Naming Lines in Geometry

Introduction for naming lines in geometry:
Lines are one dimension straight geometry figure and in solid geometry lines are used in designs.A lines are start with the one end and end with one direction then it said to be line segment.Lines are classified into many types which depends upon the line projection.Line segment is denoted with a connected piece of line.line segments names  has two endpoints and it is named by its endpoints. In this article contains naming lines in geometry I like to share this Area of a Rhombus Formula with you all through my article.

Naming Lines in Geometry:

In naming lines geometry section we have many types of lines which has propertyof its own.Lines are classified into following types.

Parallel lines:
In geometry parallel lines are mostly aplicable in design section, two lines which does not touch each other are called parallel lines.

Perpendicular lines:
In geometry Perpendicular lines are mostly aplicable drawing section,Two line segment  that form a L shape are called perpendicular lines.

Intersecting lines:
If two lines intersect at a point, these lines are called intersecting lines.

Concurrent lines:
The three or more lines passing through the same point are called concurrent lines. Understanding math help live is always challenging for me but thanks to all math help websites to help me out.

Problems in Naming Lines Geometry:

Example 1:
Find co-ordinate of the mid point of the line segment joining given points A(-4,1) and B(5,4)

Solution:
The required mid point is
Formul a   `((x_1+x_2)/2 ,(y_1+y_2)/2)` here,  (x1, y1) = (-4,1),(x2, y2) = (5,4)

=  `((-4+5)/(2))``((1 +4)/(2)) `

= `(-1/2) ` ,  ` (5/2)`


Example 2:
Find the slope of the lines given (1,-3) and (-1,3)

Solution:
(x1,y1)= (1,-3), (x2,y2)= (-1,3).
We know to find slope of line,m=` (y_2-y_1) /(x_2-x_1)`

=`(3+3)/(-1-1)`

m =`6/-2` = -3

Example 3:
Find the equation of the line having slope 5 and y-intercept -1.

Solution:
Applying the equation of the line is y = mx + c
Given,       m =5 ,c = -1
y = 5 x -1

or  y = 5x - 1
or  5x- y +1 = 0.

Plot Points

Introduction:

A rectangular co-ordinate system, or Cartesian plane, is a set of two intersect and vertical axes forming a xy plane. The horizontal axis is generally labelled the x-axis and the vertical axis is generally labelled the y-axis. The two axes split the planes into four parts known as quadrants. Any point on the plane communicate to an ordered pair (x, y) of valid numbers x and y.

Types of Plot Points:

Line plot
Scatter plot
Stem and Leaf Plot
Box plot
Line plot: A line graph plots constant data as points and then joins them with a line. Multiple data sets can be graphed simultaneously, but a key have to be used.

Scatter plot: A scatter plot defined as the organization between the two factors of the testing. A line which is used to find the positive, negative, or no correlation.

Stem and Leaf Plot: Stem and leaf plot points are defined as the documentation data values in rows, and can easily be made into a histogram. Large information sets can be accommodated by splitting stems.

Box plot: A box plot points are defined as a concise graph screening the five point abstract. Multiple box plots can be drawn side by side to evaluate more than one information set.

Advantages and Disadvantages of Plot Points:

Line plot

Advantages:

Immediate analysis of data.

Shows variety, minimum & maximum, gaps & clusters, and outliers simply.

Accurate values retained.

Disadvantages:

Not as visually attractive.

Top for below 50 data values.

Desires small range of data.

Scatter plot

Advantages:

Shows a movement in the data connection.

Retains accurate data ideals and example size.

Shows lowest/highest and outliers.

Disadvantages:

Hard to imagine outcome in huge data sets.

Flat drift line gives indecisive results.

Stem and Leaf Plot


Advantages:

Concise symbol of data.

Shows range, smallest & highest, gaps & clusters, and outliers simple.

Can hold very large data set.

Disadvantages:

Not visually attractive.

