Triangular Prism Net

A triangular prism is a solid has two parallel faces which are congruent triangle at both ends. These faces form the bases of the prism

The others faces are in the shape of parallellogram.They are called lateral faces.
A right triangular prism that has its bases perpendicular to its lateral surfaces.

Once you've gone through these, take a look at our Volume of Rectangular Prism for more refernce.

Surface area of triangular prism=2base area+peripeter of base x height prism
Volume of triangular prism = Base area of prism x height of prism.

Under this concept of triangular prism we draw the triangular prism net.
 Prepare the model of triangular prism net ,we need a paper , a pencil or pen, a scissor and gum.

Method of preparing triangular prism net

 First we draw the rectangle and two triangles to two ends of the rectangle on the paper. Then cut out along solid sides.
 Fold along dotted lines.
 Use clear tape to fasten .
 If you want to draw or colour the net,do it before you tape it together.
 If you want to decorate it by gluing on decorations, tape it together first.

Quadrilateral Prism Shape

In geometry, an n-sided prism is a polyhedron made of an n-sided polygonal base, a translated copy, and n faces joining corresponding sides. Thus these joining faces are parallelograms. All cross-sections parallel to the base faces are the same (source-Wikipedia). If the prism has four sides then it is called quadrilateral prism. in this article we shall discus about some quadrilateral prism.

Quadrilateral prim shapes:

Cube:

It is one type of quadrilateral prism. Its all dimensions are equal in measure (all side are equal).it has eight vertices. The shape of the cube is shown in below.
Formula for Volume of the cube V= a3 cubic units
Where,
            V = volume of the cube
 a =side length (or edge) of the cube.
Formula for surface area of the cube (A) = 6 a² square units

Rectangle prism:

It is one type of quadrilateral prism. The shape of the cube is shown in below.
Volume of rectangular solid (V) = length x width x height Cubic units
     = l x w x h cubic unit.
Total surface area of rectangular prism
                                    = 2 (length x breadth + breadth x width + length x height)
                                    =2(lb + bh+ lh) square units

Trapezoidal prism:

It is one type of quadrilateral prism.it has one set of parallel sides. The shape of the trapezoidal prism is shown in the figure.
Volume of trapezoidal prism = l x area of the base cubic units
                                                       l – Length of trapezoidal prism
                              Area of the base= 1/2 x (a + b) x h square units
                                                       a and b are length of parallel sides.
                                                       h – Height

Quadrilateral prism shape - Example problems:

1. Find the volume and surface area of the cube whose side length is 17 cm.
Solution:
            Given:
                        Side length (a) = 17 cm
Formula:
            Volume of the cube V= a³ cubic units
                                                = 17³
                                                = 17 x 17 x 17
                                                = 4913 cm³
            Surface area of the cube (A) = 6 a² square units
                                                          = 6 x 17²
                                                          = 6 x 289
                                                          = 1734 cm²

2. Find the lateral surface area and total surface area of a rectangular whose length is 6 cm, width 3cm and height 6cm.
Solution:
Given:
                              Length (l) = 6 cm
                         Width (w) = 3 cm
                           Height (h) = 6 cm
Total surface area of rectangular prism =2(wh + lw+ lh) square units
                                                               = 2(3 x 6+ 6 x 3 +6 x 6)
                                                               = 2 (18 + 18 + 36)
                                                               = 2 (72)
                                                               = 144
Total surface area of rectangular prism = 144 cm²
            Volume of the rectangular prism (V) = l x w x h cubic units
                                                                           = 6 x 3 x 6
                                                                           = 108
                                                                           = 108 cm³
3.figure out the volume trapezoidal prism whose length 14 cm, height 5cm, length of parallel sides a=7 cm and b=4cm.
Solution:
Given:
                  Length (l) = 20 cm
                  Height (h) = 5 cm
                  Parallel sides a=7 cm and b=4 cm
Formula:
Volume of trapezoidal prism = l x area of the base cubic units

Area of the base:
Area of the base= 1/2 x (a + b) x h
                                                   = 1/2x (7 + 4) x 5
                                                   =1/2 x 11 x 5
                                                     = 27.5 cm²
Volume of trapezoidal prism = 20 x 27.5
                                                  = 550

Volume of trapezoidal prism = 550 cm³

Figuring Volume of a Structure

Volume of structure
Volume is how much three-dimensional space a substance (solid, liquid, gas, or plasma) or shape occupies. The volume of a container is generally understood to be the capacity of the container, i. e. the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces.
Volume is measured in "cubic" units.


