Line Segment Practice for Free

Introduction for line segment:

Line segment is the important topic in geometrical chapter and math subject. Generally in math subject, a free long line is divided with two definite end points on both directions is known as line segments. But the line segment is a small part of a long line. In this article we are going to brief explain about line segment and line segment practice problems. And solve the some of example problems for free.

Line Segment Definition:

In math, a line segment can be defined as one small divided part of a lengthy line.
Line segment is also known as ‘a straight line’, which is joining the two points with coordinates, and the line is infinity after that, the end points.
So in this word segment is being an important role in line segment.
Example figure for line segment:



In this figure xy is the infinity line, and A, B are the two end points, and the line segments are bar (AB).
The length of the line segments AB would be written as bar (AB). And the line segments have used the name as to be two end points AB


Line segment formulas for solve practice problems:

The distance of two given points is called line segments length. The line segment length is calculated using formula for distance.
The line segments having two end points, this is have x and y coordinates. X(x1, y1) and Y(x2, y2)...
Line segment distance formula, XY = (sqrt((x2-x1)^(2)+ (y2-y1)^(2))
The line segment mid point formula, M = midpoint, M(x, y) = ((1/2(x1+x2), (1/2(y1+y2)))
These two are the important formulas for solve line segment problems.

Example Free Problems in Line Segment:

Example 1 for practice:

Using the distance formula, solve the length of line segment between two end points. End points are M = (3, 2), N = (5, 4)

Solution:

Line segment distance formula,

` MN= sqrt ((x2-x1) ^ (2) + (y2-y1) ^ (2))`

x1= 3   y1= 2   x2=5    y2=4

` MN = sqrt ((5-3) ^ (2) + (4-2) ^ (2))`

` MN = sqrt ((2) ^ (2) + (2) ^ (2))`

`MN = sqrt (4+ 4)`

`MN = sqrt (8)`

` MN = 2 sqrt (2)`

Answer is `2 sqrt (2)` .

Example 2 for practice:

What is the mid point between two points given? Solve (-3, -1), (4, 2)

Solution:

Using the mid point formula,

Midpoint formula = `((x1+x2) / 2), ((y1+y2) /2)`

And substitute (x1, y1) = (-3, -1) and (x2, y2) = (4, 2)

Midpoint = `((-3+4) /2), ((-1+2) /2) ` Simplify it and get,

And get midpoint = `(1/2), (1/2)`

The mid point between the two points is `(1/2), (1/2)` .

Those two example problems are free for understanding about line segments. Here we are giving some practice problems.I have recently faced lot of problem while learning cbse previous years question papers 12th, But thank to online resources of math which helped me to learn myself easily on net.

Free practice problems in line segments:

Free practice problem 1:

Using the distance formula, solve the length of line segment between two end points. End points are M = (6, 2), N = (5, 4)

Answer is `sqrt (5)` .

Free practice problem 2:

Find the mid point between two points given? Solve (-5, -1), (6, 2)

Answer is (`1/2` , `1/2` )

Those all the above example problems and free practice problems are useful for understand the line segment practices.

Geometry Acute Triangle

Introduction to geometry acute triangle:

The triangle is a 3 sided figure where all the three sides are enclosed to form the shape. The triangle can be classified according to its angle measurements and the length of the side. The geometry acute triangle is nothing but the triangle where all the angles in the triangle are less than 90 degrees. So, this triangle can be called as acute triangle. Now we are going to see about the geometry acute triangle.

About Geometry Acute Triangle:

The geometry acute triangle may also call as the acute equilateral triangle because each of the angle measures in the triangle will be 60 degrees which is less than 90 degrees. There are also types of acute triangle. They are acute obtuse angle, acute scalene triangle. In geometry acute triangle the angles will be less than 90 degrees. If it is not an acute triangle then it will be some other triangle types such as obtuse or right triangle.

The formula to find the area of a geometry acute triangle can be given as follows,

Area of triangle = `(1/2)` * b * h



Problems for Geometry Acute Triangles:

Example 1:

Find the angle of an acute triangle whose measurements are 40 degrees and 50 degrees.

Solution:

Now we calculate for the acute triangle as follows,

x + 50 + 60 = 180

Just add all the angles of the triangle,

x + 110 = 180

Now subtract the angle from 180 we get,

x = 180 - 110

x = 70

Thus, the acute angle of the triangle is 70 degrees.

