Area of Circle Worksheet

Introduction to area of circle worksheet:
A circle is formed by the set of points which are at an equal distance from a center point.

Let us know about the two important terms of the circle.

Circumference :

Total length of the curved figure is called as circumference.

Distance covered by a wheel in one revolution is the circumference.

Area:

Area of the circle is the region covered by the circumference on the plane surface.

Diameter:

It is the line segment whose end points are on the circumference  and passes through the origin.

Radius:

Radius is the distance between the center point and a point on the circumference of the circle.

Greek letter p:

The ratio between the value of circumference to the diameter is always a constant .

It is denoted by the Greek letter p.

p = 3.14 or 22/7


Area of Circle Worksheet-formulas

Let C denotes circumference, A denotes the area, r denotes the radius, d denotes the diameter.

r = d/2

C = pd units

C = 2pr units

A = pr^2 square units

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Area of Circle Worksheet-finding Area of Circle:

We could find the area of the circle using a graph paper by drawing the circle knowing the radius of the circle and then count the number of squares enclosed by the circumference of the circle.

We can estimate the area by counting the number of squares enclosed in the first quadrant or in any of the quadrant and multiply it with 4.

Do you think that it is an easy method?

No

It is very tedious to calculate the number of squares enclosed exactly.

Area of Semi Circle and Quadrant:

Area of semi circle =  area of the circle/2

= pr^2  /2

Area of quadrant =  area of the circle/4

= pr^2/4

Area of Circle Worksheet-problems on Area of Circles:

A circular pool is of the radius 10m. Find the area of the pool.
Solution:

Radius,r = 10 m

Area of the pool = pr^2 square units

= 3.14 x 10 x 10

= 314 m^2

The area of the circle is 16p cm^2. What is its circumference?
Solution:

Area of the circle = pr^2  = 16p

So, r^2 = 16

r = `sqrt(16)`

r = 4 cm.

Circumference of the circle = 2pr

= 2 x 3.14 x 4

= 25.1 cm

Find the area of circle whose circumference measures 66 cm.
Solution:

Circumference of the circle = 2pr = 66

2 x 3.14 x r = 66

r = $\frac{66}{6.28}$

r = 10.5 cm

Area of the circle = pr^2 square units

= 3.14 x 10.5 x 10.5

= 346.185 cm^2

Find the area and radius of the circle, given that the area of the semicircle is 77 cm.
Solution:

Area of the semi circle = 77 cm^2

So, area of circle = 2 x 77

= 154 cm^2

Area of the circle = pr^2 square units

Area of the circle = pr^2  = 154

So, r^2 = 49

r = `sqrt(49)`

r = 7 cm.

Parallel Lines Meet

Introduction to parallel lines:

The lines are the straight curve that never ends. We are having two numbers of lines. They are parallel lines as well as perpendicular lines. The distance of the two parallel lines is always the same. They should not meet at any point. When the two lines should meet at a place which should not be  parallel.


More about Parallel Lines:

The parallel lines are represented in pictorial form should be shown below:

One more example for parallel lines that must not meet.

In first diagram, we are having the three lines named as line A, line B and line C. The distance between the line A and line B should not be changed often. They are always in the same distance only but never meet. This is same for the distance between line B and line C as well as line A and line C.

The parallel lines could never meet at any point. Therefore, there is no meeting point for the lines that are in parallel.

Slope of parallel lines:

The parallel line is having same distance apart. Therefore, the slope of the parallel lines could be equal.

Let us consider, the two lines line X and line Y are in parallel. Then, their slope will be equivalent.

That is, Slope of line X = slope of line Y.

