Descriptive Geometry

The descriptive geometry is a branch of geometry, which deals three dimensional objects in two dimensions by using the lines, curves, solids, surfaces and points in space. Geo means “earth” and metron  means “measurement”. ”Euclid, is a 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 descriptive geometry example problems and practice problems are given below. 

Example problems for descriptive geometry:

Example problem 1:
          Show that the straight lines 2x + y − 9 = 0 and 2x + y − 10 = 0 are parallel.
Solution:
           Slope of the straight line 2x + y − 9 = 0 is m₁ = − 2
           Slope of the straight line 2x + y − 10 = 0 is m₂ = − 2 ∴ m₁ = m₂
           The given straight lines are parallel.

Example problem 2:
      Find the co-ordinates of orthocentre of the triangle formed by the straight lines x − y − 5 = 0, 2x − y − 8 = 0 and 3x − y − 9 = 0
Solution:
      Let the equations of sides AB, BC and CA of a ΔABC be represented by
                x − y − 5 = 0 … (1)
               2x − y − 8 = 0 … (2)
               3x − y − 9 = 0 … (3)
      Solving (1) and (3), we get A as (2, − 3)
       The equation of the straight line BC is 2x − y − 8 = 0. The straight line perpendicular to it is of the form
             x + 2y + k = 0
       A(2, − 3) satisfies the equation (4) ∴ 2 − 6 + k = 0 ⇒ k = 4
       The equation of AD is x + 2y = − 4 … (5)
       Solving the equations (1) and (2), we get B as (3, − 2)
       The straight line perpendicular to 3x − y − 9 = 0 is of the form x + 3y + k = 0
       But B(3, − 2) lies on this straight line ∴ 3 − 6 + k = 0 ⇒ k = 3
       The equation of BE is x + 3y = − 3 … (6)
       Solving (5) and (6), we get the orthocentre O as (− 6, 1).

Example problem 3:
         Find the values of a and b if the equation (a − 4)x² + by² + (b − 3)xy + 4x + 4y − 1 = 0 represents a circle.
Solution:
   The given equation is (a − 4)x² + by² + (b − 3)xy + 4x + 4y − 1 = 0
            (i) coefficient of xy = 0 ⇒ b − 3 = 0 ∴ b = 3
            (ii) coefficient of x² = co-efficient of y² ⇒ a − 4 = b
                        a = 7
                Thus a = 7, b = 3

Practice problems for descriptive geometry:

Practice problem 1:
          If the equation 12x² − 10xy + 2y² + 14x − 5y + c = 0 represents a pair of straight lines, find the value of c. Find the separate equations of the straight lines and also the angle between them.
     Answer: C = 2 ; 6x − 2y + 1 = 0 and 2x − y + 2 = 0 ; tan⁻¹ (1/7)
Practice problem 2:
          Find the equation of the straight line which passes through the given intersection of the straight lines 2x + y = 8 and 3x − 2y + 7 = 0 and is parallel to the straight line 4x + y − 11 = 0
      Answer: 28x + 7y − 74 = 0

Algebra is a subdivision in math, which comprises of infinite number of operations on equations, polynomials, inequalities, radicals, rational numbers, logarithms, etc. Graphing algebra equations or function is also a part of algebra. Graphing is nothing but the pictorial view of the given function or equation, it may be a line, parabola, hyperbola, curve, circle, etc.The most commonly used graphing functions with their graph is given in the following sections.

Common graph functions:

The most commonly used graph functions are,
Linear function which gives a straight line
                f(x) = ax +b
Functions with degree 2
               f(x) = ax² +b
Functions with degree 3,
               f(x) = ax³ + bx
Functions of square root,
              f(x) =` sqrt(x)`

Graph for the common functions:

The graphical view of common functions is shown below,
1. Linear function.

Linear functions are in  the format where the function has degree 1. So that the function varies linearly. The general format for linear functions is shown below,

 f(x) = ax +b

where, x power is 1. b is a constant.

The graph for the linear function is a line and the graph is shown below,

2. Function with degree 2,

The function for degree 2 is in the format,

f(x) = ax² . where, a is a constant

The graph for this function looks as shown below,

3. Function with degree 3 is as follows,

f(x) = ax³

where, a is a constant.

The graph looks as shown below,

4. Function with degree 4 is in th format,

f(x) = ax⁴

where, a is a constant.
The graph is as follows,

5. The function with square root is as follows,

f(x) = sqrt (x)

The graph is as follows,

The above shown functions are functions of x, if it is changed to function of y, then the graph changes to the y axis.

Non Linear Data

Non linear equation solver involves solving non linear equation with the help of formulas and it is mainly used to find the unknown variable value. The quadratic equation is one of the non linear equation in which it transforms the polynomial function into normal linear function. Non linear equation is also related with the families of vectors called vector spaces or non linear spaces. The following are the example problem deals with non linear equation solved with detailed solution.

Non linear equation solver example problems:

Example 1:

Solve the non linear quadratic equation.

