How do Parallel Lines Form

When two lines are placed on a plane in same distance without intersection is called as parallel line. The term parallel in math can be represented as ||. For example line XY || to line AB. The statement states that the line XY is parallel to the line AB. The line XY are parallel to the line AB, if they don’t have any common point or center point.

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How do parallel lines form:

How do parallel lines form:

The parallel lines are never cross each other. When two lines are placed on a plane in same distance without intersection is called as parallel line. The following rules are used to form the parallel lines.

Step 1:

The line AB is drew in the paper or board.

Step 2:

The point P is blotted in the space of plane (other than the line).

Step 3

The next line is drowned through the point P. The new line makes an angle to the line AB. Blot the intersection point as Q in the line AB on the plane.

Step 4:

Draw an arc from the point Q. Among the help of compass sketch an arc width as regards half PQ. This arc must cut cross or intersect the line AB and P.

Step 5:

Acquire the same measurement (similar compass width) and blot an arc in the line P from the point P.

Step 6:

Put the compass width to the lower arc.

Step 7:

Shift the compass to the upper arc. Create off an arc to make a point R.

Step 8:

Draw a straight line. The straight lime must travel through the line PR.

Now the line RP is parallel to the line AB.

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Properties of Parallel Lines:

Properties of parallel lines:

• The parallel lines will be in the same planes.
• Parallel lines are never crossed.
• The parallel lines are moved in same distance in the whole plane. 

Online Trigonometry Equation Solver

In online, many trigonometry solver are available for solving various trigonometry problems. It is very simple to work on the online trigonometry equation solver. If we enter the input, the online trigonometry equation solver automatically generates the solution. For example, if we want to find the solution for the equation sin 3x = 0 using online trigonometry solver, we have to just enter the problem and press calculate button, then the online solver will automatically generate the output as x = `(npi)/3`

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Example problems of online trigonometry equation solver:

Example 1:
Find the solution for the equation cos x = `1/sqrt2`
Solution:
The given equation is cos x = `1/sqrt2`
We know that cos` pi/4` = `1/sqrt2` and cos (2∏ -` pi/4` ) = `1/sqrt2`
Therefore,
cos `pi/4 ` = `1/sqrt2` and cos `(7pi)/4` = `1/sqrt2`
Hence, the solutions are x = `pi/4 ` and x = `(7pi)/4

Example 2:
Find the solution for the equation cos x = - `1/2`
Solution:
       The given equation is cos x = - `1/2`
       We know that cos `pi/3` = `1/2`
       Therefore,
                  cos (∏ - `pi/3` ) = - cos `pi/3` = - `1/2`   and
                  cos (∏ + `pi/3)` = - cos` pi/3` = - `1/2`
                  cos `(2pi)/3` = - `1/2` and cos `(4pi)/3` = - `1/2`
         Hence, the solutions are x = `(2pi)/3` and x = `(4pi)/3`

Example 3:
Find the solution for the equation cosec x + √2 = 0
Solution:
The given equation is cosec x + √2 = 0
The above equation can be written as sin x = - 1/(sqrt2)
                                                                            = - sin `pi/4`
                                                                            = sin (∏ + `pi/4` )
                                                                            = sin `(5pi)/4`
Therefore,
sin x = sin `(5pi)/4`
x = (n∏ + (- 1)ⁿ `(5pi)/4` ), where n ε I
Hence, the solution is x = (n∏ + (- 1)ⁿ `(5pi)/4` ), where n ε I

Practice problems of online trigonometry equation solver:

1) Find the solution for the equation sin x = `1/2`
2) Find the solution for the equation sin x = - `(sqrt3)/2`
3) Find the solution for the equation sin 2x = - `1/2`
Solutions:
1) The solutions are x = `pi/6 ` and x = `(5pi)/6`
2) The solutions are x = `(4pi)/3` and x =` (5pi)/3`
3) x = (`(npi)/2` + (- 1)ⁿ `(7pi)/12` ), where n ε I

Quadrilateral Rectangle

In Euclidean plane geometry, a quadrilateral is a polygon with four sides and four vertices or corners. Quadrilaterals are simple (not self-intersecting) or complex (self-intersecting), also called crossed. Simple quadrilaterals are either convex or concave. The interior angles of a simple quadrilateral add up to 360 degrees. A parallelogram is a quadrilateral with two pairs of parallel sides. Source-Wikipedia.

Quadrilateral rectangle:

Rectangle is one type of regular quadrilateral. It has four sides. The opposite sides are equal in length. It has four vertices. It has four internal angles which are congruent. There are two diagonals in rectangle. The length of diagonals is equal in length. The sum of internal angles of rectangle is 360 degree.

Area of the rectangle (A) = length x width square unit
Area (A) = l x w
Perimeter of the rectangle = 2(length + width)
Perimeter (p) =2(l x w) unit
Length of diagonal (d) =sqrt(length²+width²)
Diagonal length    =sqrt(l²+w²) unit.

