Euclidean Geometry

Geo means “earth” and metron  means “measurement”. ”Euclid, a distinguished Greek mathematician, called the father of geometry. A point is used to represent a position in space. A plane to be a surface extending infinitely in every directions such that all points lying on the line joining any two points on the surface. The euclidean geometry example problems and practice problems are given below.
Example Problems - Euclidean Geometry:

Example problem 1:

Can the following angles are triangle or not?

(a) 50° , 70°, 60° (b) 40°, 80°, 70°

Solution :
(a) The sum of the measure of the three angles is
50° + 70° + 60° = 180°
Therefore 50°, 70°, 60° can be the measure of the angles of a triangle.
(b) The sum of the measure of the three angles is
40° + 80° + 70° = 190°.
But the sum of the measure of the angles of a triangle is 180°.
Therefore 40°, 80°, 70° cannot be the measures of the angles of a triangle.

Example problem 2:

Find the supplementary angle of 95°

Solution:

The supplementary angle of 95° = 180° – 95° = 85°.

Example problem 3:

The lengths of two sides of right triangles are 6cm and 8cm. Find its hypotenuses.

Solution:

AC = 6 cm
BC = 8 cm
AB =?
AB2 = 62+ 82
= 36 + 64
AB2 = 100
AB = √100 = 10

Thus, the hypotenuses are 10 cms in length.
Practice Problems - Euclidean Geometry:

Practice problem 1:

Two angles of a triangle are of measures 52°and 78°. Find the measure of the third angle.

Ans: 50

Practice problem 2:

Find the angles in each of the following:
(i) The angles are supplementary and the larger angles are twice the small.
(ii) The angles are complementary and the larger is 20° more than the other

Ans:    1. smaller angle = 60°, larger angle = 120°.

2. smaller angle = 35°, larger angle =  55°.

Geometry Parallelogram Proofs

Geometry proofs are basic laws of geometry figures, using these geometry proofs the geometry problems are solved. The Geometry parallelogram proofs are the proofs for parallelogram that help the rules of solving parallelogram based problems. A parallelogram proofs form an argument that establishes the truth of the parallelogram figure. Let us see basic geometry parallelogram proofs.

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Geometry Parallelogram Proofs:

Properties of parallelogram:

We know that a quadrilateral is a closed plane figure formed by four line segments and a parallelogram is a quadrilateral in which the opposite sides are parallel to each other.

Proof in geometry 1: In a parallelogram, the opposite sides are of equal length.

Parallelogram proof

Proof: Let ABCD be a parallelogram. Link up BD. take the triangles ABD and BDC.

Since AB || CD and BD is a transversal of AB and CD, m∠ABD = m∠BDC

Since AD|| BC and BD is a transversal of AD and BC, m∠ADB = m∠DBC

The side BD is common to both ABD and BDC.

Hence, by AAS property, ΔABD ≡ ΔBDC.

∴ The particular sides are equal.

AD = BC and AB = CD

Proof in geometry 2: In a parallelogram, the opposite angles are of equal measure.
Parallelogram proof
Proof: Let ABCD be the parallelogram. Link up BD.

Since AB || DC and BD is a transversal to AB and DC,
m∠ABD = m∠BDC
Since AD || BC and BD is a transversal
to AD and BC,
m∠ADB =m∠CBD
∴ m∠ABC = m∠ABD + m∠DBC
= m∠BDC+ m∠ADB
= m∠ADC
Similarly, m∠BAD=m∠BCD
Geometry Parallelogram Proofs:

Proof in geometry 3: The diagonals of a parallelogram bisect each other.
Parallelogram proof
Proof: ABCD is a parallelogram. AC and BD form diagonals.
By ASA criterion, ΔAMB ≡ CMD
∴AM = CM, BM = DM.
∴This shows diagonals bisect each other.


Proof in geometry 4: If the opposite sides of a quadrilateral are of equal length, then the quadrilateral is a parallelogram.
Parallelogram proof
Proof: Let ABCD be a quadrilateral where AD = BC,AB = CD. Link up AC.

Consider the triangles from the figure, ACB and ADC. By SSS criterion, ΔABC ≡ ΔCDA

Then m∠ BAC = m∠ACD,m∠CAD=m∠ACB
∴ AB || CD and AD || BC. Hence ABCD form a parallelogram.

