Solve Basic Math Practice

Introduction to solving basic math practice:

The basic math practice consists of the operation like addition, subtraction, multiplication, and division. In this basic math practice all the operation are easy to solve. We are living in a mathematical world. Whenever we want to purchase, choose an insurance or health plan, we rely on math understanding. The mathematical thinking introduces the new problem solving technique increased dramatically.math problem can be solve usually a rough sheet. but now all are having the system and internet connections so we can able to see in to the computer.Please express your views of this topic Derivative Division Rule by commenting on blog.


Solve addition for basic math practice:


To solve addition for basic math practice has the numbers for addition operation it can be solve by using the following operation.

Example problem 1:

Solve 4+4=?

Given that 4 + 4

4+4=8

Example problem 2:

Solve 4+3+4=?

Given that 4 + 3+4

=7+4 (4+3=7)

=11

Solve subtraction for basic math practice:

To solve subtraction for basic math practice has the numbers for subtraction operation it can be solve by using the following operation.

Example problem 1:

Solve 5 -2=?

Given that 5 -2

=5-2

=3

Example problem 2:

Solve 9 -4 -3=?

Given that 9 -4 -3

=5 - 3 (9-4=5)

=2

Solve multiplication for basic math practice:

To solve multiplication for basic math practice has the numbers for multiplication operation it can be solve by using the following operation.

Example problem 1:

Solve 2*4=?

Given that 2*4

=2*4

=8

Example problem 2:

Solve 2*3*4=?

Given that 2*3*4

=6*4 (2*3=6)

=24

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Solve division for basic math practice


To solve division for basic math practice has the numbers for division operation it can be solve by using the following operation.

Example problem 1:

Solve 3/3=?

Given that 3/3

=3/3

=1

Example problem 2:

Solve 7/3=?

Given that 7/3

=7/3

If the denominator is lesser than numerator

=7/3           (6/3=2)

After that we multiply the numerator* (10)

=7/3 =2.33

=2.33

Solve practice problem

Addition

5+7

8+2

6+3

Subtraction

3-2

5-5

8-9

Multiplication

2*1

2*2

2*3



Division

6/6

5/1

20/2

Answers

Addition

12

10

9

Subtraction

1

0

-1

Multiplication

2

4

6

Division

1

5

10

Basic Statistics Problems Tutor

Introduction to Basic Statistics Problems Tutor

Statistics is the proper science of making effective use of numerical data connecting to groups of individuals or experiments. It deals by means of all aspects of this, counting not only the collection, analysis and interpretation of such information, except the planning of the gathering of data, in terms of the design of surveys and experiments. Now we will study the basic problems given by tutor. Let us see about the basic statistics problems.


Basic Statistics Example Problems Tutor


The following examples problems are used to learn statistics basic. It is the examples are given by the tutor.

Example 1

Find out the mean, median, mode, range of the following given numbers in statistics?

4,8,12,16,19,22,27.

Solution

The given numbers are 4,8,12,16,19,22,27.

Mean

Mean is the average of the given number. First find the total value of the given numbers.

Sum of the given numbers are = 4+8+12+16+19+22+27.

= 108.

The total value is divided by 7 (7 is the total given numbers) = 108/ 7

= 15.4

Median

Middle element of the given series is known as median.

The number series is 4,8,12,16,19,22,27.

The middle element of the above series is 16.

So the value of median is 16.

Mode

Mode is a copy value of the given series. Here no copy value.

Therefore the mode value is empty.

Range

Range is the subtraction of the values form least to high of the given number series.

Range = 27-4

= 23.


Example 2


Solve the mean, median, range of the following in basic statistics?

12,17,19,24,29.

Solution

The given numbers are 12,17,19,24,29.

Mean

Mean is the average of the given number. Find the total value of the given numbers.

Sum of the given numbers are = 12+17+19+24+29

=101.

Total values are divided by 5 (Here 5 is the total given numbers) = 101/5

= 20.2.

Median

Middle element of the given series is known as median.

The number series is 12,17,19,24,29.

The middle element of the above series is 19.

So the median value is 19.

Range

Subtract the minimum value from the topmost value of the series.

Range =29-12

=17.

These examples are the basic statistics problems given by tutor.