Does not simply indicate events of centrality for huge data sets.

Box plot

Advantages:

Shows 5-point review and outliers.

Simply compare two or supplementary information sets.

Handles extremely large data sets easily.

Disadvantages:

Not as visually attractive as extra graphs.

Accurate values not retained.

Surface Area of a Box

Introduction to surface area of a box:
Box is same as the cuboid. In box the dimensions are length, width and height. If all the dimensions are equal then the box is cube and if it is different then the box is cuboid. Box is a 3 dimensional image. A box has 8 vertices, 12 edges and 6 faces. Understanding Area of a Hexagon is always challenging for me but thanks to all math help websites to help me out.

Diagram and Formula – Surface Area of a Box:

Surface area of a box = 2(1w+lh+wh)

Where w`=>` width of the box

h`=>` height of the box

l`=>` length of the box. Is this topic how to construct parallel lines hard for you? Watch out for my coming posts.

Example Problems – Surface Area of a Box:

Example 1 :

Find the surface area of a box whose length, width and height are 12cm, 14cm, 16cm.

Solution:

Given that,

Length of the box = 12cm

Width of the box = 14cm

Height of the box=16cm.

Surface area of a box = 2(lw + lh + wh)

=2((12*14) +(12*16)+(14*16))

=2(168+192+224)

=2(584)

=1168cm3.

Example 2 :

Find the surface area of a box whose length, width and height are 2cm, 4cm, 6cm.

Solution:

Given that,

Length of the box = 2cm

Width of the box = 4cm

Height of the box=6cm.

Surface area of a box = 2(lw + lh + wh)

=2((2*4) +(2*6)+( 4*6))

=2(8+12+24)

=2(44)

=88cm3.

Example 3 :

Find the surface area of a box whose length, width and height are 8cm, 6cm, 4cm.

Solution:

Given that,

Length of the box = 8cm

Width of the box = 6cm

Height of the box=4cm.

Surface area of a box = 2(lw + lh + wh)

=2((8*6) +(8*4)+(6*4))

=2(48+32+24)

=2(104)

=208cm3.

Example 4 :

Find the surface area of a box whose length, width and height are 10cm, 20cm, 30cm.

Solution:

Given that,

Length of the box = 10cm

Width of the box = 20cm

Height of the box=30cm.

Surface area of a box = 2(lw + lh + wh)

=2((10*20) +(10*30)+(20*30))

=2(200+300+600)

=2(1100)

=2200cm3.

Two Parallel Lines Cut by a Transversal

Introduction to two parallel lines cut by a transversal:

Parallel lines:

Two lines on a plane that never  intersect or meet is known as parallel lines. The distance between both the lines must be the same. And they must not intercept with each other. It can also be explained as the length between parallel lines will be exactly same at any point. Examples of parallel lines are railway track, etc.

Transversal:

A straight line is said to be transversal if the line cuts two or more parallel lines at different points. In the figure the line l cuts the parallel lines a and b. So the line l is called as a transversal line

Two Parallel Lines Cut by a Transversal:

When two parallel lines are cut by a transversal then

The corresponding angles are equal

Pair of Vertically Opposite angles is equal

Pairs of Alternate interior angles are equal

Interior angles on same side are supplementary

Conditions Satisfies when Two Parallel Lines Cut by a Transversal:

The corresponding angles formed by the transversal will be equal. For example angle 4 and angle 6 are corresponding angles, and the other pairs of corresponding angles are (5 and 3), (8 and 2) and (1 and 7).

Therefore we have:               angle 6 = angle 4

angle 5 = angle 3

angle 8 = angle 2

angle 7 = angle 1

Pair of Vertically Opposite angles is equal

The pair of vertically opposite angles is equal when a transversal line is formed. For example, angle 1 and angle 3 are vertically opposite angles and the other pairs of vertically opposite angles are (2 and 4), (5 and 7) and (6 and 8).

