Solving Geometry Triangle Problems

Introduction of solving geometry triangle problems:

A triangle is one of the basic shapes of geometry: a polygon with three corners or vertices and three sides or edges which are line segments. A triangle with vertices A, B, and C is denoted ∆ABC.

In Euclidean geometry any three non-collinear points determine an unique triangle and an unique plane (i.e. a two-dimensional Euclidean space).

Learning Angles for Solving Geometry Triangle Problems:

A geometry triangle is a form of polygon having three sides and, therefore, three angles. The geometry triangle is a blocked figure formed from three straight line segments joined at their ends. In geometry triangle the point at the tops can be called the corners, angles, or vertices of the triangle. Since any given triangle lies totally within a plane, triangles are often treated as two-dimensional geometric figures.

Problems for Solving Geometry Triangle:

solving geometry triangle problems is an essential one. problems for solving geometry triangle are given below:

1) An isosceles triangle has angle A 60 degrees greater than angle B. Find all angles of the geometry triangle.

Sol:

An isosceles triangle has the two angles equal in size. Angle A is 60 greater than angle B then A = B + 60 o. The sum of all angles in a triangle is equal to 180o.

(B+60) + B + B = 180o

•    Solve the above equation for B.

B = 40o

•    The sizes of the three angles are

A = B + 60 = 100o

C = B = 40o

2. The apex angle B of isosceles triangle ABC is 120 degrees. Find the quantity determines of each base angle on geometry triangle.Please express your views of this topic help online with math by commenting on blog.

Sol:

•    Let x = the compute of each base angle.

•    Arrangement and equation and solve for x.

•    base angle + base angle + 120 degrees = 180 degrees

x + x + 120 degrees = 180 degrees

2x + 120 = 180

2x = 180 - 120

2x = 60

x = 30 degrees

3. A geometry triangle has a perimeter of 56. If 2 of its sides are equal and the third side is 5 more than the equal sides, what is the length of the third side?

Sol:

•    Let y = length of the equal side

•    Perimeter = sum of three sides.

•    Plug in the values from the question.

56 = y + y + y + 5

•    Combine like terms

56 = 3y + 5

3y = 56 – 5

3y = 51

y =17

So, the length of third side = 17 + 5 =22

Ellipse Parabola Hyperbola

Introduction to ellipse parabola hyperbola
The equation like Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 represents either a (non–degenerate) conic or a degenerate conic.

If it is a conic, then it is

a parabola if B2 - 4AC = 0.
an ellipse if B2 - 4AC < 0
a hyperbola if B2 - 4AC > 0.

Introduction to Ellipse Hyperbola Parabola

Ellipse:

An ellipse is a conic obtained on slicing across obliquely one nappe of a cone.If P is any point on the ellipse and
F1and F2 its foci, the angle subtended by F1P and F2P with the tangent at P are equal and if a source of light or sound is
placed at one focus of an ellipsoidal reflector (surface generated by revolving an ellipse about its major axis) all the
waves will be reflected so as to pass through the other focus



Hyperbola:

A hyperbola is a conic obtained on slicing a double napped cone by a plane parallel to the axis of the cone. The lines from the foci to any point of a hyperbola make equal angles with the tangent at that point. Hence if the surface of a reflector is generated by revolving a hyperbola about its transverse axis, all rays of light converging on one focus are reflected to the other.



Parabola:

A parabola is generally defined as the conic section obtained by the operation on slicing a right circular cone by a plane parallel to the line joining vertex and any other point of the cone.Between, if you have problem on these topics Formula for Volume of a Trapezoidal Prism, please browse expert math related websites for more help on 12th cbse sample papers.

If P is any point on the parabola with focus F and vertex V, the angles subtended by FP and PX with the tangent at P are equal where PX is parallel to the axis VFA of the parabola.




Problems Based on Ellipse Hyperbola Parabola



If e < 1, then the conic is an ellipse.  From the figure 1we observe that F2Pi is always less than PiMi.
i.e., F2Pi / PiMi =  e < 1, (i = 1, 2, 3…)

If e = 1, then the conic is a parabola. From the figure 2  we observe that FPi is always equal to PiMi.
i.e., FPi / PiMi = e = 1, (i = 1, 2, 3…)

If e > 1, then the conic is a hyperbola. From the figure 3 we observe that F1Pi is always greater than PiMi.
i.e., F1Pi / PiMi = e > 1, (i = 1, 2, 3…)

Arc Length and Sector Area

Introduction to arc length and sector area:

Consider a circle whose center is O and radius r. Let A and B be two points on the circle in such a way that arc AB = r.

The length of the arc AB of the circle is equal to the product of the radius of the circle and angle in radians subtended at the center of the circle.

That is, arc length AB = r × ?, here ? is in radians.

The area of the sector AOB is `(1/2) r^2theta` .

We can also write the above two formula as follows:

(i) arc length AB = 2?r × ` theta/360`

(ii) The area of the sector AOB is `(theta/360)` `Pir^2` .