Example 2:

Find the area of an acute triangle where the base of the triangle measurement is 5 cm and the height is about 6 cm.Is this topic math homework help free online hard for you? Watch out for my coming posts.

Solution:

Now we are going to see about the calculation part

The area of an acute triangle can be given as,

Area of triangle = `(1/2)` * Base * Height

= `(1/2)` * 5 * 6

= `(1/2)` * 30

= `30/2`

Area of acute triangle = 15 cm2

Derivative of Quadratic Functions

Introduction to derivative of quadratic functions:

In calculus the derivative is a measure of how a function changes as its input changes. The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. A quadratic equation is a polynomial equation of the second degree. The general form is ax2 + bx + c . The constants a, b, and c, are called respectively, the quadratic coefficient, the linear coefficient and the constant term or free term.

Source Wikipedia.

Derivative Formulas:

1.  ` d / dx` (x n ) = n xn-1

2.  ` d/dx` (ex)  = ex

3.  ` d/dx` `(uv)` = `u(dv)/(dx)` + `v(du)/(dx)`
4. .`d/dx` `(u/v)` = `(v(du)/(dx) - u(dv)/(dx))/ v^2`

5.  `d/dx` (sin x) = cos x

6.  `d/dx ` (cos x) = -sin x

7.  `d/dx` (tan x) = sec2x

Derivatives of Quadratic Function Problems:

Derivatives of quadratic function problem 1:

Find the derivative of quadratic function (2x3 - 3x4 + x2) with respect to x.

Solution:

Given Quadratic function is  (2x3 - 3x4 + x2)

` d/dx` (2x3 - 3x4 + x2)  = `d/dx` (2x3) - ` d/dx` (3x4) + ` d/dx ` (x2) .

=  `d/dx` (2x3) - `d/dx` (3x4)  + `d/dx` (x2).

= (2 × 3) x(3-1) - (3 × 4) x(4-1) - 2x(2-1)

= 6 x2 - 12 x3 + 2 x

`d/dx` (2x3 - 3x4 + x2) =  6 x2 - 12 x3 + 2 x

Answer: The derivative of given quadratic function is  `d/dx` (2x3 - 3x4 + x2) =  6x2 - 12 x3 + 2 x

Derivatives of quadratic function problem 2:

Find the derivative of quadratic function  y = 3t2 + t3 - 30t. with respect to 't'.Is this topic free algebra help hard for you? Watch out for my coming posts.

Solution:

Given Quadratic function is y = 3t2 + t3 - 30t.

dy  = (3 × 2)t(2 - 1) dt + (1 × 3)t(3 - 1) dt - 30t(1 - 1) dt.

= 6t dt + 3t2 dt - 30dt.

= (6t + 3t2 - 30) dt

`dy/(dt)` = (6t + 3t2 - 30)

Answer:  The derivative of given quadratic function is   `d/dt` (3 t2 + t3 - 30t) =  3t2 + 6t - 30.

Derivatives of quadratic function problem 3:

Determine the second derivative of given quadratic function f(x) = 15x2 + 10x + 5 with respect to x.

Solution:

Given quadratic function is, f(x) = 15x2 + 10x + 5

The First derivative of quadratic function is f'

f' =` (df )/ (dx)`  = (15 × 2)x(2 - 1) + (10 × 1)x (1-1)

= 30x + 10

The Second derivative of quadratic function is, f''

` d^2/dx` f(x) = `d^2/dx` ( 15x2 + 10x + 5)

= `d/dx` (f')

= `d/dx` ( 30x + 10)

= ` d/dx` (30x) +` d/dx` (10)

= 30 `d/dx` (x) + `d/dx` (10)

= 30 + 0

= 30

Answer: The Second derivative of given quadratic function is 30

Sides of a Polygon Formula

Introduction to sides of a polygon formula:

In geometry, polygon is two dimensional shapes. It has more than two sides. All sides are straight and connected to another side. The number of vertices is equal to number of sides. Vertices are nothing but corner points of the shape. In this article we shall see how to calculate the area of regular polygon.