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Example Problem:

Check whether the two equations are in parallel. x-2y+5=0 and x-2y+9=0

Solution:

Equation 1: x-2y+5=0

Subtract 5 on both sides, we get, x-2y = -5

Subtract x on both sides, -2y = -x-5 => -2y = -(x+5)

=>  2y = x+5

Divide 2 on both sides, y = `(x)/(2)` + `(5)/(2)`

y = `(1)/(2)` x + `(5)/(2)`

Therefore, slope = `(1)/(2)`

Equation 2: x-2y+9=0

Subtract 9 on both sides, x-2y = -9

Subtract x on both sides, -2y = -x-9 = - (x+9)

2y = x + 9

Divide 2 on both the sides, y = `(x)/(2)` + `(9)/(2)`

y = `(1)/(2)` x + `(9)/(2)`

Thus, the slope = `(1)/(2)`

The slope of both the equations is in equal.

So, the two lines should be in parallel.

Arc Length of Circle Formula

Introduction to arc length of circle formula:
Arc:

An arc is a portion of the circumference of a circle.The straight line joining the end points of the arc is termed as chord of the circle.If the arc length is exactly half the circle then it is called a semicircular arc.

Arc Length Of the Circle:

The arc length is the measure of the curved line making up the arc on the circumference of the circle.The length of the arc is longer than the straight line distance between its endpoints which is a chord.

Formulas Useful for Solving Arc Length of Circle

The parameters used while solving problems are listed below:

Where `C` = circumference of the circle.

` K` = Area of the circle.

` r ` = radius of the circle.

` d ` = diameter of the circle.

Some Solved Problems on Arc Length of Circle

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Gaussian Elimination Tutor

Introduction 

A system of equations in which all the given quantities are present only in the first degree is called a linear system of equations. The system of equations can be solved by various methods , one of the methods called gaussian elimination method.Here tutor will help you on gaussian elimination

Let us consider the following set of m linear equations with n unknowns
b11 x1 + b12 x2 + ....................+ b1n xn = c1
b21 x1 + b22 x2 + ....................+ b2n xn = c2
............................................................................
............................................................................
bm1 x1 +bm2 x2 + ....................+ bmn xn = cm
Here all b's and c's are constants.
If c1, c2 upto cm.... are all zero, then the system is called homogeneous. The set of values x1, x2, upto ..xn which satisfy all the equations simultaneously is called a solution to the system of equations. Let us discuss how to find this solution online by gaussian elimination method.

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Gaussian Elimination Tutor : Steps

Step 1: Place the leftmost column in the matrix not consisting totally of zeros.

Step 2: If there is zero entry present , then substitute the top row by any another row to get non-zero entry to the top of this column.
Step 3: If this non-zero way used in a multiply the first row through 1/a to bring in a top 1.
Step 4: Add the suitable possible multiples of the top rows to each other row in arrange to contain only zero entries in the column below this important step1.
Step 5: Now remove the top row of the matrix and create once more with step 1 applied to the secondary matrix that remains. Carry on pending the whole matrix is in row-echelon structure.
Step 6: Then create with the last non-zero row and effective upwards, add suitable multiples of every row to the rows over in order to carry in zeroes over the leading 1s.Consider a system of three independent equations with 3 unknowns.

b11 x1 + b12 x2 + b13 x3 = c1
b21 x1 + b22 x2 + b23 x3 = c2
b31 x1 + b32 x2 + b33 x3 = c3
The system of equations is equivalent to the matrix equation BX  =  C
where B = `[[b11,b12,b13],[b21,b22,b23],[b31,b32,b33]]`    ,        X = `[[x1],[x2],[x3]]`,        C = `[[c1],[c2],[c3]]`
Now we have to reduce the coefficient of matrix B to a possible upper triangular matrix
[ A : B ] = `[[b11,b12,b13,:,c1],[b21,b22,b23,:,c2],[b31,b32,b33,:,c3]]`