Now let us see few problems on this topic “arc length and area sector”.

Example1 on Arc Length and Area Sector

An arc of a circle subtends 150 at the center. If the radius of the circle is 4cms, find the length of the arc and area of the sector formed.

Sol: Here ? = 15^0, r = 4cms.

Let us convert the degree value of ? to radians.

? = 15^0 = `(( Pi xx 15)/180)`

= `Pi/12` .

The length of the arc is given by r × ? = 4 × `pi/12`

= `pi/3` cm.

The area of the sector formed = `(1/2)`` r^2` ?

= `(1/2)` ` 4^2` `(pi/12)`

= `(2 pi )/3 ` sq cm.I like to share this solve linear system of equations with you all through my article.

Example2 on Arc Length and Area Sector

A wire of length 10cm is bent so as to form an arc of a circle of radius 4cm. What is the angle at the center in radians?

Sol: Given: The length of the arc is given by r × ? = 10

? = `10/r`

? = `10/4` = 2.5 radians.

Example3 on Arc Length and Area Sector

Ex 3: If the distance between the sun and the earth is 1.495 × 1068 km and the angle subtended by the sun at a point on the earth is half a degree, find approximately the diameter of the sun.

Sol: Given: ? = (1/2)^0 = (`Pi/180` ) × `(1/2)`

= `Pi/360` rad.

If d km be the diameter of the sun, then we have: `(s/r)` = ?

` (d/ (1.495 xx 1068))` = `Pi/360`

Therefore, d = (1.495 × 1068) × `Pi /360`

= 1.304637 × 10^6 km.

Fraction Math Worksheets

Introduction on fraction math worksheet:

In Math, Fraction is a part of an object.

Fraction Math Worksheets:

Worksheet contains questions. So the fraction worksheet contains only questions related to fractions.

Worksheet 1:

Let us see equivalence fraction and the simplifying fractions in this worksheet no 1.

1) `(1)/(4)` = `(?)/(12)`

2)`(3)/(18)` = `(1)/(?)`

3)`(2)/(3)` = `(?)/(6)`

4) `(6)/(12)` = `(1)/(?)`

5) `(12)/(15)` = `(?)/(5)`

Answer:

1) 3

2) 6

3) 4

4) 2

5) 4

Fraction Math Worksheet 2:

Worksheet 2 contains the questions on comparing fractions.

1) Which is greater 5/2 or 4/2?

2) Which is smaller 12/2 or 14/7?

3) Which is greater 32/43 or 13/12?

4) Which is smaller 9/2 or 5/3?

5) Which is greater 4/1 or 3/4?

6)Which is smaller 23/12 or 14/15?

7) Which is greater 1/2 or 1/3?

8) Which is smaller 4/2 or 4/3?

9) Which is greater 4/5 or 4/6?

10) Which is smaller 3/2 or 3/4?

Answer:

1) 5/2

2) 14/7

3) 13/12

4) 5/3

5) 4/1

6) 14/15

7) 1/2

8) 4/3

9) 4/5

10) 3/4

Please express your views of this topic free math solver by commenting on blog.

Fraction Math Worksheet 3:

Worksheet 3 contains the questions on addition,subtraction,multiplication and division of fractions:

1) Add 1/2 and 1/2.

2)Add 3/2 and 3/4.

3) Add 4/5 and 1/5.

4)subtract 1/2 and 1/2.

5)subtract 3/2 and 3/4.

6)subtract 4/5 and 1/5.

7)multiply 1/2 and 1/2.

8)multiply 3/2 and 3/4.

9)multiply 4/5 and 1/5.

10)divide 4/5 by 1/5.

Answers:

1) 1

2) 9/4

3)1

4) 0

5) 3/4

6)3/5

7)1/4

8)9/8

9)4/25

10) 4

These are the fraction math worksheets.

Modified Box Plots

Introduction to box plots:

In statistics a box plot can be defined as a diagram explains the summary of data sets. For easy understanding the plots are modified in such a manner that a common man can examine the box plots. Generally creation of box plots involves in finding out the parameters such as median, quartiles, greatest and least observations.

Modified box plots explains the step by step procedure for creating box plots and with the help of it we can evaluate where the observations are concentrated and where they are spread out. As modified box plots are simple to understand they are widely used by many persons in different profession.

Examining Box Plots

Examining box plots is nothing but a process of reading the elements of the plot.

Actually, there are several parameters are available. Namely,

Median
Quartiles
Minimum and maximum values


The following figure shows a typical modified box plots.

Modified box plots



Median:

The median is the value that divides a data set into two parts and it is denoted as Q2.

Quartiles:

As said earlier the median divides the data set into two parts, the first part is known as Lower quartile and later is called as upper quartile.

These quartiles are considered as separate data sets and the median is evaluated.

Lower median is denoted as Q1 whereas upper median is denoted as Q3.

Maximum & Minimum values:

The greatest and the least values of the data set are called as maximum and minimum values respectively.

