Two Equivalent Fractions

Introduction :

Fraction is a number which has two parts. One part is in upper and the other part is lower. The upper part is known as the numerator and the lower part is known as the denominator.  It is possible to find equivalent fraction for the given fraction. Two equivalent fractions means finding two equivalent fractions for the given fraction.

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Example Problems – Two Equivalent Fractions:

1) Find two equivalent fractions for the fraction `(1)/(2)` .

Solution:Given , `(1)/(2)`

To find the two equivalent fraction, multiply and divide the given fraction by 2.

The given fraction would be  `(1 * 2)/(2 * 2)`    =  `(2)/(4)`

which is one of the equivalent fractions.

To find the second equivalent fraction, multiply and divide the given fraction by 3.

The given fraction would be  `(1 * 3)/(2 * 3)`  =   `(3)/(6)`

which is second  equivalent fractions.

The two equivalent fraction of the fraction `(1)/(2)`  is `(2)/(4)`  and `(3)/(6)` .

2) Find two equivalent fractions for the fraction `(2)/(4)` .

Solution:Given , `(2)/(4)`

To find the two equivalent fraction, multiply and divide the given fraction by 2.

The given fraction would be  2 * `(2)/(4)`  * 2  =  `(4)/(8)`

which is one of the equivalent fractions.

To find the second equivalent fraction, multiply and divide the given fraction by 3.

The given fraction would be  2 * `(3)/(4)`  * 3  =   `(6)/(12)`

which is second  equivalent fractions.

The two equivalent fraction of the fraction `(2)/(4)`  is `(4)/(8)`  and `(6)/(12)`

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More Examples on Equivalent Fractions:

3)Find two equivalent fractions for the fraction `(1)/(3)`.

Solution:Given , `(1)/(3)`

To find the two equivalent fraction, multiply and divide the given fraction by 2.

The given fraction would be  1 * `(2)/(3)`  * 2 = `(2)/(6)`

which is one of the equivalent fractions.

To find the second equivalent fraction, multiply and divide the given fraction by 3.

The given fraction would be  1 *`(3)/(3)`  * 3  = `(3)/(9)`

which is second  equivalent fractions.

The two equivalent fraction of the fraction `(1)/(3)` is `(2)/(6)` and `(3)/(9)` .

4) Find two equivalent fractions for the fraction `(1)/(4)`

Solution:Given , `(1)/(4)` .

To find the two equivalent fraction, multiply and divide the given fraction by 2.

The given fraction would be  1 * `(2)/(4)`  * 2 = `(2)/(8)`

which is one of the equivalent fractions.

To find the second equivalent fraction, multiply and divide the given fraction by 3.

The given fraction would be  1 * `(3)/(4)`  * 3 = `(3)/(12)`

which is second  equivalent fractions.

The two equivalent fraction of the fraction `(1)/(4)`  is `(2)/(8)`  and `(3)/(12)`

Multiplying 2x2 Matrices

Introduction to Multiplication of 2by 2 Matrix:

There are two types of multiplications is there.

Scalar multiplication
Matrix Multiplication
Scalar multiplication:

It is an ordinary multiplication.the regular numbers are called the scalar.

Matrix multiplication:

It is  the operation of multiplying  a matrix with another matris or multiplying matrix with scalar.

Ordinary matrix product:

It is mostly used to Multiplying Matrices. The multiplication is defined between two matrices If only the breadth of the first matrix is equals to the height of the next matrix that  means the second matrix.

If we multiplying   mx a matrix with  aXn matrix means the result  will be the  m x n matrix..

Properties of Matrix Multiplication:

Properties of matrix multiplication:

It is not  usually  commutative. that means  AB  BA.
If A and B are the two n x n matrices , then the determinant of its product  is independent for the      order of the matrices in their products.
If both the matrices are  diagonal matrices then their product is commutative.
The matrix multiplication is satisfies the  Associative property.
A(BC)=(AB)C
It is distributive over the addition:
A(B+C)=AB+AC
(A+B)C=AC+BC
It is compatible with the scalar multiplication.
It does not  satisfied the commutative property.

How to multiply matrices

Consider  two matrices A and B .

step 1:The first row first element of the matrix A is multiply by the First column First element of the matrix B.

step 2: And add to the  The first row second  element of the matrix A is multiply by the first column second element of the matrix B.

Step 3:The second row first element of the matrix A is multiply by the second column First element of the matrix B.

step 4: And add to the  second row second  element of the matrix A is multiply by the second  column  second element of the matrix B.