Volume of Cylinder
A cylinder is a solid that has two parallel faces which are congruent circles. These faces form the bases of the cylinder. The cylinder has one curved surface. The height of the cylinder is the perpendicular distance between the two bases.
The volume of a cylinder is given by the formula:
Volume = Area of base × height

V =   r²h where r = radius of cylinder and h is the height or length of cylinder.

Volume of hollow cylinder

Volume of hollow cylinder
V=πR²h-πr²h
Where R is the radius of the outer surface and r is the radius of the inner surface.

Volume of Cone

A cone is a solid with a circular base. It has a curved surface which tapers (i.e. decreases in size) to a vertex at the top. The height of the cone is the perpendicular distance from the base to the vertex.

Volume of cone = 1/3 Area of base × height
V = 1/3πr²h where r is the radius of the base and h is the height of the prism.

Volume of Pyramid

A pyramid is a solid with a polygonal base and several triangular lateral faces. The pyramid is named after the shape of its base. For example, rectangular pyramid, triangular pyramid.
The lateral faces meet at a common vertex. The height of the pyramid is the perpendicular distance from the base to the vertex.

The volume of a pyramid is given by the formula:

Volume of pyramid =  1/3Area of base × height

V = 1/3Axh where A is the area of the base and h is the height of the pyramid.

Algebra Data Analysis Help

he Algebra data analysis is one of the important branch of algebra concerning the learn of rules of operations and dealings. The constructions and concepts arising from algebra data analysis. Algebra data analysis including terms, polynomials, expressions, quadratic formula and algebraic structures or pre-algebra structure. We will solve examples in polynomial equations and quadratic formula. This article has information about algebra data analysis help as well as examples.

Polynomial equations – Algebra Data Analysis Help:

In algebra data analysis, the polynomial equations are one of the most significant factors.
Now we will solve the polynomial problem in algebra data analysis with help of example problem.

Example Problem 1:
Solve given polynomial equations.
6x² + 4 + 9x + 3x² + 3x + 9 + 7x
Solution:
Step 1:
First we have to mingle terms x²
6x²+3x²=9x²
Step 2:
Now combine the terms x
9x + 3x + 7x = 19x
Step 3:
Then join the constants terms
4 +9 =13
Step 4:
Finally, combine all the terms
9x² + 19x +13
So, the final answer is 9x² + 19x +13

Example Problem 2:
Solve following polynomial expression.
11x² + 2 + 9x + 13x² + 5x + 4 + 3x
Solution:
Step 1:
First we have to mingle terms x²
11x²+13x²=24x²
Step 2:
Now combine the terms x
9x + 5x + 3x = 17x
Step 3:
Then join the constants terms
2 + 4 =6
Step 4:
Finally, combine all the terms
24x² + 17x +6
So, the final answer is 24x² + 17x +6

Quadratic formula - Algebra Data Analysis Help:

In algebra data analysis, quadratic equation with real or complex coefficients contains two solutions, is known as roots.
The roots are given by the quadratic formula

Example of quadratic formula:
Example Problem1:
Solve 7x² + 5x = -3.
Solution:
Now we will find the standard form for the quadratic equation and calculate a, b, and c.
7x² + 5x + 3 = 0
a = 7
b = 5
c = 3

Now applying the formula,


= `(-5 +- sqrt(5^2 - 4 xx 7 xx 3))/(2 xx 7)`
=`(-5+-(sqrt(25-84)))/(2xx7)`
=`(-5 +-(sqrt(-59)))/(14)`
=`(-5+- i 7.68)/14`
=2.68 `+-i` -7.68

Answer:
2.68 `+-i` -7.68
These are examples in algebra data analysis help.