Quadratic Equation x Intercept

In coordinate geometry, the x intercept means the x-value of the point where the graph of a function or relation intercepts the x-axis of the coordinate system. It means the point at which the line cuts the x-axis.

The x intercept of a line is denoted as (x, 0).

In this article, we are going to learn finding x intercept for the quadratic equation.


Example Problems to Find X Intercept for Quadratic Equation:

Example 1:

Find x intercept for the quadratic equation y = x2 + x - 12

Solution:

Step 1: Given function

y = x2 + x - 12

Step 2: Plug y = 0 in the given function

0 = x2 + x - 12

We can write it as

x2 + x - 12 = 0

Step 3: Solve the above function for x to find x intercept

Find sum of two numbers and its product for x and - 12x2

x = 4x - 3x

-12x2 = (4x)(-3x)

Replace the term x by 4x -3x

x2 + 4x - 3x - 12 = 0

Take the common factor outside,

x(x + 4) - 3(x + 4) = 0

(x + 4)(x - 3) = 0

Factor each term,

x = - 4   and x = 3

Step 4: Solution

X intercepts for the given equation are (- 4, 0) and (3, 0)

Example 2:

Find x intercept for the quadratic equation y = 2x2 + 10x + 8

Solution:

Step 1: Given function

y = 2x2 + 10x + 8

Step 2: Plug y = 0 in the given function

0 = 2x2 + 10x + 8

We can write it as

2x2 + 10x + 8 = 0

Step 3: Solve the above function for x to find x intercept

Find sum of two numbers and its product for 10x and 16x2

10x = 8x + 2x

16x2 = (8x)(2x)

Replace the term 10x by 8x + 2x

2x2 + 8x + 2x + 8 = 0

Take the common factor outside,

2x(x + 4) + 2(x + 4) = 0

(2x + 2)(x + 4) = 0

Factor each term,

x = - 1   and x = - 4

Step 4: Solution

X intercepts for the given equation are (- 1, 0) and (- 4, 0)




Practice Problems to Find X Intercept for Quadratic Equation:

1) Find x intercepts for the quadratic equation y = x2 + 18x + 32

2) Find x intercepts for the quadratic equation y =2 x2 + 14x + 12

2) Find x intercepts for the quadratic equation y = 3x2 + 6x - 8

Solutions:

1) x intercepts are (-2, 0) and (-16, 0)

2) x intercepts are (-1, 0) and (-6, 0)

3) x intercepts are (0.914, 0) and (-2.914, 0)

Histogram Unequal Intervals

The Histogram Definition unequal intervals of applications are mostly used in statistics. The histogram unequal intervals are most effective way  to analysis the frequency distribution of grouped data. The concepts of histogram unequal intervals to show the quick way of understanding the grouped data in the graphical representations. The analyzing histograms diagrams are constructed from the two ways are depending on the uniform width, and varying widths.
Constructions for Histogram Unequal Intervals:

To analyzing  the histogram unequal intervals , we draw two perpendicular axes and choose a suitable scale for each axis. We blot class or group intervals of the continuous grouped or class of data on the horizontal axis and the respective number of  occurrence  of each class on the vertical axis. For each class, a rectangle is constructed with class interval as the base and height determined from the class frequency so that areas of the rectangles are proportional to the frequencies.

Let us analyzing histogram unequal intervals through an example, how a grouped frequency distribution of number of teachers in a city can be represented in histogram.


Histogram

From the above histogram unequal intervals diagrams, it is clear that the maximum number of students in the group 50 - 60 and the minimum number of students in the group 70 - 90.

Notes: In the diagram  (kink) before the class interval 30 - 40 on the horizontal axis. It shows that the full distance 0 - 30 is not shown.
Example of Histogram Unequal Intervals:

Another example of histogram unequal intervals for a math tutor wanted to analyse the performance of two class of students in a math test of 100 marks. Looking at their performances,  math tutor found that a few students got under 20 marks and a few got 70 marks or above. So the math tutor decided to analyzing histogram unequal intervals to group them into intervals of varying sizes as follows: 0 - 20, 20-30, ...., 60 - 70, 70 - 100. Then she formed the following table.
Marks                          Number of students
0 - 20                                         9
20 - 30                                       14
30 - 40                                       14
40 - 50                                       18
50 - 60                                       18
60 - 70                                       12
70 - above                                  4

Negative Fraction Calculator

Negative fraction is nothing but, the given input fractions are in negative values, and using negative fraction calculator will produce the output with in a second. Numerator and denominator to created the fraction, the numerator representing a number of equal parts and the denominator telling how many of those parts make up a whole.