Therefore we have:                angle 1 = angle 3

angle 2 = angle 4

angle 5 = angle 7

angle 6 = angle 8

Pairs of Alternate interior angles are equal

The pair of alternate angles is equal when the transversal line is formed. Here the pairs (2 and 6) and (3 and 7) are alternate interior angles. I have recently faced lot of problem while learning simple math problems for kids, But thank to online resources of math which helped me to learn myself easily on net.

Therefore we have:                angle 2 = angle 6

angle 3 = angle 7

Interior angles on same side are supplementary

Interior angles on same side are supplementary when a transversal is formed. The angles on the same side of the transversal are (6 and 3) and (2 and 7).

Therefore we have:                angle 3+ angle 6 = 180

angle 2 + angle 7 = 180

Radius from Circumference

Introduction to radius from circumference:
In day to day life, we often came across some unique math terms. Radius is one of the special math terms that falls under this category.

Radius of a circle is nothing but the line segment from the center of the circle to its perimeter. In other terms, half the diameter is the radius.

In this article of radius from circumference, we are going to find the radius of the circle from the circumference formula.

Formula for Radius from Circumference:

The Circumference of the circle is given by the following formula:

Circumference =  2`pi`r

If the Circumference C is given, the radius can be calculated by the following formula:

radius  =  `C/(2 pi)`

Example Problems for Finding Radius:
Example 1:

Find the radius, if the circumference of the circle is 100 cm

Solution:

Radius of circle, r  =  `C / ( 2 pi )`

=  `100 / (2 * 3.14)`

=  `100 / 6.28`

=  15.92 cm

Example 2:

Find the radius, if the circumference of the circle is 94 cm

Solution:

Radius of circle, r  =  `C / ( 2 pi )`

=  `94 / (2 * 3.14)`

=  `94 / 6.28`

=  14.97 cm

Example 3:

Find the radius, if the circumference of the circle is 60 mm

Solution:

Radius of circle, r  =  `C / ( 2 pi )`

=  `60 / (2 * 3.14)`

=  `60 / 6.28`

=  9.55 mm

Example 4:

Find the radius, if the circumference of the circle is 35 cm

Solution:

Radius of circle, r  =  `C / ( 2 pi )`

=  `35 / (2 * 3.14)`

=  `35 / 6.28`

=  5.57 cm

Example 5:

Find the radius, if the circumference of the circle is 20 mm

Solution:

Radius of circle, r  =  `C / ( 2 pi )`

=  `20 / (2 * 3.14)`

=  `20 / 6.28`

=  3.18 mm

Example 6:

Find the radius, if the circumference of the circle is 4 m

Solution:

Radius of circle, r  =  `C / ( 2 pi )`

=  `4 / (2 * 3.14)`

=  `4 / 6.28`

=  0.64 m


Practice Problems for Finding Radius:

1) Find the radius, if the circumference of the circle is 50 cm

Answer: 7.96 cm

2) Find the radius, if the circumference of the circle is 20 m

Answer: 3.18 m

3) Find the radius, if the circumference of the circle is 110 mm

Answer: 17.52 mm

4) Find the radius, if the circumference of the circle is 75 cm

Answer: 11.94 cm

5) Find the radius, if the circumference of the circle is 72 mm

Answer: 11.46 mm

Geometry Edge of Rectangular

Introduction to Geometry edge of rectangular:

Rectangular shape is one of the geometry two dimensional objects. Geom`etry edge of rectangular properties are the crossed quadrilateral which consists of two opposite sides of a rectangle along with the two diagonals. Its angles are not right angles. Opposite sides are parallel and congruent . The diagonal bisect each other The diagonals are congruent. A four -sided plane figure with four right angles. Understanding Volume of a Rectangular Prism Formula is always challenging for me but thanks to all math help websites to help me out.

Basic Concepts of Geometry Edge of Rectangular:

Geometry edge of rectangular:

Each and every object should have edges. Two edges make the angle of geometry object.And also edges to make the corners and vertices of object .Rectangle have  four  edges and also have four vertices or corners .Each corner make the angle of 90degree.