Basic Math Definitions Solve Online

Introduction to Mathematical definition solving online:

The basic mathematical definitions are which very useful in our daily life. For every incident we use mathematics in one or the other kind. The mathematical definitions are like addition, subtraction, multiplication, division, average, place value, fractions and so on. There are many basic technique is used to solve math problems with some example problems and practice problems. The student can view and solve math definitions from any point of the world using online. So now let us see the mathematical definitions solving in this article.I like to share this Find the Average Rate of Change with you all through my article.

Addition
Subtraction
Multiplication
Division
Fraction
Place value
Average

Basic math definitions solving online


Solving Addition online: Addition is the sum of two or sometimes more numbers and symbol for addition is ”+”.

For example: Add 33 and 52?

Sol : Given data 33 ,52

= 33 + 52

= 85

Answer: 85

Solving Subtraction online: Subtraction is the difference between two numbers and symbol for subtraction is “-”

For example:  Sub 20 from 40?

Sol:      Given data 20, 40

= 40 - 20

= 20

Answer: 20

Solving Multiplication online: Multiplication is a Product of two or more numbers or it is also called as repeated addition. The symbol for the multiplication are “x, (), * and also .”

For example:  Multiply the numbers 3 from 3

Sol:      Given 3, 3

= 3 x 3

= 9

Answer is: 9

Solving Division online: Division is a arithmetic operation and its operation is to Split of numbers into equal parts. The symbol for division is ( / )

For example: divide 2 from 20

Sol:   Given 20, 2

Divide 20 by 2

= 20 / 2

= 10

Answer is: 5

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Solving Fraction online: Fraction is a part of a whole number.

For example: fractions 1/3, 4/5

Solving Place value online: place value is a digit value in a number.

For example: Find the place value of 2 in this number 321?

the value of 2 in the given number is tens place 2x10.

Solving Average online: Average is the sum of total numbers divided by to total count of those numbers

For example: find the average of these numbers 2, 2, 2, 2,

Sum of the above numbers are 8

Total numbers are 4

So average of these numbers becomes 8/4 = 2

Study About Basic Geometry Test

Introduction to study about basic geometry test:

Geometry is the one of the fundamental concept in mathematics. In this section we will see the study about basic geometry test, which is include several set of test questions along with answer keys. And also geometry test helps the students to recognize the geometry concepts with neat and clear. It also help students to self approximation their ability to perform test in geometry. Let us see the study about basic geometry test.

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Test questions for study about basic geometry test:


Test question 1: The measure of an angle is 45 degree. Find the measure of a complementary angle.

Test question 2: The measure of an angle is 32 degree. Find the measure of a supplementary angle.

Test question 3: A triangle has angle measurements of 100 degree, 60 degree, and 20 degree. Find the kind of triangle is it.

Test question 4: A triangle has angle measurements of 110 degree, 40 degree, and 30 degree. Find the kind of triangle is it.

Test question 5: Do calculation to find the area of the triangle whose base is 10 inches and height 14 inches.

Test question 6: The area of the triangle is 30 sq cm. The base of the triangle is 5 cm. Do the calculation for find the length of the triangle.

Test question 7: The diameter of the circle is 50 cm. Do the calculation for find the radius of the circle.

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Answer keys for study about basic geometry test:


Answer key 1: The complementary angle of 45 degree is 45 degree.

Answer key 2: The supplementary angle of 32 degree is 148 degree.

Answer key 3: Measurement of the triangle is obtuse.

Answer key 4: Measurement of the triangle is scalene.

Answer key 5: Area of the triangle is 70 sq inches.

Answer key 6: Length of the triangle is 12 cm.

Answer key 7: Radius of the circle is 25 cm.

Introduction to Basic Algebra

Introduction to basic algebra:

Basic algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with geometry, analysis, topology, combinatorial, and number theory, basic algebra is one of the main branches of pure mathematics.

Source: Wikipedia


Basic algebra rules:


Commutative Property of Addition.

a + b = b + a

Example:

5+6=6+5

Commutative Property of Multiplication.

a * b = b * a

Example:

7*3 = 3*7

Associative Property of Addition.

(a + b) + c = a + (b + c)

Example:

(5+8) + 2=5 + (8+2)

Associative Property of Multiplication.