Sides of a Polygon Formula - Formulas:

Triangle:

Formula to find area of triangle:

Area of triangle (A) = `1/2` (b x h) square units

b – Base

h – Height

Square:



Formula to find area of square:

Area of square (A) =a2 square units

a – side length                            

Pentagon:



Formula to find the pentagon:

Area of the pentagon (A) = t2 1.72 square units

t – Side length

Hexagon:



Formula to find area of hexagon:

Area (A) = t2 2.6 square units.

t - Side length

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Sides of a Polygon Formula – Example Problems:

1. Find the area of triangle whose base is 40 cm and height is 25 cm.

Solution:

Given:

Base (b) = 40 cm

Height (h) = 25 cm

Substitute the given value in the formula

Area of the triangle (A) = `1/2` (b x h) square units

= `1/2` (40 x 25)

= `1/2` (1000)

= 500

Area of the triangle (A) = 500 cm2

2. Find the area of the square, whose side length is 27 cm

Solution:

Given:

Side (a) = 27 cm

Substitute the given value in the formula

Area of square = a2 square units

= 27 x 27

Area of square = 729 cm

3. The side length of pentagon is 13.5 cm. Find the area and perimeter of the pentagon.

Solution:

Given:

Side length (t) = 13.5 cm

Substitute the given value in the formula

Area of the regular pentagon (A) = 1.72 t2 square units

= 1.72 x 13.52

= 1.72 x 182.25

= 313.47

Area of the regular pentagon (A) = 313.47 square units

4. The side length of hexagon is 14.7 cm. find the area of the hexagon.

Solution:

Given:

Side length (t) = 14.7 cm

Substitute the given value in the formula

Formula:

Area of the hexagon (A) = t2 2.6 square units

= 14.72 x 2.6

= 216.09 x 2.6

= 561.834

Area of the hexagon = 561.834 cm2

Perpendicular Line Segments

Introduction to perpendicular line segments:

Perpendicular line:

In geometry, two line segments or planes (or a line and a plane), are considered perpendicular (or orthogonal) to each other if they form congruent adjacent angles (a T-shape).  Thus, referring to Figure 1, the line AB is the perpendicular to CD through the point B.



Fig(i) Perpendicular line

Line segments:

The line segments are the the part of the line that has two end points.It include all the points between its end points.For example,consider the following figure,

Fig(ii) Line segment

In the above figure AB is a line and CD is a line segment.C and D are the two end points of a line segment.We are going to see about the perpendicular line and line segments.

Examples for Perpendicular Line Segments:

The following are the geometry examples for line segments,

Sides of the triangle

Sides of the Rectangle

Sides of the Square

Properties of perpendicular line:

The multiplication of slopes of the perpendicular lines is equal to -1.

That is m1=slope of perpendicular line 1

m2 = Slope of perpendicular line 2

m1 × m2 = -1

Distance formula for line segments:

The following formula used to find the lengh of the line segments ,

length = `sqrt((x2-x1)^2+(y2-y1)^2))`

Here (x1,y1) and (x2,y2) are the two end points of the line segments.

Problems on Perpendicular Line Segments:

Problem 1:

Find the equation of a line which is perpendicular to 4y - x = 20 and passes through the  point (2, -3).

Solution:

Given 4y - x = 20 and the point (2,-3)

To find the perpendicular line we need to find the slope.

To find the slope we need to change the given equation into slope intercept form.

4y - x = 20

Add x on both side,

4y - x  = 20

+ x =  +x

4y = x + 20

Divide by 4 on both side,

y = (x/4) + 5

y = mx + b form

So the slope m = 1/4

We know that multiple of slopes of the perpendicular lines equals to -1

1/4 * m = -1

Multiply by 4 on both sides,

m = -4

The line equation is,

(y-y1) = m(x-x1)

(y - (-3)) = -4(x-2)

y+4 = -4x+8

Subtract  4 on both sides,

y = -4x +4

Answer : The line perpendicular to the given line is    y = -4x +4

Problem 2:

By using the line segments distance formula find the length of the line segmen xy with x(5 ,3) y(1,2)

Solution:

length = `sqrt((x2-x1)^2+(y2-y1)^2))`

x1 = 5    y1 =3    x2 = 1    y2 = 2

=` sqrt( ( 1-5)^2 + ( 2-3)^2)`

=` sqrt( (-4)^2 + (-1)^2)`

= `sqrt (16+1)`

= `sqrt (17)`

Answer :The length of the line segment xy = `sqrt (17)`

Solving Vector Cartesian Coordinates

Introduction on solving vector Cartesian coordinates:

This article is about solving vector Cartesian coordinates. Solving vector Cartesian coordinates is very simple. The tutors of tutor vista helps the students in solving vector Cartesian coordinate any time. The vector coordinates are x,y and z. Three simple methods are there to describe a vector. Directions, specific lengths, angles and projections or components are used to describe vectors. The simplest method of these is Cartesian or rectangular co-ordinate system. Below we can see about solving vector Cartesian coordinates.