From this diagram:

AB, BD, DC, AC are edges of the rectangle.

AB edge is parallel to CD edge

AC edge is parallel to BD edge

AB || CD, AC || BD (opposite sides are equal in rectangle)

Each edge should make the angles are


Please express your views of this topic how many edges does a rectangular prism have by commenting on blog.

Area and Perimeter of the Geometry Edge of Rectangular:

Area and perimeter:

Area of rectangle= Length * Width

Perimeter of the rectangle=2(Length + Width)

Both area and perimeter are depends on the rectangle edges.

A rectangle has two Length edges and two width edges.

Length edges are AB, CD

Width edges are BD, AC

Two length edges are equal AB=CD

Two width edges are equal BD=AC

Example problems in Geometry edge of rectangular:

1.From given diagram Find the area and perimeter of the rectangle  and name of the rectangular edges?

Solution:

Given data PQRS is closed rectangle

PQ= 13cm(PQ||SRsO P)

PQ=SR=13

QR=5.5cm(QR||PS)

So QR=PS=5.5

Length   of the rectangle=13

Width of the rectangle =5.5

Finding the area:

(1  )Area of the rectangle=Length x Width

= 13x5.5

Area =71.5cm2

(2) Finding the perimeter:

Perimeter of the rectangle=2( L+W)

=2(13+5.5)

=2(18.5)

=37cm

Perimeter=37

(3)Finding the name of the edges:

From given diagram edges are PQ,QR,RS,SP

Volume of Equilateral Triangle

Introduction

In geometry, an equilateral triangle is a triangle in which all three sides are equal. In traditional or Euclidean geometry, equilateral triangles are also equiangular; that is, all three internal angles are also congruent to each other and are each 60°. They are regular polygons, and can therefore also be referred to as regular triangles.As, equilateral triangle is two dimensional, it is impossible to determine its volume.So, lets know how to determine volume of the prism with equilateral triangular base.

(Source -wiki)

Formula for Volume of Equilateral Triangle Base Prism:

The volume of equilateral triangle, ` v = 1/3 B xx H`

Example for Volume of Equilateral Triangle Base :

Example 1. Find the volume of equilateral triangle base prism of side 10 cm and height `17/2 cm`

Solution:

Let, H = `17/2 cm` , B = 10cm

Formula used:  ` V = 1/3 B H`

Therefore, volume of equilateral triangle base prism ,

`V = 1/3 xx 10 xx 17/2` = `85/3 cm3`

Example2. Find the volume of equilateral triangle base prism of side 100 m and height 85 m

Solution:

Let, `H = 17/2 m` , B = 10 m

Formula used: ` V = 1/3 B H`

Therefore, volume of equilateral triangle base prism,

`V = 1/3 xx 10 xx 17/2`

= 2833.33 m3

Example3. Find the volume of equilateral triangle base prism of side 9 cm and height `15/2` cm

Solution:

Let, `H = 15/2 cm` , B = 9 cm

Formula used:   `V = 1/3 B H`

Therefore, volume of equilateral triangle base prism ,

`V = 1/3 xx 9 xx 15/2`

= `45/2` cm3

Example4. Find the volume of equilateral triangle base prism of side 90 cm and height 75 cm

Solution:

Let, H = 75cm, B = 90 cm

Formula used:      `V = 1/3 B H`

Therefore, volume of equilateral triangle base prism ,

`V = 1/3 xx 90 xx 75`

= 2250 cm3



Practice Problems Volume of Equilateral Triangle Base Prism

1. Find the volume of equilateral triangle base prism of side 180 m and height 156 m

2. Find the volume of equilateral triangle base prism of side 450 cm and height 390 cm

Solution for practice problems of volume of equilateral triangle base prism :

1. 9360 m3

2. 58500cm3