(a * b) * c = a * (b * c)

Example:

(5*9) * 4=5 * (9*4)



Distributive Properties of Addition over Multiplication.

a * (b + c) = a * b + a * c

(a + b) * c = a * c + b * c

Example:

4 * (5+7) =4*5 + 4*7

(2+7) * 9=2 * 9+7 * 9

The reciprocal of a non-zero real number a is `1/a`.

a*(`1/a`) = 1

Example:

reciprocal of 7 is `1/7` and 7*(`1/7`) = 1

The additive inverse of a is -a.

a + (-a) = 0

Example:

additive inverse of -7 is -(-7) = 7 and - 7+(7) = 0

The additive identity is 0.

a + 0 = 0 + a = a

The multiplicative identity is 1.

a * 1 = 1 * a = a

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More about basic algebra:


Basic algebra expressions:

An expression in algebra 1 is a grouping of numbers, variables, constants, operators and parentheses or grouping brackets that are stated as an entity. An equation has an expression on each area of the equals sign. Some expressions are prepared of sub-expressions. Much of algebra concern simplifies expressions to facilitate the solution of equations.

Steps for simplify basic algebra expressions:

Step 1: assembly the terms containing the same variable collectively in algebra 1 expressions.

Step 2: execute the operation inside the parentheses for the variable and other.

Step 3: revise the expressions and simplifying the algebra expressions.

Step 4: To verify the equation, if there is capable to simplify the expression, then recur the step 1 to 4.

Introduction for basic algebra equation:

An algebra equation is mathematical statements that assert the equality of two expressions. Basic algebra equations consist of the expressions that are to be equal on opposite sides of an equal sign, as in

x+3=5

One use of basic algebra equation is in mathematical identities, assertions that are true independent of the values of any variables contained within them.

Laws of addition in basic algebra:

For the addition of positive and negative number, the follow rules, established in the First Course, apply make easy to solve algebra equation:

A number represented by a letter is called a literal number, and any number expression in which more than numbers of symbols are variable is called a literal expression.

Laws for addition in basic algebra:

1)  To add two numbers with similar signs, find the sum of their total values, and prefix the sign common to both.

2)  To add two numbers with different signs, find the diversity of their total values, and prefix the sign of the one with the larger total value.

How to Solve Basic Integers

Introduction to basic integers:
In mathematics, basic integers formed by natural numbers and including zero (0). The natural numbers like as, (1, 2, 3…) and 0 not including the natural numbers. Integers formed with positive numbers and negative numbers. Negative numbers are like this symbol, (-1, -2, -3 …). Negative numbers not including zero. Fractions and decimals are all not integers. Solve basic integers used to find the sum of basic integers. Here we are going to solve about  basic integers.


Examples problem for solve basic integers:


Rule 1: Sum of two or more negative integers is a negative integer.

Problem 1:

Solve the sum of basic integers, given integers is -6 and -2.

Solution:

Given integers is -6 and -2.

= -6 + -2

= -8

Sum of the integers is -8.

Problem 2:

Solve the sum of basic integers, given integers is -12 and -52.

Solution:

Given integers is -12 and -52.

= -12 + -52

= -74

Sum of the integers is -74.

Problem 3:

Solve the sum of basic integers, given integers is -16 and -12.

Solution:

Given integers is -16 and -12.

= -16 + -12

= -28

Sum of the integers is -28.

Rule 2: Sum of two or more positive integers is a positive integer.

Problem 4:

Solve the sum of basic integers, given integers is +6 and +6.

Solution:

Given integers is +6 and +6.

= +6 + +6

= +12

Sum of the integers is +12.

Problem 5:

Solve the sum of basic integers, given integers is +20 and +20.

Solution:

Given integers is +20 and +20.

= +20 + +20

= +40

Sum of the integers is +40.

Problem 6:

Solve the sum of basic integers, given integers is +14 and +12.

Solution:

Given integers is +14 and +12.

= +14 + +12

= +26

Sum of the integers is +26.

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Practice problems for solve basic integers:


1. Solve the sum of basic integers, given integers is -4 and -5.

Answer is -9.

2. Solve the sum of basic integers, given integers is -8 and -12.

Answer is -20.

3. Solve the sum of basic integers, given integers is -50 and -15.

Answer is -65.

4. Solve the sum of basic integers, given integers is +5 and +2.

Answer is +7.

5. Solve the sum of basic integers, given integers is +25 and +22.

Answer is +47.

6. Solve the sum of basic integers, given integers is +11 and +11.

Answer is +22.

Basic Probability Calculations

Introduction to basic probability calculations:
The basic probability calculations are very helpful to the students. Calculations are the best tool to study what you want without disturbing the time. Calculation can be more interesting and interact than books and lecturing. Probability is result of examine Probability is number of possible actions divide by total quantity of proceedings. If the event arises it represents like one, or else not occurs it must symbolize zero. Now we will see basic probability calculations.