Gaussian Elimination Tutor : Examples

Answer the given set of  linear system using Gaussian elimination Method:

x - 3y + z = 4,
2x + 8y - 8z = -2,
-6x + 3y + 15z = 9.
Solution:
The increased matrix for the method is
1        -3      1        |        4
2           8     -8      |        -2
-6        3       15      |        9
Add -2*row 1 to row 2 to get a zero directly below the leading 1. Also add 6*row 1 to row 3 to get a second zero.
1        -3        1     |          4
0        -2       6      |        -10
0        -15   -9      |        33
Consequently multiply row 2 by 1/2 to get a leading 1 and multiply row 3 by 1/3.
1        -3           1  |              4
0        -1          3   |             -5
0        -5         -3  |             11
Add -5*row 2 to row 3 to get a zero under this leading 1.
1        -3           1      |        4
0          -1         3      |       -5
0           0           -18 |      36
We at the present twist around the change procedure and turn the increased matrix into a system of equations we have
x - 3y + z = 4,
0x - y + 3z = -5,
0x + 0y - 18z = 36.
- 18 * z = 36.
z = - 2. (Answer)
- y + 3(-2) = -5
- y - 6 = -5
y = - 1
x - 3y + z = 4,
x - 3(-1) - 2 = 4,
x + 3 – 2 = 4
x + 1 = 4
x = 3
Then the solution is (x, y, z) = (3, -1, -2)

Partial Differentiation Versus Implicit Differentiation

Differentiation Definition
Differentiation is the rate of change of a function with respect to one of the variables. It is something like finding the slope of the tangent line of a function at a point. We can define differentiation of a function f(x) with respect to x as f’(x) = lim(h->0)[f(x+h) – f(x)]/h

Partial Differentiation 
A function in two variables given by f(x,y) when differentiated by keeping one of the variables constant, like differentiating y with respect to x keeping y constant we get what is called Partial differentiation of the function, it is denoted as doe(f)/doe(x) or fx when it is with respect to x and when the function in two variables is differentiated with respect to y keeping x constant it is denoted as doe(f)/doe(y) or fy.
For instance, let us find the partial differentiation of the function, f(x,y) = 2x^2y+ 3x +y
Solution: Keeping y constant let us differentiate first with respect to x, we get
fx = doe(f)/doe(x) = doe/doe(x) [2x^2y+3x+y]
   = doe/doe(x)[2x^2y] + doe/doe(x) [3x] + doe/doe(x) [y]
   = 4xy +3 +0 = 4xy + 3
Now keeping x constant let us differentiate the function with respect to y, we get
fy = doe(f)/doe(y) = d/dy [2x^2y+3x+y]
    = doe/doe(y)[2x^2y] = d/dy[3x] + d/dy [y]
    = 2x^2 + 0 + 1 = 2x^2 + 1
So, the partial differentiation of the funcntion f(x,y) = 2x^2y+3x + y is fx= 4xy =3 and fy =2x^2+1

Implicit Differentiation Solver
In Implicit Differentiation the differentiation is done with respect to x when the given function involves two variables, x and y. But here we do not assume the variable y to be constant, y treated as it is.  Usually x is assumed to be the independent variable and the variable y is assumed to be the function of x, the independent variable. For differentiation, chain rule is applied with respect to x followed by some algebraic solving steps to get dy/dx. The formula for Implicit Differentiation solver is, given a function F(x,y) equals zero, which defines a differential relationship of the two variables x and y, then,

dy/dx = - [doe(F)/doe(x)]/[doe(F)/doe(y)] here, doe(F)/doe(x) is the partial derivative of F with respect to x and doe(F)/doe(y) is the partial derivative of F with respect to y.

Let us solve the Implicit differentiation of the function, x^3y^2- xy = 5
Solution: Implicit differentiation, dy/dx [x^3y^2 – xy] = dy/dx [5]
Applying chain rule, we get
dy/dx [x^3y^2] – dy/dx [xy] =0
3x^2y^2 + x^3. 2y. dy/dx – 1.y – x. 1. dy/dx = 0
(2x^3y- x) dy/dx = y – 3x^2y^2
dy/dx = [y-3x^2y^2]/[2x^3y – x]

Know more information on Differentiation and Differential Equations.