That's all about algebra data analysis help.

Measurement of Angle

• Line joins two positions and which are nonstop in two directions. Lines have one endpoint of two lines and infinite in one direction is called angle.
•  The grouping point of two lines is called the vertex.

               

•  Angles are denoted as degrees. In measurement of angles can diverge in between 0 and 180 angles.

Concept -measurement Angle:

Concept of angles:

      There are three major categories of angles: right angle, obtuse angle, and acute angle.

• A right angle – a measurement of an angle is 90 degrees.
• An acute angle - a measurement of an angle in between 0 to 90 degrees.
• An obtuse angle - a measurement of an angle in between 90 and 180 degrees.

More about angles and measurement of angle:

More about angles:

1)      Vertical angle measurements are two angles with a common vertex and with sides that are two pairs of opposite of opposite rays. (That is, the union of the two pairs of sides is two lines.)  

                              

2)       A right angle measurements has a measure of 90^o.    

     

                       

3)      An acute angle measurement is an angle whose measure is larger than 0^o and smaller than 90^o.  

          

                

4)      An obtuse angle measurement has a measure larger than 90^o and less than 180^o.  

           

               

5)     Straight angle measurement has a measure of 180^o. The rays of a straight angle lie on a line.

6)      Complementary angle measurements are two angles, the addition of whose measures is 90^o.

               

7)      Supplementary angle measurements are two angles, the addition of whose measures is 180^o.

               

Step to measurement of angle:

          

Step 1: To find the angle is obtuse or acute before using the protractor.

          

Step 2: Position the protractor on the center mark on the vertex of the angle

          

Step 3: position the protractor, line up with one ray of the angle and another ray of the angle cross the protractor’s scale

          

Step 4: to check the measure of the angle of the line that crosses the protractor scale.

          

How to calculate volume

Example problems:

Here are some solved examples on How to calculate volume:
Problem 1:

Problem 2:
Calculate the volume of the cone with base radius = 5cm and height at the apex is 4 cm.
Solution:
Volume of a cone: (13)(π)(r2)h
                                 = (13)(π)(52)4
                                = 104.666 cm^3
Problem 3:

Problem 4:
Calculate the length of the rectangular prism, when the volume is 36 cm^3 , width = 3 m, height is = 4m
Solution:
Volume of the rectangular prism = l * w * h.
                                                     36 =  l* 4*3
                                          Therefore,  l= 3m
Students can understand How to calculate volume from the above examples and learn to solve on their own.

Parametric t Test

Conventional statistical procedures are also called parametric tests. In a parametric test a sample statistic is obtained to estimate the population parameter. Because this estimation process involves a sample, a sampling distribution, and a population, certain parametric assumptions are required to ensure all components are compatible with each other. 

For example, in Analysis of Variance (ANOVA) there are three assumptions:

• Observations are independent.
• The sample data have a normal distribution.
• Scores in different groups have homogeneous variances.

In a repeated measure design, it is assumed that the data structure conforms to the compound symmetry. A regression model assumes the absence of collinearity, the absence of auto correlation, random residuals, linearity...etc. In structural equation modeling, the data should be multivariate normal.

Why are they important? Take ANOVA as an example. ANOVA is a procedure of comparing means in terms of variance with reference to a normal distribution. The inventor of ANOVA, Sir R. A. Fisher (1935) clearly explained the relationship among the mean, the variance, and the normal distribution: "The normal distribution has only two characteristics, its mean and its variance. The mean determines the bias of our estimate, and the variance determines its precision." (p.42) It is generally known that the estimation is more precise as the variance becomes smaller and smaller.

Put it in another way: the purpose of ANOVA is to extract precise information out of bias, or to filter signal out of noise. When the data are skewed (non-normal), the means can no longer reflect the central location and thus the signal is biased. When the variances are unequal, not every group has the same level of noise and thus the comparison is invalid. More importantly, the purpose of parametric test is to make inferences from the sample statistic to the population parameter through sampling distributions.