Example problems for basic probability calculations:


Example 1:- Using basic probability calculations

In school there are 100 students available. In those students, there are 50 students are science group, 25 students are maths group and 25 students are the English group. Find the probability if,

i) Science group students are leaving first from the school.

ii) English group students are leaving second from the school.

Solution:

Given,

Total students n(S)=100

Science group students n(A)=50

Maths group students n(B)=25

English group students n(C)=25

i) Let P(A) is the event of the science group students are leaving first.

So the P(A)=`(n(A))/(n(S))`

=`50/100`

=1/2

ii)Let P(B) is the event of the English group students are leaving first.

P(B)=`(n(B))/(n(S))`

=`25/100`

=1/4.

Example 2:- Using basic probability calculations

The bag has the 60 kerchiefs. In those kerchiefs, there are 15 is the blue color, 20 is the yellow color and 25 is the white color. If John select the white color kerchief means solve the probability?

Solution:

First list the given information,

Total kerchiefs n(S)=60

Blue colour kerchiefs n(A)= 15

Yellow colour kerchiefs n(B)= 20

White colour kerchiefs n(C) = 25

Assume P(E) is the probability for john select the white colour kerchief.

So P(E)=`(n(C ))/(n(S))`

=`25/60`

=`5/12` .

In the above way we can solve the problems.

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Example 3:- Using basic probability calculations


Two coins are spinning. What is the probability for if two head are receiving?

Answer

Two coins flipped means the feasible outcomes are {(H,T),(H,H),(T,H),(T,T)}

So the total numbers of outcomes are 4.

We get the two head possibility is 1.

Probability = 1/4.

Example 4:- Using basic probability calculations

From the following scores are 41,13,46,27,79,25,15. We can find the probability for above 25 scores? 2.

Solution:

Probability for finding above 25 scores:

Above 25 scores=41, 46, 27, 79.

In the given scores the above 35scores probability values are (41, 46, 27, 79)

Probability for above 25 scores=41+46+27+79

=193.

Total scores values = (41+13+46+27+79+25+15) =246.

Probability=above 25/Total given scores.

Total scores=246

Probability for above 25 scores=`193/246` .

=`193/246` .

Formulas for finding probabilities for below 25 scores are,

Basic Geometry Concepts

Introduction to basic geometry concepts:

In basic geometry concepts, the geometry is all on the subject of shapes. Geometry was enormously major to prehistoric societies and was used for surveying, astronomy, direction-finding, and structure. Geometry is the study of angles and triangles, outer limits, area and quantity. It differs from algebra in that one develops a reasonable formation where mathematical relations are prove and functional.

Basic Geometry Terms:

Some of the basic terms in geometry are line, rays, angles, and plane.
Now we will discuss about basic concepts in geometry.
Line segment:

In these basic geometry concepts, a line segment is a straight line up segment which is division of the straight line among two points.
To organize a line section, one can inscribe AB.
The points on each surface of the line division are referred to as the finish points.
Ray:

In basic geometry concepts, a ray is the element of the line which consists of the arranged point and locates of all points on one elevation of the closing stages point.
Angle:

In basic geometry concepts, angle can be discrete as two rays or two line segments having a common finale position.
The endpoint becomes predictable as the vertex.
Angles come about while two rays accumulate or mix up at the matching endpoint.

Plane:

A plane is often instead of by a blackboard, communication board, a plane of a box or the peak of a table.
These 'plane' surfaces are damaged to connect any two or added points in a straight line.
A plane is a parallel frontage.

Types of Angles:

An angle is different as wherever two emission or two-line segment join mutually at a general endpoint is known as vertex.
In basic geometry concepts having some types of angles. these are,
Acute angle.
Right angle.
Obtuse angle.
Acute angle:

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An acute angle procedure a lesser of that 90°
Right Angle:

A right angle connections correctly 90°
Obtuse Angle:

An obtuse angle measures not required than 90° but less than 180°

Math Factors of 59

Introduction to math factors of 59:

Math factors are the numbers that we do product in show to get another number:

For example the number 6 has the factors 1, 2 and 3 as shown as follows,

6 =1 x 2 x 3

In this article we are going to learn how to factorize the numbers and all about the math factors of 59.


Math factors of 59:


In order to get the factors we have to divide with the numbers starting with the number 1.

Let us do the math factors of 59.

Step 1: Divide the number 59 by 1 we get 59.