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Example for 2 X 2 Matrix Multiplication

Example for 2 x 2 matrix multiplication

Ex 1 :   `[[2,3],[1,3]]`     `xx``[[0,1],[2,1]]`

Sol :     Step 1:

(2  3 )`((0),(2))`

= (2 x 0)+(3 x2)

=6

Step 2:        (2  3)`((1),(1))`

=(2 `xx` 1)+ (3`xx`

=2+3

=5

Like wise we have to do the  remaining steps.

= `[[6,5],[6,4]]`

Semicircle Shape

Introduction:

Semicircle shape:

Semicircle shape is the two dimensional geometrical figure that is formed from the half of the circle. The total angle in the circle is 360 degrees since the semicircle is half of the circle, the total angle in the semi circle is 180 degrees.

If the triangle is inscribed inside the semi circle is always right angled triangle.

Formula Related to Semicircle Shape:

The diameter of the semicircle shape is denoted by D and the radius of the semicircle is denoted by r.Diameter of the semicircle shape is twice the radius of the semicircle shape.

D= 2r units

Radius of the semicircle shape = `((diameter)/ 2) ` units

Area of the semicircle shape:

Area of the semicircle shape = `(180/360) (pi*r^2) ` square units

= `(1/2) (pi*r^2)`

Here r is the radius of the semicircle

Circumference of the semicircle shape:

Circumference of the semicircle shape = `(180/360) (2*pi*r)` units

= `(1/2) (2*pi*r)`

= `(pi*r)` units.     

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Model Problems for Semicircle Shape:

Problem 1:

Find the area of the semicircle shape whose diameter is 10cm?

Solution:

Diameter of the semicircle shape = 10 cm

Radius of the semicircle shape = `(10/2)`

= 5cm

Area of the semicircle shape:

Area of the semicircle shape = `(180/360) (pi*r^2)` square units

= `(1/2) (pi*r^2)`

Here r is the radius of the semicircle

= `(1/2) (3.14*5^2)`

= `((78.5)/2)`

= 39.25

Area of the semicircle shape is 39.25cm^2

Problem 2:

Find the circumference of the semicircle shape whose diameter is 10cm?

Solution:

Diameter of the semicircle shape = 10 cm

Radius of the semicircle shape = `(10/2)`

= 5cm

Circumference of the semicircle shape:

Circumference of the semicircle shape= `(180/360) (2*pi*r)` units

= `(1/2) (2*pi*r)`

= `(pi*r)` units.       

Here r is the radius of the semicircle

= (3.14*5)

= (15.7)

= 15.7

Circumference of the semicircle shape is 15.7 cm

Problem 3:

Find the area of the semicircle shape whose diameter is 40cm?

Solution:

Diameter of the semicircle shape = 40 cm

Radius of the semicircle shape = `(40/2)`

= 20cm

Area of the semicircle shape:

Area of the semicircle shape = `(180/360) (pi*r^2)` square units

= `(1/2) (pi*r^2)`

Here r is the radius of the semicircle

= `(1/2) (3.14*20^2)`

= `(1256/2)`

= 628

Area of the semicircle shape is 628cm^2

Arc of a Circle

Introduction:

Arc refers to a part of a circle or a curve. In case of the circle, it contains minor arc and major arc. The arc length of minor is the multiplication of circumference of a circle and fraction of an angle`theta`, (`theta` /360).

Geometry is a theoretical subject, but easy to understand, and it has many real practical applications. Eventually, geometry has evolved into a skillfully arranged and sensibly organized body of knowledge.

Arc Length of a Circle

Arc length of a circle:

In the above figure shows the description of arc length of a circle.

The length of an arc of a circle can be represented with the radius r and an angle ? at the center of the circle.

That is, L/circumference = ? / (2p)

Here we can substitute the circumference formula and then we get,

L/ (2pr) = ? / (2p)

Using cross multiplication, we get the following

L= r?

Substitute the value of ?= (`theta` p) / 180

Therefore, the length of a circle,

L= (`theta` p r) /180

Arc Area and Arc Segment Area of Circle

Arc Area:     

In the above figure shows the description of arc of the circle.

In the above circle  AB is called minor arc and major arc as indicated. If A and B are end points of a diameter then it is named as semicircle.

The area between the center and an arc of a circle is:

A=1/2(r2?)

A/ (pr2) =?/2 p

We can get rid of a p on both sides:

A/r2=?/2

And then, by multiplying both sides by r2, we get the Final Formula

A=1/2(r2?)

Arc Segment Area:

The area of the shape restricted by the arc and a straight line between the given two end points is

((1/2) r2 (?-sin?))

To get the area of segment area we can subtract the values of arc area of triangle and the sine value of center of circle.

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