When the assumptions are not met in the sample data, the statistic may not be a good estimation to the parameter. It is incorrect to say that the population is assumed to be normal and equal in variance, therefore the researcher demands the same properties in the sample. Actually, the population is infinite and unknown. It may or may not possess those attributes. The required assumptions are imposed on the data because those attributes are found in sampling distributions. However, very often the acquired data do not meet these assumptions. There are several alternatives to rectify this situation.

Free Sample Trinomials

Trinomials:

In elementary algebra, a trinomial is a polynomial consisting of three terms or monomials.(source : WIKIPEDIA)

The trinomial must be one of the following form .

Examples for free sample trinomial:

1. 9x + 3y + 5z , where x , y, z are variables.

2. xy + x + 2y, Where x, y are variables.

3. x2+x-8, where x is variable.

4. ax + by + c = 0 , Where a,b,c are constants and x,y are variables.

Our  tutor vista website provide opportunity to learn about sample trinomials with free of cost. In this article we are going to see some sample problems on solving and factoring trinomials.

Free sample problems on trinomials:

Problem 1:

Square the following trinomial,

x+3y-2

Solution:

Given, x+3y-2

We need to find the square for the given trinomial,

That is (x+3y-2)2

We can apply the following formukla to find the square for the above trinomial,

( a + b + c)2 = a2+b2+c2+2ab+2bc+2ca

(x+3y-2)2  = x2+(3y)2+22+2(x)(3y)+2(3y)(-2)+2(-2)(x)

= x2+9y2+4+6xy-12y-4x

= x2+ 9y2+ 6xy - 4x -12y + 4

Answer: (x+3y-2)2 = x2+ 9y2+ 6xy - 4x -12y + 4

Problem 2:

Factor the trinomail x2 + x – 156 .

Solution:

Given , x2 + x – 156 .

- 156 (product)

/     \

-12     13

\    /

1 (sum)

= x2 - 12x + 13x -156

= x ( x - 12 ) + 13 ( x -12 )

= (x+13) (x-12)

Answer: (x+13) and  (x-12) are the factors of the given trinomial.

Problem 3:

Solve the trinomial 2x2 - 2x = 12.

Solution:

Given, 2x2 - 2x = 12.

Subtract 12 on both sides,

2x2 - 2x - 12 = 12 - 12

2x2 - 2x - 12 = 0

2(x2 - x - 6 ) =0

Divide by 2 on both sides,

x2 - x - 6 = 0

-6

/  \

-3  2

\  /

-1

x2 -3x + 2x - 6 = 0

x(x - 3) + 2(x-3) = 0

(x+2) (x-3) = 0

(x+2) = 0

x = -2

(x-3) = 0

x =3

Answer: x = 3 , -2

Practice problems on free sample trinomials:

Problems:

1.Find the roots of the trinomial x2 - 5x = -6

2.Factor the trinomial 2x2 + 12 x - 14

Answer Key:

1. x = -1 , x = 6

2. ( x - 1) and (x + 7)

Examples Functions in Math

Functions in math deals with finding unknown variable from the given expression with the help of known values. In algebraic expression the variable are represented in alphabetic letters. Functions in math, the numbers are consider as constants. Algebraic expression deals with real number, complex number, and polynomials. In algebraic expression several identities to find the x values by using this we can easily find the algebraic expression of the particular function. The example function in math  may include the function of p(x), q(x),… to find the x value of the functions.

Examples function in math:

Q(y) = 4y2+12y + 40. In this equation we need to find the variable of Q(2) functions in math her y is 2.

Problems using examples functions in math

Examples functions in math

Problem 1: Examples functions in math using p(y) = y2 +2y +4

p(y) = y2 +2y +4 find the f(6).

Solution :

Given the function of p(y) there is y value is given function in math

p(y) = y2 +2y +4 find the p(6)

The value of x is 2 is given

p(6) = 62 +2*6 +4

p(6) = 36 +12 +4 In this step 6 square is 36 it is calculate and 2*6 is 12 be added

p(6) = 52.

The functions in math p(6) = y2 +2y +4 find the p(6) is 52.