Step 2: We can’t divide 59 further as it is a prime and it gives the result with the decimal

Point.

Step 3: Finally the factors of 59 are 1 and 59.

That is 59 = 1 x 59.

That is 59 = 59 x 1.

Note: prime number is a number which is divided by 1 and the number itself.


Example problems- Math factors of 59:


The following examples will make you clear about how to factorize the numbers in math.

Example 1:

Find the factors of 159

Solution:

Step 1: Divide the number 159 by 1 we get 159.

Step 2: As it is a odd number we cant divide it by 2 therefore it is better to go with 3

Step 3: Divide 159 by 3 :

159/3 = `(3*53)/3`

= 53 (where 3 cancels each other)

Step 4: As 53 includes in the above step it is also a factor of 159.

Finally the factors of 159 are 1, 3, 53 and 159.

That is 159 = 1 x 159;

That is 159 = 3 x 53;

That is 159 = 53 x3;

That is 159 = 159 x 1.

Example 2:

Find the factors of 259

Solution:

Step 1: Divide the number 259 by 1 we get 259.

Step 2: As it is a odd number we cant divide it by 2 therefore it is better to go with 3

Step 3: Dividing by 3 is also not possible so Divide 159 by 7 :

259/7 = `(7*37)/7`

= 37 (where 7 cancels each other)

Step 4: As 37 includes in the above step it is also a factor of 259.

Finally the factors of 259 are 1, 7, 37 and 259.

That is 259 = 1 x 259;

That is 259 = 7 x 37;

That is 259 = 37 x 7;

That is 259 = 259 x 1.

Example 3:

Find the factors of 359

Solution:

Step 1: Divide the number 359 by 1 we get 359.

Step 2: We can’t divide 359 further as it is a prime and it gives the result with the decimal

Point.

Step 3: Finally the factors of 359 are 1 and 359.

That is 359 = 1 x 359.

That is 359 = 359 x 1.

Practice problems- Math factors of 59:

Problem 1:

Find the factors of 559.

Solution:

The factors of 559 are 1 , 13 ,43 , and 559.

Problem 2:

Find the factors of 659.

Solution:

The factors of 659 are 1 and 659

Problem 3:

Find the factors of 759.

Solution:

The factors of 759 are 1 , 3 , 11 , 23 , 33 , 69 , 253 , 759.

Basic Notes on Algebra

Introduction to basic notes on algebra:

Algebra is defined as the major part of mathematics which deals with the study of laws of the operations and those concepts which involves various polynomials, algebraic structures and the equations. There are various kinds of algebra which includes elementary algebra, abstract algebra, etc.

In this article we are going to see about the basic notes of algebra which includes various laws and some example problems.

Basic notes on the laws of algebra:

laws of algebra:

There are various laws that are used in algebra for solving the equations and various algebraic structures.

The first basic notes on algebra are addition which involves the commutative law.

It states that a + b = b + a

Example:  5+6 = 6 + 5

11 = 11

The commutative law can also be applied for multiplication.

The basic law for multiplication is given by

a * b = b *a

Example:    5* 6 = 6* 5

30 = 30

We can also use the associative law for both the addition and the multiplication.

The basic notes for the associative law in algebra for addition is given by

(a + b) + c = a+ (b + c)

Example: (1+2) + 3 = 1 + (2+3)

3 + 3 = 1 + 5

6 = 6

The associative law for multiplication is given by

(a * b) * c = a*(b * c)


Example: (1*2) * 3 = 1* (2*3)

2 * 3 = 1 * 6

6 = 6

The next important basic notes of algebra include the distributive law.

(a * b) + c = (a*b) + (a * c)

Example: (1 * 2) + 3 = (1 *2) + (1 * 3)

2 + 3 = 2 + 3

5 = 5

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Notes on basic rules of algebra:

The additive inverse of a is given by –a for any constant or a variable.

Example :a + (-a) = 0 and 5 + (-5) = 0

The multiplicative inverse of a is 1/a for any constant or a variable.

Example: a * 1/a = 1 and 5 * 1/5 = 1

In case of addition the identity element is zero for any constant or variable.

Example: a + 0 = a and 5 + 0 = 5

In case of multiplication the identity element is one for any constant or variable.

Example: a * 1 = 1 and 5 * 1 = 5.

Do Basic Division

Introduction to do basic division:

Division is one of the arithmetic operations. Division is the inverse form of multiplication. The term division is sometimes said to be sharing. Division is denoted by the symbol `-:`. For example: When M is divided by N.