Example is 4/5, the numerator value, 4, tells us that the fraction represents 4 equal parts, and the denominator, 5, tells us that 5 parts make up a total. An afterward growth be the common or "vulgar" fractions which are still used today like this (½, ⅝, ¾, etc.).
Examples Problems for Using Negative Fraction Calculator:

1. Subtract the fraction of -5/6 - 8/6, using negative fraction calculator.

Solution:

First enter the two fractions values in the negative fraction calculator box.
And we click the enter the equal to button in calculator.
Then we get the answer with in a second like this,

fraction calculator

Reduce button used to reduce the exact value in calculator.

Steps to calculate the negative fraction manually,

These two fractions have the same denominators, so we can directly subtract the numerators alone by keeping the denominator as it is.

Given value is,

= -5/5 - 2/6

= (-5 - 2) / 6

We get the answer is,

= -7 / 6.

2. Subtract the fraction of -9/2 - 8/2, using negative fraction calculator.

Solution:

First enter the two fractions values in the negative fraction calculator box.
And we click enter the equal to button in calculator.
Then we get the answer with in a second like this,

fraction calculator

Reduce button used to reduce the exact value in calculator.

Steps to calculate the negative fraction manually,

These two fractions have the same denominators, so we can directly subtract the numerators alone by keeping the denominator as it is.

Given value is,

= -9/2 - 8/2

= (-9 - 8) / 2

We get the answer is,

= -17 / 2.

More Examples Problems for Using Negative Fraction Calculator:

3. Subtract the fraction of -2/7 - 4/7, using negative fraction calculator.

Solution:

First enter the two fractions values in the negative fraction calculator box.
And we click enter the equal to button in calculator.
Then we get the answer with in a second like this,

fraction calculator

Reduce button used to reduce the exact value in calculator.

Steps to calculate the negative fraction manually,

These two fractions have the same denominators, so we can directly subtract the numerators alone by keeping the denominator as it is.

Given value is,

= -2/7 - 4/7

= (-2- 4) / 7

We get the answer is,

= -7 / 6.

4. Subtract the fraction of -6/3 - 5/3, using negative fraction calculator.

Solution:

First enter the two fractions values in the negative fraction calculator box.
And we click enter the equal to button in calculator.
Then we get the answer with in a second like this,



fraction calculator

Reduce button used to reduce the exact value in calculator.

Steps to calculate the negative fraction manually,

These two fractions have the same denominators, so we can directly subtract the numerators alone by keeping the denominator as it is.

Given value is,

= -6/3 - 5/3

= (-6 - 5) / 3

We get the answer is,

= -11 / 3.

Linear Functions and Matrices

Definition of Matrices:

A matrix is a group of numbers arranged into a fixed quantity of rows and columns. Usually a matrix is a two dimensional range of numbers or terms arranged in a set of rows and columns. m × n matrices A have m rows and n columns and it is written,

A = `[[a_(11),a_(12),a_(1n)],[a_(21),a_(22),a_(2n)],[a_(m1),a_(m2),a_(mn)]]`

Definition of Linear Functions:

A polynomial functions of single degree is defined as a linear functions. It relates a dependent variable with an independent variable in a simple way. Mathematical equation in which there is no independent-variable is raised to a power greater than one. A simple linear function with one independent variable traces a straight line when plotted on a graph. It is also called as linear equation.
Solving Linear Functions Using Matrices:

Steps to solve the system of linear equations using matrices,

Step 1: Let us consider the equation,

a1x + b1y = c1,

a2x + b2y = c2,

Step 2: If, A = `[[a_1,b_1],[a_2,b_2]]` , X = `[[x],[y]]` , and C = `[[c_1],[c_2]]` .

then, AX = C.

Step 3: Now multiply both sides by, A−1

=> A−1AX = A−1C

Step 4: We know that, A−1A = I, where I is identity matrices,

=> IX = A−1C.

Step 5: Now, Multiplication of any matrices with identity matrices gives the same matrices. So,

Hence, X = A−1C.
Example for Solving Linear Functions Using Matrices:

Example: 5x – y + 11 = 0 and x – y – 5 = 0, solve the linear equations for x and y using matrices.