Problem 2 Examples functions in math using q(x) = x2 +2x +40 find the q(3).

q(x) = x2 +2x +40 find the q(3).

Solution :

Given the examples functions in math of q(x) there is x value is given functions in math

q(x) = x2 +2x +40 find the f(3)

The value of x is 2 is given

q(3) = 32 +2*3 +40

q(3) = 9 + 6 +40 In this step 3 square is 9 it is calculate with 2*3 is 6 be added to 40 to find the example function in math q(3).

q(3) = 55.

The examples functions in math q(3) = x2 +2x +40 find the q(3) is 55

Example functions in math using cubic equation

Problems1: examples functions in math using f(x) = x3 +2x2 + 2x + 4 to find the f(3).

F(x) = x3 +2x2 + 2x + 4 find the functions in math f(3).

Solution

Given the function in math of  f(x) there is x value is given as 3. Find example function in math.

f(x) = x3 +2x2 + 2x + 4 find the f(3).

Here the value of x is given as 3

f(3) = 33 + 2*32 + 2*3 +4

f(3) = 27 +18+ +6 +4 In this step 3cube is calculated  as 27 and  3square is 9.

f(3) = 55.

The math functions of f(x) = x3 +2x2 + 2x + 4 find the f(3) = 55.

Problems 2: Examples functions in math using q(x) = x3 +2x2 + 4x + 25 to find the q(3).

q(x) = x3 +2x2 + 4x + 25 find the functions in math q(3).

q(x) = x3 +2x2 + 4x + 25 find the math function in q(3).

Solution Given the examples functions in math of  q(x) there is x value is given as 3. Find example function in math.

q(x) = x3 +2x2 + 2x + 25 find the f(3).

Here the value of x is given as 3

q(3) = 33 + 2*32 + 2*3 +25

q(3) = 27 +18+ 6 + 25 In this step 3cube is calculated  as 27 and  3 square is 9 multiplied with 2 and add 25 to find the example function in math of q(3).

q(3) = 76 .

The examples functions in math of q(x) = x3 +2x2 + 2x + 25  is the q(3) = 76.

Length of Line Segment

The line segment is the straight line and it has two points. They are starting point and ending point. The starting point is in the starting place of the line and the ending point is in the ending place of the line. The length of line segment is the distance between the starting point and ending point of a line.

Diagram and formula - Length of line segment:

The formula to find the length of the line segment with two points (x1,y1) and (x2, y2) is

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

Example problems - Length of line segment:

Find the length of line segment and the points are (1,1) and (2, 2).

Solution:

Given , (1,1) and (2, 2).

Let us take (1 ,1 ) as (x1, y1) and (2, 2) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((2-1)^2 + (2-1)^2)`

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

= `sqrt(1 + 1)`

= `sqrt(2)`

Find the length of line segment and the points are (1,1) and (3, 3).

Solution:

Given , (1,1) and (3, 3).

Let us take (1 ,1 ) as (x1, y1) and (3, 3) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((3-1)^2 + (3-1)^2)`

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

= `sqrt(4+4)`

= `sqrt(8)` .

= `sqrt(4 * 2)`

= `sqrt(4)` `xx` `sqrt(2)`

= 2 `xx` `sqrt(2)` .

Find the length of line segment and the points are (3,3) and (2, 2).

Solution:

Given , (3,3) and (2, 2).

Let us take (3 ,3 ) as (x1, y1) and (2, 2) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((2-3)^2 +(2-3)^2)`

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

= `sqrt(1 + 1)`

= `sqrt(2)` .

Find the length of line segment and the points are (2,2) and (5, 5).

Solution:

Given , (2,2) and (5, 5).

