 feet.

         Heigh t= 10feet

                    = 1/2 (5x 10)

                    =1/2(50)

            Area =25 ft²

6. What is the area of triangle with base 7 and height 14 feet?

Solution:

Area = `(1)/(2)bh`

Given: Base= 7 feet.

         Height=14feet

                    =1/2 (7 x 14)

           Area =1/2 (98)

            Area =49 ft²

Geometric area of Circle

Area of the circle is calculated by multiplying pi ( π = 3.14) by the square of the radius

Area of the circle= π x r²

7. The radius(r) of a circle is 3 inches. Find the area of that circle?

Solution:

Area of the circle= π x r²

π=3.14

Given: r=3inches

          A=3.14 x (3)²

                   =3.14 x 9 inche^2.

    Area= 28.26 in^2.

8. The radius(r) of a circle is 7.5 meters. Find that area of the circle.

Solution:

Area of the circle is calculated by multiplying pi ( π = 3.14) by the square of the radius

Area of the circle= π x r²

π=3.14

Given: r= 7.5 meters

          A= 3.14 x (7.5)²

            =3.14 x 56.25

  Area =176.625 m²

circumference circle

In our real life, we could see many circular objects. For example, wheel of a vehicle, shape of a pizza, tennikoit ring,.. Take a rope or an inelastic thread and measure the circular path of the given object and that forms the circumference of the circular object. As the size of the circle increases, the circumference of circle also increases.

Introduction on circumference of a circle:

Circle is one of the most important shapes in geometry. Circumference is defined as the measure needed to make the closed curve. Circle has a fixed center. Distance between the center and a point on the circumference of the circle is called as radius. A line passing through center and any two points on the circumference of the circle is called as diameter.  If the radius or diameter is given then we can easily calculate the circumference of the circle. This article helps us to find the circumference of the circle with the given details.


Circumference of a circle:


The diagram below shows the circumference of circle.



The Circumference of a circle can be calculated using the following formula:

Circumference = 2  `pi`r  or `pi`

where r ----> radius of circle

d ----> diameter of circle

π---- > constant and value of π is  3.14 or 22/7

Diameter is twice the radius. So d = 2r and   r = `d/2`


Problems on finding circumference of the circle:

Ex 1: Find the Circumference of the circle with radius 20 cm.

Sol:

Step 1: Write the formula

Circumference of a circle  = 2  `pi`r units

Step 2: Plug the known values and calculate circumference

2π r               = 2 (22/7) 20

= 2 (3.14) 20

= 125.6 cm

Ex 2: Find the Circumference of the circle with radius 32 cm.

Sol:

Step 1: Formula for Circumference of a Circle

Circumference of a circle  = 2  `pi`r units

Step 2: Plug the known values and calculate circumference

2πr                  = 2 (22/7) 32

= 2 (3.14) 32

= 200.96 cm

Ex 3: Find the Circumference of the circle with diameter 52 in.

Sol:

Step 1: Write the formula

Circumference of a circle  = `pi` d

Step 2: Plug the known values and calculate circumference

πd             = (22/7) 52

= (3.14) 52

= 163.28 in

Ex 4: Find the Circumference of the circle with diameter 74 in.

Sol:

Step 1: Write the formula

Circumference of a circle  = `pi` d

Step 2: Plug the known values and calculate circumference

πd            = (22/7) 74

= (3.14) 74

= 232.36 in

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Practice problems on circumference of the circle:


Find the Circumference of the circle with radius 18 cm. .
Answer: 113.04 cm

Find the Circumference of the circle with radius 28 cm.
Answer: 175.84 cm

Find the Circumference of the circle with diameter 63 in.
Answer: 197.82 in

Find the Circumference of the circle with diameter 81 in.
Answer: 254.34 in

Different Number Combinations

In combinatorial mathematics, a k-combination of a finite set S is a subset of k different numbers of S. Specifying a subset does not arrange them in a particular order; by contrast, producing the k different numbers in a specific order defines a sequence without repetition, also called k-permutation (but which is not a permutation of S in the usual sense of that term. SOURCE: WIKIPEDIA



Example problems of different number combinations:

Different number combinations example 1:

How many lines can you draw using THREE non collinear (not in a single line) points X, Y and Z on a plane?

Solution:

You need two points to draw a line. The order is not important. Line XY is the same as line YX. The problem is to select TWO points out of THREE to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

XY, XZ

YX, YZ

ZX, ZY

There is a problem: line XY is the same as line YX, same for lines XZ and ZX and YZ and ZY.

The lines are: XY, YZ and XZ; three lines only.

So in fact we can draw THREE lines and not SIX and that's because in this problem the order of the points X, Y and Z is not important.

This is a combination problem: combining TWO items out of THREE and is written the general form as follows:

n C r = n! / [ (n - r)! r! ]

Special case:

n C 0 = n C n = 1

The number of combinations (nCr) is equal to the number of permutations divided by r! to removes those counted more than once because the order is not mainly use.

Different number combinations example 2:

Calculate 4C3

Solution:

We can find the combination value by using the following formula:

n C r = n! / [(n - r)! r!]

Substitute the value of n and r into the above formula, then we get

4C3= 4! / [(4-3)! 3!]