Standard Deviation Examples

What is Standard Deviation?
Standard deviation is used in probability theory and statistics.  It shows how much variation exists in the data set from the average. The average can also be called as mean or expected value. If the variance of a data set is known, then the standard deviation can be obtained by finding the square root of it. The standard deviation can be calculated for a data set, random variable, probability distribution or statistical population.  An important property of standard deviation that makes it the preferred method of statistical measure is that the standard deviation computed for a data set has the same unit as that of the data set.

Standard Deviation in Statistics
Statistics Standard Deviation is used to measure the dispersion of the set of data given.   Standard Deviation Statistics shows how some data has high standard deviation and some have low standard deviation. Standard Deviation examples such as volatile stock and blue chip stock demonstrate this difference. The high standard deviation in volatile stock and low standard deviation in stable blue chip stock shows how standard deviation is used for statistical analysis of the given data sets. If the standard deviation is low, it indicates that the data values in the data set are very closely spread around the mean. If the standard deviation is high, it shows the wide dispersal of the data values over a larger range.

Steps to compute Standard Deviation
The first step in computing standard deviation is to find the mean of the data values of the given data set. The next step is to find out how far each data value is deviated from the mean. To compute this deviation, the mean is subtracted from each data value of the given data set and then the values are squared. Then find out the mean for the squared values. Then find the square root of this computed mean. Thus, by finding the square root of the mean of the squared values, you get the standard deviation.

Example of Standard Deviation 
Let us consider an example for standard deviation: The students of a school have participated in Mathematics Olympiad exam. To know their level of knowledge in Mathematics, marks of few students is taken as a sample from the whole population of students who attended the exam and the standard deviation is determined for this sample. This is demonstrated below:

Here is the sample: the marks of 6 students are 91, 100, 100, 98, 96, and 100
Step 1: The mean of these marks will be (92+100+100+99+97+100)/6 which will be 98
Step 2: The next step is to find the squared differences from the mean and sum it: (92-98)^2 +(100-98)^2 +(100-98)^2 +(99-98)^2 +(97-98)^2 +(100-98)^2  = 50
Step 3: Find the square root of the above sum divided by one less than the number of students. Thus you will end up in Square root (50/5), which is equivalent to 3.1623. This resultant value 3.1623 is the standard deviation.

As this standard deviation is low, we can understand that the actual marks scored by the students are very near to their expected marks i.e. 100. As this is a sample data set, it implies that all the students who have attended the Mathematics Olympiad exam have scored good marks.

An overview of the fundamental theorem of calculus

The second fundamental problem addressed by calculus is the area of a region of the plane bounded by various curves. Many practical problems in various disciplines require the evaluation of areas for their solution, and the solution to the problem of areas necessarily involves the notion of limits. On the surface the problem of areas appears unrelated to the problem of tangents. However, the two problems are very closely related. One is the inverse of the other. Finding area is equivalent to finding the anti derivative or finding the integral. The relationship between derivatives and integrals is called the fundamental calculus theorem.

We shall now demonstrate the relationship between definite integral and indefinite integral. A consequence of this relationship is that we will be able to calculate definite integrals of functions whose anti derivatives we can find.
1^st fundamental theorem of calculus:
Suppose that a function f is continuous on an interval I that contains the point a. Let the function F be defined on I  by:

Then F is differentiable on I and F’(x) = f(x) there. Thus, F is an anti derivative of f on I:

2nd fundamental theorem of calculus:
If G(x) is any anti derivative of f(x) on l, so that G’(x) is equal to f(x) on l, then for any b in l we have,

Fundamental theorem of calculus proof for part I:

Using the definition of derivatives we can calculate

Second fundamental theorem of calculus proof:
Continuing from the proof above, if G’(x) = f(x), then F(x) = G(x) + C on l for some constant C. Hence,

Let x = a and obtain 0 = G(a) + C, so C = -G(a). Now let x= b, then we get,

For fundamental theorem of calculus example, anywhere else, we can the t with x or any other variable as the variable of integration on the left hand side.