Let us take (2 ,2 ) as (x1, y1) and (5, 5) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((5-2)^2 + (5-2)^2)`

= `sqrt(3^2 + 3^2)`

= `sqrt(9+9)`

= `sqrt(18)`

= `sqrt(9 * 2)`

= `sqrt(9)` `xx` `sqrt(2)`

= 3 `xx` `sqrt(2)`

Mean Difference Standard Deviation

Definition of mean difference:

Mean difference is defined as the measure of the difference between the given data set and mean value.For finding the mean difference first have to find the mean,

Formula for finding the mean,

`barx = (sumx) / n`

Using the mean value mean difference have to be found Formula for mean difference is,

`x-barx`

Definition of Standard Deviation:

Standard Deviation is the determination of describing the variability and spread of the Data set in the given total values in data set. It is used to take the measurement for the average of numbers in the given Data set. Standard Deviation given by the square root for the summation of the total squared mean difference and it is divided by the total number of values minus one.

Formula for standard deviation,

S =` sqrt(((sum(x - barx))) / (n-1))`

Steps for calculating mean difference and standard deviation:

Get the mean for the given n numbers in the given data set.
Get mean difference of each given numbers in the Data set from the mean.
Take Square for all each deviations. It is called as the squared mean deviation.
Calculate the summation for standard  mean deviations.
Now apply the Standard Deviation formula for finding the Standard deviation form the mean.


Mean difference standard Deviation - Example Problems:

Mean difference standard Deviation - Problem1:

Calculate the mean difference and standard deviation in the following data set.56, 52, 54, 57, 58.

Solution:

Mean:

Formula For finding the mean.

`barx = (sum x) / n`

` barx = (56+ 52+ 54+ 57+ 58) / 5`        

`barx = 277 / 5`

`barx = 55.4`

Mean difference is given below

x                             `(x - barx)`

56                         56 - 55.4 =   0.6

52                         52 - 55.4 =  -3.4

54                        54 - 55.4 =   -1.4

57                        57 - 55.4 =    1.6

58                        58 - 55.4 =    2.6


Standard Deviation: Standard Deviation is given by,

S =` sqrt(((sum(x - barx))) / (n-1))`

S = `sqrt(((0.6)^2+(-3.4)^2+(-1.4)^2+(1.6)^2+(2.6)^2 )/ (5-1))`

S = `sqrt(23.2 / 4)`

S = `sqrt(5.8)`

S = 2.40831892

Mean difference standard Deviation - Problem 2:

Calculate the mean difference and standard Deviation for the given data set. 23, 25, 24, 26.

Mean:

Formula For finding the mean.

`barx = (sum x) / n`

` barx = (23+ 25+ 24+ 26) /4`        

`barx = 98 / 4`

`barx = 24.5`

Mean difference is given below

x                                  `(x-barx)`

23                        23 - 24.5 =   0.5

25                        25 - 24.5 = - 0.5

24                        24 - 24.5 =   1.5

26                        26 - 24.5 = - 1.5

Standard Deviation: Standard Deviation is given by,

S =` sqrt(((sum(x - barx))) / (n-1))`

S = `sqrt(((0.5)^2+(-0.5)^2+(1.5)^2+(-1.5)^2)/ (4-1))`

S = `sqrt(5 /3)`

S = `sqrt(1.66666667)`



S = 1.29099445

Finding Orthocenter

Orthocentre is point of intersection of all the three altitudes of a triangle.

The altitude of a triangle is the line perpendicular from one vertex to the opposite side.

Steps to find the orthocentre of triangle ABC algebrically :-

1) Find the equation of any two altitudes of the triangle

The slope of the line joining the opposite side BC  is calculated ,

Then,  the perpendicular slope is then calculated.

slope of a line is `(y2 - y1 )/(x2-x1)` =  m ,

perpendicular slope is  `(-1)/(m)`

Given are vertex A and slope of line,

we plug the values in y = mx +b we get the y intercept b

The equation of the altitude is y = mx + b

Same way we find the equation of the other altitude

2) Find point of intersection of the two altitudes is the orthocentre of the triangle.

Steps to find the orthocentre of a triangle graphically:

1) First we plot all the points of the triangle

2) We draw two altitudes of the trianlge

3) The point of intersection of the altitudes is the Orthocentre

Examples of Solving Orthocentre

Find the orthocentre of the of the triangle ABC whose vertices are A(-2,-1) , B ( -1,-4) and C (0,-5) .