Solution:           5x – y + 11 = 0     => 5x – y = – 11,

x – y – 5 = 0         => x – y = 5.

Let, A =` [[5,-1],[1,-1]]` , X = `[[x],[y]]` , and C = `[[-11],[5]]`

To find the value of A−1:

A = ` [[5,-1],[1,-1]]`

Swap the leading diagonal values,

=> ` [[-1,-1],[1,5]]`

Change the signs of the other two elements in the matrices,

=>` [[-1,1],[-1,5]]`

Now, find the value of |A|.

|A| = (–1 × 5) – ((–1) × 1)

=> |A| = –5 + 1

=> |A| = –4

Hence, A−1 = `1/(|A|) xx [[-1,1],[-1,5]]`

=> A−1 = `1/(-4) xx [[-1,1],[-1,5]]`

=> A−1 =` [[(-1)/(-4),1/(-4)],[(-1)/(-4),5/(-4)]]`

=> A−1 =` [[1/4,(-1)/4],[1/4,(-5)/4]]`

Therefore the solution is,

X = A–1C,

=> X =` [[1/4,(-1)/4],[1/4,(-5)/4]] xx [[-11],[5]]`

=> X =`[[(-5)/(15) -(10)/(15)],[(10)/5 - 5/5]]`

=> X = `[[(-15)/(15)],[5/5]]`

=> X =` [[-1],[1]] = [[x],[y]]`

Hence the solutions are, x = – 1, y = 1.

Solve Herons Formula

We are familiar with the general formula for finding the area of a triangle when the base and height of the triangle is given. For scalene traingle we do not have any area formula. Heron's formula proof is a general formuls used to find area of triangles of all types.

Heron a famous mathematician gave simple formula for finding the area of any triangle based on its three sides. Thus, this formula for area is known as Heron's formula and is stated as below,

If p, q, r, denote the lengths of sides of a `Delta` PQR, then,

Heron's formula of a triangle

Area of `Delta` PQR = `sqrt(s(s-p)(s-q)(s-r))` , where   s = `(p+q+r)/(2) = ("perimeter of triangle")/(2)`

Here, s = semiperimeter of `Delta` PQR.

Heron's formula can used to find area of all types of triangles, quadrilateral, trapezoid, etc.
Solved Examples Using Heron's Formula

Ex 1: Find the area of a triangle whose sides are respectively 150 cm, 120 cm, and 200 cm.

Sol: Given three sides, p = 150 cm, q = 120 cm and r = 200 cm.

Step 1:  s = `(p + q + r)/(2) = (150 + 120 + 200)/(2)`

= `(470)/(2) = 235`

Step 2:  (s-p) = (235 -150) = 85

(s-q) = (235 - 120) = 115

(s -r ) = (235 - 200) = 35

Step 3: Area of triangle = `sqrt(s(s-p)(s-q)(s-r))`

= `sqrt(235xx 85xx 115xx 35)`

= `sqrt(5xx 47 xx 5xx 17 xx 5 xx 23 xx 5 xx 7)`

=  25`sqrt(47 xx 17 xx 23 xx 7)`

= 8966.56 cm2

Ex 2: Find the area of a triangle whose two sides are 18 cm and 10 cm and the perimeter is 42 cm.

Sol:  Given p = 18 cm, q = 10 cm and perimeter = 42 cm

Step 1:  perimeter = p + q + r

42 = 18 + 10 + r

42 = 28 + r

subtract 28 on both sides

42 - 28 = 28 - 28 + r

14 = r

Step 2 :  s = `("perimeter)/(2) = (42)/(2)`

s = 21

Step 3:  (s - p) = 21 - 18 = 3

(s - q) = 21 - 10 = 11

(s - r)  = 21 - 14 = 7

Step 4:  Area of triangle = `sqrt(s(s-p)(s-q)(s-r))`

= `sqrt ( 21 xx 3 xx 11 xx 7)`

= `sqrt( 3 xx 7 xx 3 xx 7 xx 11)`

= 3 x 7 `sqrt(11)`

= 21`sqrt(11)` cm2
Practice Problems on Heron's Formula

Pro 1: Find the area of a triangle whose sides are respectively 100 cm, 80 cm, and 60 cm

Ans: 2400 cm2

Pro 2: Find the area of a triangle whose two sides are 12 cm and 8 cm and the perimeter is 30 cm.

Ans: 39.68 cm2