=4! / [1!*3!]

=24/[1*6]

=24/6

=4

Answer: 4

Different number combinations example 3:

Calculate 6C6

Solution:

We can find the combination value by using the following formula:

n C r = n! / [(n - r)! r!]

Substitute the value of n and r into the above formula, then we get

6C6= 6! / [(6-6)! 6!]

=6! / [0!*6!]

=6! / [1*(6!)]

=6! / 6!

= 1

Answer: 1

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Practice problems of different number combinations:


1) In how many ways can you select a committee of 2 students out of 5 students?

Answer: 20 (5C2)

2)  How many triangles can you make using 4 non collinear points on a plane?

Answer: 24 (4C3)

Learn Area of a Rhombus

Rhombus is a quadrilateral in which all the sides are equal. Rhombus has all the properties of a parallelogram and its diagonals bisect each other at right angles. The area of a rhombus can be calculated using the conventional formula used to calculate the area of a parallelogram b*h where "b" is the base and "h" is the perpendicular distance between the base and the opposite parallel line.



ABCD is a rhombus and we shall derive the formula for finding out the area given its diagonals AC and BD


learning how to find area of a rhombus


Let the length AC be d1 and that of BD be d2. The formula for calculating the area of a triangle is 1/2 bh where 'b" is the base and "h" is height or altitude.

We shall consider the rhombus as the combinations of two triangles ABC and ADC.If we add the areas of triangles ABC and ADC we are sure to get the area of the rhombus ABCD.

The diagonals of rhombus bisect each other at right angles the height of triangles of ABC and ADC will be half of the diagonal BD which is equal to d2/2. We shall now calculate the area separately and add.


Formula to learn area of a rhombus


Area of triangle ABC =1/2  (d1* d2/ 2) =(d1 d2) /4

Area of triangle ADC=1/2  (d1* d2/ 2) =(d1 d2) /4

Area ABC +Area ADC = (d1 d2 )/4 + (d1 d2 ) /4 =d1 d2 /2

The area of the rhombus  = 1/2 product of its diagonals

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learn examples on area of a rhombus


We shall illustrate this by two examples. We are given the diagonals of a rhombus as 10 cms and 12 cms. We are asked to find the area of the rhombus. The area of a rhombus= 1/2 the product of its diagonals= 1/2 *10*12 = 60 square cm.

This is an example to find the area of the rhombus using the conventional formula. Suppose the length of the side of a rhombus is given as 12cm and the perpendicular distance between the sides is given as 10 cm the area of the rhombus =side* perpendicular distance = 10 * 12 =120square cm.

Rational Algebraic Expressions

A rational expression is an algebraic expression of the `(P)/(Q)` from ,where P and Q are simpler expressions (usually polynomials), and the denominator Q is not zero.



This rational expressions whose numerators and denominators are (or written as) polynomials. Like polynomials, rational expressions appear frequently in the Algebra II and higher mathematics. We must understand the how to perform basic operations with rational expressions and how to solve rational equations.


Two Types:-


1.Adding and Subtracting Rational Expressions

2.Multiplying and Dividing Rational Expressions



1. Adding and Subtracting Rational Expressions

Adding Subtracting Rational Expressions with the Like Denominators

Add and subtract their numerators and then write the results over the denominator. Then, simplify the numerator factor, and write the expression in lowest terms.

Adding and Subtracting Rational Expressions with the Unlike Denominators

Rational expressions can be added and subtracted if and only if they have the same denominator. Thus, we must rewrite them as expressions with a common denominator.



2.Multiplying and Dividing Rational Expressions

Multiplying Rational Expressions

To multiply two rational expressions, factor them. Then multiply their numerators and denominators, crossing out the any factors that appear in both the numerator and the denominator

Dividing Rational Expressions

To divide by rational expression,and multiply by its reciprocal.


Examples:-


Example 1:

Simplify the rational expression

4x - 2
--------
2x - 1
Solution :-

Factor both the numerator and denominator completely.

2(2x - 1)
---------
2x - 1
Cancel common factors to reduce and then simplify the given expression.

2(2x - 1)
= ------------ = 2 , with x not equal to 1
2x - 1


Example 2:

Simplify the rational expression

4x + 16
---------
x2 - 16
Solution:-

Factor both the numerator and denominator completely.

4x + 16         4(x + 4)
--------- = ----------------
x2  - 16      (x + 4)(x - 4)
Cancel common factors to simplify the expression.

4(x + 4)
= ---------------
(x - 4)(x + 4)


4
= --------    , with x not equal to -4.
(x - 4)

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Example 3:

Simplify the rational expression

x2 + x - 2
-------------
-x2 -2x + 3

Solution:-

Factor both the numerator and denominator completely.

x2 + x - 2      (x + 2)(x - 1)
------------- = ----------------
-x2 -2x + 3      (x + 3)(-x + 1)


Note that - x + 1 = - (x - 1) in the denominator.

(x + 2)(x - 1)
= -----------------
-(x + 3)(x - 1)


Cancel common factors.

(x + 2)(x - 1)
=--------------------
-(x + 3)(x - 1)


(x + 2)
= -------------   , with x not equal to 1
(x + 3)