Solution:    {i) Equation of altitude AD which is perpendicular to BC

x 2 = -1, y2 = -4 , x 1 = 0 , y1 = -5

slope of a line BC is  `(-4 +5)/(-1-0)` =  `(1)/(-1)`  =  -1

slope of line AD is `(-1)/(-1)` =  1

Line AD passes through A(-2,-1) and slope m = 1,

Equation of line is y = 1x +b

we plug in x = -2, y = -1,  we get,

-1 = -2 + b

-1 +2  = b

b = 1

Equation of AD is   y = x + 1 ------------------> 1
(ii) Equation of Altitude BE perpendicular to AC

x2 = 0, y2 = -5, x1 = -2, y1 = -1

slope of AC =  `(-5+1)/(0+2)` = `(-4)/(2)`  =  -2

Perpendicular slope is  `(-1)/(-2)`  =  `1/(1/2)`  = slope of BE

Equation of BE is y = `(1)/(2)` x + b

B (-1,-4)  Plug in x = -1, y = -4              ,

-4   = -1/2 +b

-4 + 1/2 = b ,  b = 7/2

Altitude BE is        y = `(1)/(2)` x + `(7)/(2)`  ---------------> 2

(iii) Point of intersection of AD , BE

Solve 1 and 2,    x +1  = `(1)/(2)` x + `(7)/(2)`

we get (-9,-8)

Solving Orthocentre of Different Types of Triangles using Graphs

Orthocentre of a right angled triangle is the vertex which is the right angle.

Orthocentre of a an obtuse angled triangle is  outside the triangle.

For an equilateral triangle , the orthocenter lies on the perpendicular bisector of each side of the triangle.

Some graphs to illustrate the above facts:-

Does Geometry use Integers

The geometry generally use the integers and the variables. The answer for the question "Does geometry use integers?" is yes, the geometry uses the integers.For specific purpose of using the geometry with integers, the separate topic available as algebraic geometry. The integers are used for to represent the co-ordinates in vertex, some equations of the line in the 2d geometry. The examples and practice problems are given below for the question "Does geometry use integers or not?".

some examples to explain "does geometry use integers"

• Consider the given points  (4,4), (2,3), (8,3), (6, 2 ), (5, 1) and plot them in the graph. And also we have to denote the quadrant in which each of the point lies.

Solution:

The plotted points in the coordinate plane are shown in the graph. All the co-ordinates of the points are in positive so all will be in the first quadrant itself. And also that  the co-ordinates of the points are not having the zero term, they are all having only the non-zero values. Therefore the points are not lie on the x-axis and the y-axis.

• Consider the points (−3, −4) and (−9, 11) and find the horizontal and the vertical distances between them.

Solution:

For the points (−3, −4) and (−9, 11) The horizontal distance between the two points is a distance between the point corresponding to x coordinates−3 and −9 on the number line x- axis; i.e., (−3) − (−9) = 9 − 3 = 6. and the vertical distances between the two points is the distance between the points corresponding to y co-ordinates −4 and 11 on the number line y axis; i.e., (11) − (−4) = 15.

some more problems to explain "does geometry use integers"

• Consider the line passing through (5,6) and (15,9) and state whether the line is rising up or falling down find the slope.

Solution:

Take (5,6) as (x₁, y₁) and (15, 9) as (x₂, y₂). Then the slope of the line is

m = y₂-y₁ / x₂-x₁

  = `(9-6)/(15-5)`

  = `3/10`

The slope is a positive(+ve) number and so the line is rising up. Here the geometry is used to determine(find) the direction of the lines using the integers.

• Consider the line passing through (−16, 29) and (40, −6) and state whether the line is rising up or falling down and find the slope.
  

Solution:

The slope of the line is

m = y₂-y₁ / x₂-x₁

  = `(-6-29)/(40-(-16))`

  = `-35/56` 

  = `-5/8`

Here m is a negative number, the line will be falling down

In this problem the geometry is used to determine the direction of the lines using the integers.

Isosceles Triangle

Isosceles triangle is a three sided closed polygon with two equal sides and two equal angles.