√ (x ² – 11x+28) = 2

Solution:

Given function is

√ (x ² – 11x+28) = 2

Squaring on both sides, then the above equation becomes

[√ (x ² – 11x+28)] ² = (2) ²

And simplify.

x ² – 11x+28= 4

Make the above equation in factor form.

x ² – 11x + 24 =0

The above equation is in quadratic form with two solutions.

x = 3 and x = 8

Conclusion:

The given equation has two real values as solution x = 3 and x = 8.

Example 2:

Solve the non linear quadratic equation.

         y ² – 4y + 13 = 0
Solution:

Given equation is

y ² – 4y + 13 = 0

The discriminant is given as

D = b² - 4ac

= (-4)2 - 4(1)(13) = -36

Since the discriminant results negative, the square root of the discriminant value is a pure imaginary number.

√(D) = √(-36) = √(-1)√(36) = 6i

where i is the imaginary part defined as i = √(-1).

To find the solution we use b / 2a formula.

y₁ = (4 + 6i)) / (2*1) = 2 + 3i

y₂ = (4 - 6i) / (2*1) = 2 - 3i

Conclusion:

So the given equation has two imaginary solutions 2 + 3i and 2 – 3i.

Non linear equation solver practice problems:

1) Solve the non linear quadratic equation.
                      y ² + 9 = 0
Answer: (y – 3i) (y + 3i) is the solution.
2) Solve the non linear quadratic equation.
                         -y ² + 2y = -3
Answer: (y + 1) (y – 3) is the solution.

Bisector in Geometry

In geometry, bisector is a line dividing something into two equal or congruent parts. The most often considered types of bisectors are the segment bisector (a line that passes through the midpoint of a given segment) and the angle bisector (a line that passes through the apex of an angle, that divides it into two equal angles). (Source: From Wikipedia).

A plane is used as a bisector in bisecting three dimensional shapes. It is called as bisector plane. Bisectors are used to divide line segments, shapes, and solids in to two equal, similar and congruent parts.

In this article we are going to learn, drawing bisector to a line segment.

Drawing bisector to a line segment

Here we are going to learn how to draw bisector of a line segment step by step.

Step 1

Draw a line segment PQ

Step 2

Fix a compass in point P, and expand it over half the length of the line segment.

Step 3

Now, draw two arcs, on each side of the line segment.

Step 4

Without changing or adjusting the measure of the compass, draw two another arcs on each side of the line segment from the point Q.

Step 5

Now we have drawn two set of arcs on each side of the line segment PQ, from points P and Q. And we have two points on either side of the equations formed by the arcs drawn from the points P and Q.

Now join those two points of intersection between the arcs either sides of the line segment by a ruler.


Step 6

The line joined by the points of intersection between arcs is called as the bisector of the line segment PQ. The bisector is perpendicular to the line segment PQ. The point, where the bisector cuts the line segment PQ, is the mid point of the line segment PQ. So, the line segments PJ and JQ are equal in length. PJ = JQ.

Trigonometric Functions of Sum

trigonometric functions of sum and difference of two angles:

The trigonometric functions of sum and difference of two angles A are B are listed below:
Sin (A+B) = Sin (A) Cos (B) + Cos (A) Sin (B) .................. (1)
Sin (A-B) = Sin (A) Cos (B) - Cos (A) Sin (B) .................... (2)
Cos (A+B) = Cos (A) Cos (B) - Sin (A) Sin (B) ................... (3)
Cos (A-B) = Cos (A) Cos (B) + Sin (A) Sin (B) ................... (4)
Adding (1) and (2) we get, 
Sin (A+B) + Sin (A-B) = 2 Sin (A) Cos (B)
Subtracting (1) and (2), we get,
Sin (A+B) - Sin (A-B) = 2 Cos (A) Sin (B)
Similarly, adding (3) and (4), we get,
Cos (A+B) + Cos (A-B) = 2 Cos (A) Cos (B)
Subtracting (3) and (4), we get,
Cos (A+B) - Cos (A-B) = - 2 sin (A) sin (B)
These can be written in the forms
Sin (A+B) + Sin (A-B) = 2 Sin (A) Cos (B) ............................ (5)
Sin (A+B) - Sin (A-B) = 2 Cos (A) Sin (B) ............................. (6)
Cos (A+B) + Cos (A-B) = 2 Cos (A) Cos (B) ........................ (7)
Cos (A - B) - Cos (A+B) = 2 sin (A) sin (B) ........................... (8)
Note: The order on the right-hand side of (8) must be carefully noted
The sum and difference of two angles A and B of a Tangent function are given below:
Tan (A+B) = [Tan (A) + Tan (B)] / [1 - Tan (A) Tan (B)] .................. (9)
Tan (A-B) = [Tan (A) - Tan (B)] / [1 + Tan (A) Tan (B)] ................... (10)

Example for trigonometric functions of sum and difference of two angles

 Express as the sum of two trigonometrical ratios 
sin (5θ) cos (3θ)
Sol : Using the Identity:  2 Sin (A) Cos (B) = Sin (A+B) + Sin (A-B)  
On substitution, we get,
sin (5θ) cos (3θ) = (1/2) {Sin (5θ+3θ) + Sin (5θ-3θ)  
= (1/2) {Sin (8θ) + Sin (2θ) 

