Study about Unit Circle

In Euclidean geometry the simple shape is called circle and it consisting points in a plane which is middle from a given point called the center. Radius is the distance of the points from the center.

Unit circle:

In geometry, a unit circle is a circle with a radius of one. The unit circle is the radius one centered at the origin (0, 0) in the Cartesian coordinate system in the Euclidean plane except the trigonometry. The higher dimensions is the unit sphere is denoted by s1.

If (x,y ) is the first quadrant of the unit circle, then p and q are the lengths of a right triangle whose hypotenuse has length 1. That is, from the Pythagorean Theorem satisfy the following equation with x and y,

x2 + y2 = 1.

Explanation:


1. Length of circumference:

The circumference’s length is related to the radius (r) by

c = 2`pi` r

diameter d  = 2r

r = 2 / d

c = `pi`d

2. Area enclosed the circle:

Area of the circle =  `pi ` × area of the shaded square

The area of the circle is π multiplied with the radius squared:

A =` pi` r2

Area of the circle interms of diameter

A  = ` pi`(d/2)2

Area =` pi` d2 / 4


Trigonometric functions on the unit circle:




In geometrical term the trigonometric functions of the angle θ can be modified to a unit circle centered at O.

If  a point of the unit circle is (x, y) , and if the origin (0, 0) to (x, y) makes an angle t from the x-axis, in the trigonometric functions like sine, cosine terms.

cos(t) =x

sin(t) = y

The equation x2 + y2 = 1 gives the relation

cos2(t) + sin2(t) = 1

Solution:

Finally from properties of circle we conclude that the circle does not have any number of sides. That is circle have 0 sides.

x and y intercept solver

Introduction to x and y intercept solver:

X-intercept:
The x-intercept is the x-value of the point where the graph of a function or relation intersect the x-axis of the coordinate system.

Y-intercept:
The y-intercept is the y-value of the point where the graph of a function or relation intersect  the y-axis of the coordinate system. (Source: Wikipedia)
In this article we are going to see the problems on  x and y intercept by using the x and y intercept solver.

Methods to find x and y intercept using solver:

X-intercept solver:

• We can find the x intercept of the line by substitute y = 0 in the line equation.
• The x intercept can be represented as (x,0)

Y-intercept solver:

• We can find the y intercept of the line by substitute x = 0 in the line equation.
• The y intercept can be represented as (y,0)

Problems on x and y intercept using intercept solver :

Problem 1:
Find the x intercept of the line 5x-3y+15 = 0

Solution:
Given, the line 5x-3y+15 = 0
To find the x intercept we need to substitute y = 0 in the equation of line.
5x-3y+15 = 0
Substitute y = 0 in the above line equation,
5x – 3(0) + 15 =0
5x – 0 + 15 = 0
5x + 15 = 0
Subtract 15 on both sides,
5x - 15 - 15 = 0 - 15
5 x = - 15
Divide by 5 on both sides,
x = - 3
The x intercept of the line is (-3,0)

Problem 2:
Find the x-intercept of the line y = x -12

Solution:

Given, the line y =x -12
To find the x intercept we need to substitute y = 0 in the equation of line.
y = x -12
Substitute y = 0 in the above line equation,
y =x -12
0 = x -12
Add 12 on both sides,
12 = x -12 + 12
12 = x
x = 12
The x intercept of the line is ( 12 , 0)

Problem 3:
Find the y-intercept of the line 3y =6x -15

Solution:
Given, the line 3y =6x -15
To find the y intercept we need to substitute x = 0 in the equation of line.
3y =6x -15
Substitute x = 0 in the above line equation,
3y =6(0) -15
3y = - 15
Divide by 3 on both sides,
y = -15/3
y =-5
The y intercept of the line is ( 0 , -5)

Problem 4:
Find the y-intercept of the line y = 2x - 10

Solution:
Given, the line y = 2x - 10
To find the y intercept we need to substitute x = 0 in the equation of line.
y = 2x - 10
Substitute x = 0 in the above line equation,
y =2(0) -10
y = - 10
The y intercept of the line is ( 0 , -10)

How to find area

Students can learn How to Find Area of a Shape. They can learn to solve problems assocaited to it. Geometry deals with the study of 2 dimensional shapes and 3 dimensional solids and their properties like area, volume, length, surface area etc,.