M/N = O

M – Dividend.
N – Divisor.
O – Quotient.
For example:

The value 120 divided by12.

120/12 = 10

In this,

120 – Dividend.

12 – Divisor.

10 – Quotient.


Do basic division


In this division there are divided, divisor, remainder and quotient will appear in the basic way. It is very easy to solve. This division is very easily understandable by the kids. Let see some examples of the basic division and do some practice problems for this division method.

Example problems for basic division:

Calculate the value 245 is divided by 15.

Solution:

Steps to solve the given problem

First step:

To find the number of times do 15 go into the 240?

Second step:

24 is divided by 15

24/15 = 1

The remainder is 9

Third step:

Then the remainder is put in front of the value 0.

90 is divided by 15

90/15 = 6.

The remainder is said to be zero (0).

Solution:

15) 240 (16

15

_______

90

90

_______

0

________

Divisor = 15

Dividend = 240

Quotient = 16

Remainder = 0

Another example for basic division:

The value 418 is divided by 16.

Dividend = 418.

Divisor = 16.

Solution:

16) 418 (24

32

_________

99

90

__________

9

_________

Answer:

Quotient = 24

Remainder = 9.

Another example for basic division:

The value 16724 is divided by 12.

The dividend = 16724

The divisor = 12.

Solution:

12) 16732 (1311

12

_______

47

36

_______

13

12

_________

12

12

_________

0

__________

Answer:

Quotient = 1311

Remainder = 0.


Some practice problems for the basic division

Problem 1:

Find out value 450 is divided by 5.

Answer:

The correct answer is 90.

Problem 2:

Find the value of 43978 is divided by 14.

Answer:

The correct answer is

Quotient = 3141

Remainder = 4

Problem 3:

Find the value of 12450 is divided by 10

Answer:

The correct answer is

Quotient = 1245

Remainder = 0.

Problem 4:

Find the value of 29726 is divided by 2.

Answer:

The correct answer is

Quotient = 14863

Remainder = 0

Area Math 4th Grade

Introduction to 4th grade math area:

Area is a quantity expressing the two-dimensional size of a defined part of a surface, typically a region bounded by a closed curve. The surface area of a 3-dimensional solid is the total area of the exposed surface, such as the sum of the areas of the exposed sides of a polyhedron. Area is an important invariant in the differential geometry of surfaces. In this article we shall discuss about 4th grade math area problem. (Source: Wikipedia)

4th grade math area example problem

Here we are going to discuss some 4th grade math area problems with detailed solutions.

Example 1:

Find the area of the rectangle, the rectangle base value 23m and height of rectangle is 10 m.

Solution:

Let base b = 23 and rectangle height h = 10.

Then the area of the rectangle formula = b × h

The area of the rectangle = 23 × 10

= 230 sq. meters.

Therefore the area of rectangle = 230 sq. meters.

Example 2:

Find the area of the circle, the circle radius value 12.

Solution:

Area of the circle formula = pi r2

Pi = 22/7 or 3.14

The area of the circle = 22/7× 12x12

= 452

Answer:

Therefore the area of circle = 452

Example 3:

Find the area of the square; the square one side of length value is 22units.

Solution:

Formula for finding area of square = a2 of (Side) 2

The side value of the area = 22

Therefore (22) 2

That is 22 * 22 = 484

Area of the square = 484units.

Example 4:

Find the area of triangle, the triangle base value=20cm and the height of the triangle = 12cm

Solution:

Area of the triangle formula = `1/2` b x h

Height of the triangle = 12

Base of the triangle = 20

Therefore area of triangle = `1/2` x 20 x 12

The area of triangle = 120cm

Example 5:

Find the area of the rectangle, the rectangle base value 22m and height of rectangle is 15 m.

Solution:

Let base b = 22 and rectangle height h = 15.

Then the area of the rectangle formula = b × h

The area of the rectangle = 22 × 15

= 330 sq. meters.

Therefore the area of rectangle = 330 sq. meters.

Understanding right triangle trigonometry is always challenging for me but thanks to all math help websites to help me out.

4th grade math area practice problem

Problem 1:

Find the area of the rectangle, the rectangle base value 20m and height of rectangle is 10 m.

Answer:

The area of rectangle = 200 sq. meters.

Problem 2:

Find the area of the square; the square one side of length value is 20 units.

Answer:

Area of square = 400 units