Areas related to Circles

Areas related to Circles:

Perimeter and Area of a Circle — A Review

The distance covered by travelling once around a circle is its perimeter,
usually called its circumference.

In other words, circumference diameter = π
or, circumference = π × diameter
= π × 2r (where r is the radius of the circle)
= 2πr

Let us recall the concepts learnt in earlier classes, through an example.
Example 1 : The cost of fencing a circular field at the rate of Rs 24 per metre is
Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m2. Find the cost of
ploughing the field (Take π =22/7).
Solution : Length of the fence (in metres) =
Total cost Rate = 5280 = 22024
So, circumference of the field = 220 m
Therefore, if r metres is the radius of the field, then
2πr = 220
or, 2 ×
22
7 × r = 220
or, r =
220 × 7
2 × 22 = 35
i.e., radius of the field is 35 m.
Therefore, area of the field = πr2 =272
× 35 × 35 m2 = 22 × 5 × 35 m2
Now, cost of ploughing 1 m2 of the field = Rs 0.50
So, total cost of ploughing the field = Rs 22 × 5 × 35 × 0.50 = Rs 1925

Areas of Sector and Segment of a Circle:

The circular region enclosed
by two radii and the corresponding arc is called a
sector of the circle and the portion (or part) of the
circular region enclosed between a chord and the
corresponding arc is called a segment of the circle.

Word problem on Time and Work

Some Important Facts about Time and work problems from answers to math problems

• If A can do a piece of work in n days, then A's 1 day work = 1/n

• If A’s 1 day’s work=1/n, then A can finish the work in n days

• If A is thrice as good workman as B,then: Ratio of work done by A and B =3:1. Ratio of time taken by A and B to finish a work=1:3 , 

Let's see an example from 8th grade math worksheets

Question:-

Working together drew and joyce can put a futon frame together in 36 minutes.If it takes drew 60 minutes to put a futon frame together by himself.How long would it take joyce ?

Answer:-

Let us consider that joyce takes x min to complete it.

In 1 min he can do 1/x part of it.

Drew takes 60 min to complete the work

In 1 min he completes 1/60 part of it.

So, working together ,in 1 min they can do [1/x+1/60] part of it

It has given that together they can complete it in 36 min.

So ,together in 1 min they can do 1/36 part of it

So 1/x+1/60 = 1/36

So, x = 90 min

Which means joyce can put the frame together ,by her self in 90 min.

A word problem on sets

A set is a collection of distinct objects, considered as an object in its own right. Sets are one of the most fundamental concepts in mathematics. Although it was invented at the end of the 19th century, set theory is now a ubiquitous part of mathematics, and can be used as a foundation from which nearly all of mathematics can be derived. In mathematics education, elementary topics such as Venn diagrams are taught at a young age, while more advanced concepts are taught as part of a university degree.

Here is an example from algebra 1 word problems


Question:-

In a school there are 20 teachers who teach mathematics or physics,of these ,12 teach mathematics and 4 teach both physics and mathematics .How many teach physics

Answer:-

The total number of students =20 (this is we call what is Universal Set U)

Let M be the teachers who teach mathematics

and P be the teachers who teach physics.

the word 'or' gives a clue of union and the word 'and' gives a clue of intersection.

n(M U P)=20

n(M)=12

n(M Ω P)=4

you can see a example of Venn diagram

using the formula

n(M U P)=n(M)+n(P)-n(M Ω P)

20 = 12+n(P)-4

20 = 8+n(P)

subtract 8 on both sides

12 = n(P)

So 12 teachers teach physics

How to find critical points for an equation

In calculus, a critical point of a function of a real variable is any value in the domain where either the function is not differentiable or its derivative is 0. The value of the function at a critical is a critical value of the function. These definitions admit generalizations to functions of several variables, differentiable maps between R^m and Rⁿ, and differentiable maps between differentiable manifolds.

Lets see a problem from calculus help with probability problems,which explains us
more about this .

Question:-

f(x)=3x⁴-4x³-12x²+6

Differentiating with respect to x

Solution:-

f'(x)=12x³-12x²-24x

we know f'(x)=0 ,so

12x³-12x²-24x=0

12x(x²-x-2)=0

12x(x²-2x+x-2)=0

12x(x-2)(x+1) = 0

x(x-2)(x+1) = 0

Now we use the zero product property ,which states that
if ab=0 ,then a=0 and b=0,this property is from geometry terms and definitions .

so x=0 (x-2)=0 (x+1)=0

x=0 x=2 x=-1

Therefore,x=0,-1,2 are the critical points for y=f(x)

Simple Algebraic Equation Problem

Topic: Algebraic Equation

An equation is a mathematical statement, in symbols, that two things are exactly the same . Equations are written with an equal sign, as in
2 + 3 = 5
9 - 2 = 7
The equations above are examples of an equality: a proposition which states that two constants are equal. Equalities may be true or false
How to solve Equations ?
Its simple.Here is a example of 7th grade math equations :

Solving a integral problem by using substitution method

Topic:- Integration

Integration is an important concept together with differentiation, forms one of the main operations in calculus help
Given a function ƒ of a real variable x and an interval [a, b] of the real line, the definite integral

is defined informally to be the net signed area of the region in the xy-plane bounded by the graph of ƒ, the x-axis, and the vertical lines x = a and x = b.

Let's work out a simple example on this.
Question:-

solve ∫ ( x / √x²-9 )

Answer:-

We do it by substitution method.
 Let u = x2-9
    
    du
    ---- = 2x
    dx
 
   du
or ---- = xdx
   2
substituting these values ,the integral becomes

∫ (du/2) / √u

= 1/2 ∫ (u)-1/2 du

        (u)-1/2 + 1
= 1/2 ---------------
         -1/2 + 1
by equalizing the denominators
   
    -1+2
   ------- = 1/2
      2
So the integral becomes
    
        (u)1/2
= 1/2 ---------------   + c
         1/2

= (x2-9)1/2+C (as u =x2-9)

= √( x2-9 ) +c  is the Answer

For more help on this ,you can reply me .

A problem on Integrating Factor

Topic:Integrating Factor

In simple term integrating factor in mathematics, is an function that is chosen to facilitate
the solving of a given equation involving differentials. It is generally used to solve ordinary differential equations, but is also used within multivitamin calculus,in this case often multiplying through by an integrating factor allows an inexact differential to be made into an exact differential (which can then be integrated to give a scalar field).
In simple term integrating factor is denoted as I.F.

Question:
   
           dy
solve     ____ + = ex
           dx

Answer:

   Standard form: dy
                 ____ + p(x) * y = q(x)    Here p(x) = 1
                  dx                            q(x) = ex


 Now Inergrating factor = e√pdx = e√idx = ex

 Now the equation is:

                     y = i
                        ___
                        I.F  √I.F * Q(x) dx + c

Plugging in all the values:

                     y = 1
                        ___
                         ex √(ex * ex) dx + c

                     y = 1
                        ___
                         ex √(ex)2 dx + c

                     y = 1 * e2x
                        ___ ___ + c
                        (ex) 2

                     y = ex
                        ___ + c
                         2        

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