Outliers:

Upper outlier is the value which is more than 3/2 times of upper quartile and Lower quartile is the value which is less than 3/2 times of lower quartile.Is this topic Greatest Common Factor Word Problems hard for you? Watch out for my coming posts.

Example

Consider the following data set.

Data set => (12, 4, 9, 66, 8, 53, 37)

Solution:

Step 1:

Rearrange the data set in order.

(4, 8, 9, 12, 37, 53, 66)

As there are 7 elements in the data set. The fourth element lies exactly in the center.

So, median Q2 = 12

Step 2:

Quartiles:

Lower quartile = (4, 8, 9)

Median of lower quartile Q1= 8

Upper quartile = (37, 53, 66)

Median of the upper quartile Q3= 53

Step 3:

Minimum value= 4

Maximum value= 66

So, min=4, Q1=8, Q2=12, Q3=53, max=66

Step 4:

Draw a straight line which extends in both directions.

Step 5:

Draw the modified box plots using the parameters.

Modified box plot

Solving Basic Statistics Problems

Introduction to solving basic statistics problems:   

Everyday we come across a wide variety of information’s in the form of facts, numerical figures, tables, graphs, etc. These figures are the numerical data collected with a definite purpose, are called data. Data is a word in a plural form of the Latin word datum. Every part of our lives utilizes data in one form or the other. This extraction of significant information is studied in a branch of mathematics called Statistics.

Solving Basic Statistics Problems-statistics Definition

Basically the word ‘statistics’ appears to have been taken from the Latin word ‘status’ meaning ‘a state (related to a political)’. In its origin, statistics is simply the collection of data on different aspects of the life of people, useful to the State. Basic Statistics deals with collection, organization, analysis and Interpretation of data. The word ‘statistics’ has held different meanings in different contexts, so that the calculation process is based on the meaning and data’s.
Between, if you have problem on these topics how to long divide with decimals, please browse expert math related websites for more help on need help with math homework.
Solving Basic Statistics Problems

Example for basic stastistics problems:

Consider the marks obtained by 10 pupils in a mathematics test as given below:

55 36 95 73 60 42 25 78 75 62

The data in this form is called raw data.

By looking in this way, can you find the highest and the lowest marks?

Did it take you some time to search for the maximum and minimum marks? Wouldn’t it be a lesser amount of time consuming if these scores were arranged in ascending or descending order? So let us arrange the marks in ascending order as

25 36 42 55 60 62 73 75 78 95

Now, we can clearly watch that the lowest marks are 25 and the highest marks are 95.

The difference of the highest and the lowest marks in the basic stastistics data is called the range of the data. So, the range in this case is 95 – 25 = 70.

Arrangement of data in ascending or descending order can be quite time consuming, particularly when the number of observations in an experiment is large.

Variance of a Data Set

Introduction to variance of a data set:

Data is nothing but a set of information or collection of facts. We can analyze the data’s with the help of charts and graphs. The method of collecting or preparing the data’s is the data collection.
Variance is the term that  explains how average value of the data set vary from the measured data.

Variance of Data Set:

Following are the steps involved in finding the variance of data set:    

Step 1: First step is to find the Average, that is the arithmetic mean for the given set of numbers.

Mean = Sum of all values / Total number of elements

Step 2: To find the sum of squares for all the given values in the given data set.

Step 3: Divide the result we obtained in Step 2 over the total number of elements in the given data set.

Step 4: Find the difference between the square of mean and the result we obtained from step 3

Step 5: For finding the variance, take the square root of the number that  we obtained as a result from step 4.

Thus, using the above steps, we can easily calculate the variance of the given data set.

Between, if you have problem on these topics Absolute Value Equations and Inequalities, please browse expert math related websites for more help on What is the Largest Prime Number.

Solved Examples for Finding Variance of a Data Set:

Ex1: Find the variance  of the given data set: { 5, 6, 9, 13, 19 }

Sol:

Step 1: Arithmetic mean             =  (5 +6 + 9+ 13 + 22) / 5

= 55 / 5

=  11

Step 2: Sum of squares               =  52 + 62 + 92 + 132 + 222

=  25 + 36 + 81 + 169 + 484

= 795

Step 3: Sum of squares / Total number of elements

= 795 / 5

= 159

Step 4: Result from Step 3 – square of mean value

= 159 – 112       

= 159 – 121

= 38

Step 5: Variance                        = square root ( 38 )

= 6.16

The variance of the given data set is 6.16

Ex 2: Find the variance  of the given data set: { 2, 3, 4, 5,16 }

Sol:

Step 1: Arithmetic mean             =  (2 +3 + 4+ 5 + 16) / 5

= 30 / 5

=  6

Step 2: Sum of squares               =  22 + 32 + 42 + 52 + 162

=  4 + 9 + 16 + 25 + 256

= 310

Step 3: Sum of squares / Total number of elements

= 310 / 5

= 62

Step 4: Result from Step 3 – square of mean value

= 62 – 62       

= 62 – 36

= 26

Step 5: Variance                        = square root ( 26 )

= 5.1

The variance of the given data set is 5.1