Propeties of an isosceles triangle:

The legs are equal in the length and the third side is the base of the triangle.
The angles formed in the base side are equal in measures.i.e. they they are congruent to each other.
The angle formed by two legs of the triangle is vertex angle.

geometry isoscles triangle

Formula related related to isosceles triangle:

1. Formula to find the vertex angle of the isosceles triangle is

Sum of the angles in triangle is 180 degrees.

Here the base angle is equal.

a+a+ vertex angle= 180

Vertex angle = 180-2a.

2.Formula for the area of the isosceles triangle is `1/2 (BH)` square.units.

3.Formula for the perimeter of the isosceles triangle is sum of three sides.

Model problems to isosceles triangle

Pro 1: Find the vertex angle of the triangle whose base angle is 70 degrees?

Sol: Step 1:   Let the base angle be 70 degrees

We know that the base angles in the isosceles traingle is equal

Step 2:  Sum of the angles in triangle is 180 degrees.

Angle 1+ Angle 2+ Angle 3 = 180 degrees

x + x + vertex angle = 180

Step 3:   Vertex angle= 180- (140)

Vertex angle is 40 degrees

Vertex angle of the geometry isosceles triangle is 40 degrees.

Pro 2. Find the area of the isosceles triangle whose base side is 10cm and the height of the triangle is 3 cm.

Sol:  Step 1:   Base side, B= 10cm

Height of the triangle, H= 3cm

Step 2:  Formula for the area of the isosceles triangle is 1/2 (BH) square.units.

= `(1/2) (10 xx3)`

= `(30/2)`

= 15cm2

Area of the isosceles triangle = 15cm2

Pro 3: Find the perimeter of the isosceles triangle whose base side is 10cm and leg side is 5cm?

Sol: Step 1: We know that leg sides are equal in length.

Perimeter of the isosceles triangle  = (10+5+5)

= 20 cm

Perimeter of the isosceles triangle is 20cm

Geometry Reflections

In geometric, the determination of reflection of a point in a line translation is performed only through a given point or line. In reflection of a point in a line, the basic ideas of reflection are the transformation of a point to a reflected point that is the equal length of the opposite side of a  line. The transformation of  reflections in two right angular axes produce a rotation of straight angle (180°), that is a half turn.

Reflection of a point:

Reflection:

The reflection is the "flip or mirror image" of it over a line.  So, the actual and mirror images are identical to each other.

The two very common reflections of polygons are given by

Horizontal reflections.

Vertical reflections.

The colored vertices used for each of the triangle.  The line of reflection is halfway from both red points, blue points, and green points. The line of reflection is directly in the center of both points.

Reflection of  a point in a line to describe the X-axis is consider the mirror line or axis of reflection. Therefore, we change the given geometric object point  into x = x  and y = -y.

geometry reflections

Reflection of  a point in a line to describe the Y-axis is consider the mirror line or axis of reflection.. Therefore, we change the given geometric object point  into x = -x  and y = y.

geometry reflections

Reflection Example:

We take a shape of triangle, Learn horizontal reflection assists, the transformation of the geometric triangle of co-ordinates are P, Q, and R. It is transformed by using the horizontal reflection, that is the horizontal line y = x.

Solution:

Step 1: To draw the triangle object from the given co-ordinates are P, Q, and R in the X-Y plane.

geometry reflections

Step 2:  To draw the horizontal line or mirror line of that triangle PQR, and draw the perpendicular line to the horizontal line or mirror line from each co-ordinates of  given triangular object. And then measure the distance of each perpendicular lines are y1, y2, y3.

geometry reflections

Step 3:  To plot the mirrored points are P', Q', R', the same vertical distance from the horizontal line or mirror line.

geometry reflections

Step 4:  To join the plotted points P', Q', R'. We get the horizontally reflected triangular object.

geometry reflections

Step 5:   This is required horizontal reflections of triangular object.

Horizontal reflection of X- axis:

Horizontal reflection assist that concepts to describe the x-axis are considering the mirror line. Then change all the coordinates of the given geometric object into the x = x and y = -y. For example,

geometry reflections