The amount of surface occupied by the plane figure is called its Area.  The SI unit of area is square meter (m2). Area of regular or irregular shape can be obtained by using basic formulas defined for basic structures like square, triangle, rectangle, circle etc.

Here are some formulas which aid students to learn how to find the area of a shape.

Area of Square = a^2     

Area of Rectangle = ab       

Area of Parallelogram = bh         

Area of Trapezoid = h/2 (b1 + b2)   

Area of Circle = `pi`r 2          

Area of Triangle  = `(1)/(2) b.h`     

In the following example students can learn how to find area of a shape. Students can follow similar steps to compute the area of various shapes.

Let us take l=10 meters and b=6 meters

For ex:  This example shows how to find the area of a shape. This is the compuation of the area of a Rectangle. The area of a rectangle is the product of its length and width. The formula to be used is A = l * w, where l = length and w = width of the rectangle



Substituting the values of the dimensions in the formula and necessary computation is to be done to obtain the final value. For rectangle, it can be calculate as follows:

A= l * b = 10 * 6 = 60

The area of the rectangle is 60 square meters


Finding area


Here are some problems on how to find area of a shape.

Let us consider another sample problems on how to find area of a shape to have a better understanding.

For example, how to find area of a shape, the shape being a Circle in this example.



The area of the circle is A= π * r 2                  , where r is the radius of the circle.

Here, the value of the constant is π=3.14

Ex:  How to find area of a circle with radius 4 m.

Sol:  Radius = 4 m.

substituting the values in the formula  A = Π r2

A = 3.14 *42

= 50.24

The area of the circle is calculated to be 50.24 square meters.

Pro 1:   How to Find area of a circle whose radius is 5meters.

Ans: 78.5 square meters

Pro 2:  How to Find area of a rectangle whose dimesions are 3meters and 9meters.

Ans: 27square meters

Pro3 :  The side of a cube is 5meters.Find its total surface area.

Ans:150 square meters

Pro 4: The radius of a sphere is 4cm.Find its area.

Ans: 200.96 square centimeters

Find area of irregular shape


First the irregular figure is to be split into known regular figures. We already know how to find the area of a shape i.e. regular shapes. Thus, the area of the irregular figure can be calculated. Students can learn about how to find area of a shape which is irregular from the following examples:

Ex: Find the area of the following irregular figure



Sol: To find its area it is to be divided into known figures. The above given irregular figure can be divided into two known figures. That can be done a shown in the above figure.

Step 1: In the figure, the dotted line splits the irregular figure into two known figures like the rectangle with length (L) 14mm and width (W) 10mm and semi- circle with diameter (D) 14mm. Now the area of the irregular figure can be calculated as follows:

Step 2; Area of irregular figure= (Area of rectangle) + (Area of semi-circle)

= (L*W) + (π * D *D*0.25)

Step 2:                                        = (14 *10) + (3.14 *14*14*0.25)

Step 3:                                        = 140 + 153.86

Step 4:                                         =293.86 square mm

Thus the area of an irregular figure can be calculated.


Problems on Irregular figures:

Find the Area of the following irregular figures


Answers:      1. 51 square cm

2. 60 square cm

3. 27 square inches

Students can learn how to find area of a shape from the above examples and practice more problems on similar lines.

learning diameter

In geometry, the diameter is a  circle is any straight line segment that passes through the center of the circle and whose endpoints are on the circle. The diameters are the longest chords of the circle. In more modern usage, the length of a diameter is also called the diameter. In this sense one speaks of the diameter rather than a diameter, because all diameters of a circle have the same length, this being twice the radius.


The diameter Formula;


The diameter of circle:

Diameter of circle = 2 * r.

where, r is radius of the circle.