More examples for trigonometric functions of sum and difference of two angles

Change in to sum of trigonometric functions: Sin (70^o) Sin (20^o)
Sol: Using,   2 sin (A) sin (B) = Cos (A - B) - Cos (A+B)
Sin (70^o) Sin (20^o) = (1/2) { Cos (70^o - 20^o) - Cos (70^o+20^o) }
                            = (1/2) { Cos (50^o) - Cos (90^o) }
= (1/2) {Cos (50^o) }           Since Cos(90^o) = 0

Square Pyramid

Pyramid is one where the surface of outer area looks triangular and it converges at that point. They are congruent .Square pyramid is a pyramid in which the base of pyramid looks to be square. The most popular form of square pyramid is Johnson square pyramid. If the apex is perpendicularly on top of the center of the square, it will have C4v symmetry.

what is square pyramid

Formulae for square pyramid:

Volume of a square pyramid V = `1/3` (Ah)

= `1/3` a2 h     (Area of square = a2)

Height of square pyramid h = (3 * V) / a2

Where, A -----> area of base of a pyramid

V -----> Volume of a pyramid

a ----->  Side length

what is square pyramid

Surface area of a square pyramid = side2 + 2 x side x length

what is square pyramid

Example problems for square pyramid:

ExampleExample 1:

Find the surface area of a square based pyramid with the given sides 6 cm, height 7 cm, and the slant height 8 cm.

Solution:

Formula:

Surface area of a square pyramid = side2 + 2 x side x length

Given:

Side = 6 cm, length = slant height = 8cm

Surface area             = (6)2 + 2 x 6 x 8

= 36 + 96

= 132

The Surface area of square pyramid is 132 square cm.

ExampleExample 2:

Find the volume of square pyramid for a given dimensions base = 5cm, height = 6cm.

Solution:

Formula:

Volume of the pyramid = `1/3` * Area of base * height.

Given:

Base = 5cm, height = 6cm.

Area of base = b2

Volume = `1/3` * 52 * 6 cm3

= `1/3` * (150) cm3

= 50 cm3

The Final answer is 50 cm3.

ExampleExample 3:

Find the volume of the square pyramid whose base area is 16 cm and height is 12 cm.

Solution:

Formula:

Volume of the pyramid = `1/3` * B*h

Given:

Base area is 16 cm and height is 12 cm.

Volume of the pyramid = `1/3` * 16 * 12

= `1/3` * 192

= `192/3`

= 64 cm3

Practice problems:

Problem 1:

Find the volume of square pyramid for a given dimensions base = 2cm, height =3cm.

Solution:

Volume of a square pyramid is 4cm3.

Problem 2:

Find the volume of the square pyramid whose base area is 15 cm and height is 13 cm.

Solution:

Volume of a square pyramid is 65cm3.

Decimal in Words

In mathematics, a decimal in words is nothing but, it is only represent the decimal numbers. The heart of our number system is considering by place value. Many ways can be represent the decimal numbers. But the important one is place value. The numbers in a base-10 numeral system is specified as decimal notation. A dot with a decimal number, like to present in 41.602. Decimal powers in decimals, (1, 10, 100, and 1000) and secondary symbols for half these values (5, 50, and 500) are contained as roman numerals.

How to write Decimal System in words

Below are examples which will show you how to write decimals in words:

1. Write the decimals in words for given decimal number, 5.32

Solution:

Given the decimal number is 5.32,

The (1.32) decimals in words is write,

Five and thirty two hundredths.

2. Write the decimals in words for given decimal number, 65.44

Solution:

Given the decimal number is 65.44,

The (65.44) decimals in words is write,

Sixty five and forty four hundredths.

3. Write the decimals in words for given decimal number, 24.65

Solution:

Given the decimal number is 24.65,

The (24.65) decimals in words is write,

Twenty four and sixty five hundredths.

4. Write the decimals in words for given decimal number, 36.875

Solution:

Given the decimal number is 36.875,

The (36.875) decimals in words is write,

Thirty six and eight hundred seventy five thousandths.

5. Write the decimals in words for given decimal number, 0.02

Solution:

Given the decimal number is 0.02,

The (0.02) decimals in words is write,

Two hundredths.

Solved Examples

1. Write the decimals in words for given decimal number, 6000.32

Solution:

Given the decimal number is 6000.32,

The (6000.32) decimals in words is write,

Six thousand and thirty two hundredths.

2. Write the decimals in words for given decimal number, 475.00055

Solution:

Given the decimal number is 475.00055,

The (475.00055) decimals in words is write,

Four hundred seventy five and fifty five hundred thousandths.

3. Write the decimals in words for given decimal number, 98.84

Solution:

Given the decimal number is 98.84,

The (98.84) decimals in words is write,

Ninety eight and eighty four hundredths.

4. Write the decimals in words for given decimal number, 637.674.

Solution:

Given the decimal number is 637.674,

The (637.674) decimals in words is write,

Six hundred thirty seven and six hundred seventy four thousandths.

5. Write the decimals in words for given decimal number, 9.69

Solution:

Given the decimal number is 9.69,

The (9.69) decimals in words is write,

Nine and sixty nine hundredths.