Solving Vector Cartesian Coordinates

In cartesian coordinate system three co-ordinate axes x,y and z are mutually right angles to each other. Consider a point P(x,y,z) in space at a distance r from the orgin. The vector r can be represented as

r = `xbara_x``+ y bara_y + z bara_z`

Where `bara_x, bara_y,` and `bara_z` are unit vectors

x,y and z are the components vectors. Components vectors have a magnitude and direction. Unit vectors have unit magnitude and directed along the co-ordinate axis.

A unit vector in a given direction is a vector in that direction divided by its magnitude. It is given by

`a_r = r/|r|`

`a_r = (xbara_x + ybara_y + z bara_z)/sqrt (x^2 + y^2 + z^2)`

Consider the points P(x,y,z)and Q(x + dx, y +dy, z+dz) in rectangul;ar co-ordinate system. The differential length dl from P to Q is the diagonal of the parallel piped is given by

dl = sqrt ((dx)^2 + (dy)^2 + (dz)^2)

The differential area ds = dx dy

= dy dz

= dz dx

The differential volume dv = dx dy dz

Solving Vector Cartesian Coordinates

Conversion of cylindrical to cartesian system

The cylinderical co-ordinates (?, F, z) can be converted into cartesian co-ordinates ( x, y, z)

Given                                                       Transform

?                                                             x = r cos ?

f                                                            y = r sin f

z                                                             z = z

conversion of spherical to cartesian co-ordinates system

The spherical co-ordinates (r, `theta` , f) can be transformed into cartesian co-ordinates (x, y, z)

Given                                                       Transform

r                                                             x = r sin ?. cosf

?                                                            y = r sin ? sin f

f                                                            z =r cos ?


Solving Vector Cartesian Coordinates

Example Problem: Give the Cartesian co-ordinates of a point whose cylinderical are ? = 1, f = 45°, z =2.

Given

Cylinderical co-ordinates are ? = 1, f = 45°, z =2.

Cartesian co-ordinates are x, y, z

x = ? cosf

= 1.cos 45 = 0.707

y = ? sinf

= 1. sin 45 = 0.707

z = z

= 2

Cartesian co-ordinates are (0.707, 0.707, 2)

Regular Irregular Polygons

Introduction for Regular irregular polygons:

Regular polygons:

When each side and angles of any polygon are similar, then it is said to be a regular polygon. A regular polygon is one with its entire sides similar and each of its angles similar, for instance: A regular polygon with 3 surfaces is said to be equilateral triangle.

Irregular Polygons:

When angles of the any polygon are not the same, then it is called as irregular polygon.

Regular Irregular Polygons-angles of Regular Polygons:

The regular polygons’ angle is grouped into two types. They are

Exterior angle
Interior angle
Interior angle of Regular Polygons:

Angle of regular polygons = [(n - 2) / n] × 180°

where n =number of sides.

Example 1:

Octagon is an eight sided polygon

Regular Polygons’ Interior Angle = [(n - 2) / n] × 180°

Here n = 8 for octagon

So the interior angle of an octagon = [(8 – 2)/ 8] * 180°

= (3 / 4) * 180°

Interior angle of octagon = 540° / 4 = 135°

Exterior angle of Regular Polygons:

360° is the exterior angle of polygons.  Hence the exterior angle can be estimated by the formula,

Regular Polygons’ Exterior angle = 360° / n

where n is the number of sides.

Example 2:

Pentagon is a polygon with five sides. Polygon’s Exterior angle = 360° / n

For Pentagon, n = 5

So the exterior angle of a pentagon = 360 / 5 = 72°.

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Regular Irregular Polygons-area of Irregular Polygons:

The steps for determining the area of irregular polygon are,

Convert every vertex into co-ordinate in graph.
Take any line segment and down to the x - axis.
Compute the average of two altitudes.
Compute the difference of two widths.
Multiply altitude with width.
Go over steps 2 to 5 until manipulating every line segment.
Then add all the areas of every line segment.
So, we will result in area of irregular shape.

Symmetry for Irregular Polygon:

Since the irregular polygons do not have equal angles, they do not posses line of symmetry..