Problems to learning diameter


Example in diameter learning 1:

Find the area, diameter and circumference of circle and with the radius of 4cm.

Solution:

1.The area of circle:

Area of circle = ∏ *r*r.

=(3.14)*4*4.

= 50.24 cm2.

2.The diameter of circle:

Diameter of circle = 2*r.

= 2* 4.

= 8cm.

3.The circumference of circle:

Circumference of circle = ∏ * d.

= (3.14)* 8.

= 25.12cm.


Example in diameter learning 2:

Find the diameter of a circle if the radius is 35centimeter.

Solution:

Given radius = 35

Diameter =?

Diameter = radius * 2

= 35 * 2

= 70

So the diameter of the given circle is 70 centimeter.

Example in diameter learning 3:

Find the diameter of a circle if the radius is 44 centimeter.

Solution:
Given radius = 44

Diameter =?

Diameter = radius * 2

= 44 * 2

= 88

So the diameter of the given circle is 88 centimeter.

Example  in diameter learning 5:

Find the diameter of a circle if the radius is 25 centimeter.

Solution:

Given radius = 25

Diameter =?

Diameter = radius * 2

= 25 * 2

= 50

So the diameter of the given circle is 46 centimeter.

Example in diameter learning 6:

Find the diameter of a circle if the radius is 70 centimeter.

Solution:

Given radius = 70

Diameter =?

Diameter = radius * 2

= 70 * 2

= 140

So the diameter of the given circle is 140 centimeter.

Example  Circumference to Diameter learning 7:

Find the diameter of a circle if a radius is 10 centimeter.

Solution:

Given radius = 10

Diameter =?

Diameter = radius * 2

= 10 * 2

= 20

So the diameter of the given circle is 8 centimeter.

Learn Online Pythagorean

Pythagorean Theorem is a very important theorem used in Trigonometry.  It gives a relationship between the sides of a right triangle. A Greek philosopher Pythagoras was the founder of Pythagorean Theorem.

Introduction to Pythagorean Theorem:

Pythagorean Theorem describes a relationship between the longest side of the right triangle and the remaining smaller sides.

Statement: In right triangle, the square of longest side (hypotenuse) is equal to the sum of square of the remaining two sides.


Explanation of Pythagorean theorem:


In a right-angled triangle,

Let ‘a’ = adjacent side, ‘b’ = opposite side, ‘c’ = hypotenuse



Using Pythagorean Theorem,

(Hypotenuse) 2 = (adjacent side) 2 + (opposite side) 2.


Examples of Pythagorean Theorem :


Ex 1: Find the hypotenuse of the right triangle when the adjacent and opposite sides of the right angled triangle is 4cm and 3cm.

Sol:   The triangle given is a right angled triangle , it obeys pythagorean theorem

Step I:         (adjacent side)2+(opposite side)2= (hypotenuse side)2

Step II:    42 +32 = x2

Step III:    16+9 =x2

Step IV:     25 = x2

Step V:    x = 5

The hypotenuse side of the triangle is 5cm

Ex 2: Find the adjacent side of the right triangle when the hypotenuse and opposite sides of the right angled triangle is 10cm and 6cm.

Sol:  The given triangle is a right angled triangle , it obeys pythagorean theorem

Step I:    (adjacent side)2+(opposite side)2= (hypotenuse side)2

Step II:    x 2 +62 = 102

Step III:    100-64 =x2

Step IV:    36= x2

Step V:        x=6

The hypotenuse side of the triangle is 6cm

Ex 3: Find the opposite side of the Right Triangles Trigonometry when the hypotenuse and adjacent sides of the right angled triangle is 10cm and 5cm.

Sol:   The given triangle is a right angled triangle , it obeys pythagorean theorem

Step I:    (adjacent side)2+(opposite side)2= (hypotenuse side)2

Step II:    52+ x 2 = 102

Step III:    100-25 =x2

Step IV:     75= x2

Step V:     x=8.7

The hypotenuse side of the triangle is 8.7cm

These examples are used to learn online pythagorean theorem.

How to Find the Median

INTRODUCTION:

Median is the middle value of a given numbers or allocation of their ascending order. Median is an average value of the two middle elements when the size of the allocation is even.

To locate the Median, place the numbers in ascending order and find the middle number.
If there are two middle numbers then average those two numbers to find median.


Median for Odd numbers


Ex 1:

Find the median for the following list of values:

9, 3, 44, 17, 15

Solution:

Find the Median of: 9, 3, 44, 17, and 15 (Odd amount of numbers)

Line up your numbers: 3, 9, 15, 17, and 44 (smallest to largest)

The Median is: 15 (The number in the middle)

Ex2:

Find the median for the following list of values:

13, 18, 13, 14, 13, 16, 14, 21, 13

Solution:

Find the Median of: 13, 18, 13, 14, 13, 16, 14, 21, and 13(Odd amount of numbers)

Line up your numbers: 13, 13, 13, 13, 14, 14, 16, 18, and 21 (smallest to largest)

The Median is: 14 (The number in the middle)


Median for Even numbers


Ex 1:

How to find the median for the following list of values:

8, 3, 44, 17, 12, and 6

Solution:

Find the Median of: 8, 3, 44, 17, 12, and 6 (Even amount of numbers)

Line up your numbers: 3, 6, 8, 12, 17, and 44(smallest to largest)

Add the 2 middles numbers and divide by 2:

= (8 + 12)/2

= 20 ÷ 2

= 10

The Median is 10.

Ex 2:

How to find the median for the following list of values:

8, 9, 10, 10, 10, 11, 11, 11, 12, 13

Solution:

Find the Median of: 8, 9, 10, 10, 10, 11, 11, 11, 12, and 13 (Even amount of numbers)

Line up your numbers: 8, 9, 10, 10, 10, 11, 11, 11, 12, and 13 (smallest to largest)

Add the 2 middles numbers and divide by 2:

= (10+11)/2

= 21/2

= 10.5

The Median is 10.5

Solving Root of a Number

 If any number is expressed as x × x, then x is the product of two same numbers. We know that 5^2 = 5× 5 =25.Here 25 is called the square of 5 and 5 is called the square root of 25. And (2/3)^2 = (2/3)× (2/3) = 2 ×2/3 ×3 = 4/9. 4/9 is called as square of 2/3 and is also called as square root of 4 / 9.

Examples of Solving Root of a Number

1. Simplify 7^2 = 7 ×7 = 49

Here 49 is called the square of 7 and 7 is called the square root of 49

2. (0.4)^2 = (0.4)× (0.4) = 0.16

Here 0.16 is called the square of 0.4 and 0.4 is called the square root of 0.16

3. Simplify 9^2 = 9 ×9 = 81

Here 81 is called the square of 9 and 9 is called the square root of 81

4. Simplify 121^2 = 121 ×121 = 14641

Here 14641 is called the square of 7 and 7 is called the square root of 14641

5. Simplify 81^2 = 81 ×81= 49

Here 6561 is called the square of 81 and 81 is called the square root of 6561

Multi Step Square Root Examples:

(1). Find the square root of 144

Solution:

Split the number into the product of prime factors.

                              144 = 3 × 3 × 2 × 2 × 2 × 2

                            √144 = √3^2 × 2^2 × 2^2

                                    =3 × 2 × 2

             Therefore √144 = 12

(2). Find the square root of 5^3 × 5^5

Solution:

                         5^3 × 5^5 = (5 × 5 × 5) × (5 × 5 × 5 × 5 × 5)

                                       =5^2 × 5^2 × 5^2 × 5^2

       Therefore √5^3 × 5^5 = √5^2 × 5^2 × 5^2 × 5^2

                                       = 5 × 5 × 5 × 5 × 5

                                      = 625

      Therefore √5^3 × 5^5 = 625

Discuss of Solving Root of Number

(1)   Find the square root of 36
                
(2)   Find the square root of 6^2×7^2

(3)   Find the square root of 8100


Answers:
1. 6    2. 42   3.  90