Example problems for rectangle:

1. Find the area and perimeter of rectangle, whose length and width are 5meters and 3 meters respectively.
Solution:
Area of rectangle = l x w square unit.
Given:    Length= 5 meters, Width =4 meters
=5 x 4

Area of rectangle = 20 m²                 
Perimeter of the rectangle   = 2(l + w)
=2(5+ 4)
= 2 (9)

\Perimeter of the rectangle = 18 meters
2. Find the area and perimeter of rectangle, whose length and width are 9meters and 5 meters respectively.
Solution:
Area of rectangle = l x w square unit.
Given:    Length= 9 meters, Width =5 meters
=9 x 5

Area of rectangle = 45 m²                 
Perimeter of the rectangle   = 2(l + w)
=2(9+ 5)
= 2 (14)
Perimeter of the rectangle = 28 meters

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3. Find the area and perimeter of rectangle, whose length and width are 12meters and 6 meters respectively.
Solution:
Area of rectangle = l x w square unit.
Given:    Length= 12 meters, Width =6 meters
=12 x 6
Area of rectangle = 72 m²                 
Perimeter of the rectangle   = 2(l + w)
                                               =2(12+ 6)
                                               = 2 (18)

Perimeter of the rectangle = 36 meters

Star Test for Geometry

Geometry is a part of mathematics concerned with questions of size, shape, relative position of figures, and the properties of space. Geometry is one of the oldest sciences. Initially a body of practical knowledge concerning lengths, areas, and volumes. The field of algebraic geometry is the modern incarnation of the Cartesian geometry of co-ordinates. The star test geometry example problems and star test practice problems are given below. Source: Wikipedia

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Example problems for star test geometry:

Example problem 1:
Find the supplementary angle of 95°
Solution:
The supplementary angle of  95° = 180° – 95° = 85°.

Example problem 2:
From the figure write

    (a) collinear points.
    (b) Concurrent lines and their point of concurrence.

Solution:
    (a) from the figure, the points A, B, C are collinear.
    (b) The lines AD, BD, CD are concurrent lines. D is the point of concurrence

Practice problems for star test geometry:

Star test question 1:
The sum of the three angles of a triangle is _________ .
              a) 90°
              b) 360°
              c) 180°

Star test question 2:
      In an equilateral triangle, the 3 sides are __________ .
              a) Unequal
              b) Equal
              c) Parallel

Star test question 3:
      The triangle in which, the two sides are equal is called an ________ triangle.
             a) Isosceles
             b) Right
             c) Perpendicular

Star test question 4:
      If a triangle has one right-angle, it is called as _________ triangle.
             a) Perpendicular
             b) Right angled triangle
             c) Isosceles

Star test question 5:
      In a triangle the sum of the measure of any two given sides are _______ than the third side.
             a) Equal
             b) Lesser
             c) Greater

Star test question 6:
      One of the angles of a triangle is 100° and the other two angles are equal. What is the measure of each of these equal angles.
             a) 40°
             b) 80°
             c) 120°

Answer key:
       1. (c) 180°
       2. (b) equal
       3. (a) isosceles
       4. (b) right angled triangle
       5. (c) lesser
       6. (a) 40°

General Quadrilaterals

A general quadrilateral is a polygon with four sides. A general Quadrilateral is the two dimensional plane figures which has four sides so it is a four sided polygon. The total angle of a general quadrilateral is 360 degrees. The number of diagonals for a general quadrilateral are two. When we have two diagonals then both the diagonals will intersect at a common point. The general quadrilateral has four vertices's and four edges. Here in this topic we are going to see about general quadrilaterals.

Types of general quadrilaterals.

Square
Rectangle
Parallelogram
Rhombus
Trapezoid
Kite

General quadrilatel - Figures:

Square
In a square all the sides are equal in length and all the angles are equal to 90 degree. The diagonals of a square are equal to each other.

Rectangle:
In a rectangle, opposite sides are equal to each other and all the angles are equal to 90 degree.

Parallelogram:
In a parallelogram opposite sides are equal in length and opposite angles are also equal to each other.

Rhombus:
Rhombus is a combination of square and parallelogram. In Rhombus all sides of the lengths are equal and the opposite angles are equal to each other
.

Trapezoid:
In a trapezoid one of the pair of opposite sides are parallel.

Kite:
It has two pair of sides and the lengths of the adjacent pair are equal in length.

General Quadrilaterals - Example problems:

1) What is the area of a square having side-length 15 cm?

Solution:

The area of the Square = L²

Where L= 15cm

Length= L*L

= 15*15

= 225cm²

2) What is the area of a rectangle having a length of 8cm and a width of 3.3cm?

Solution:

The area of the rectangle = length * width

= 8 * 3.3= 26.4 cm²

3) What is the area of a trapezoid having the length of the parallel sides as  13 and 7 and the height being 9?

Solution:

The area of the trapezoid = h/2*(b1+b2)

=9/2(13+7)

=4.5(20)

=90 square units

4) What is the area of a parallelogram having a base of 30cm and a corresponding height of14cm?

Solution:

The area of the parallelogram=b*h

= 30*14

=420cm²

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5) What is the area of a square having side-length 5.4cm?

Solution:

The area of the Square =L²

= 5.4 * 5.4

=29.16cm²

Pentagonal Pyramid

Pyramid is a solid whose base is a planar polygon with a triangular face on each side. These triangular faces meet in a point which is not in the plane fo the base. Pentagonal pyramid is pyramid whose base is a pentagon, and having 5 triangular faces on each side.

 Pentagonal Pyramid Formula:
 Let,   a = apothem length, s,b = side, h = height and l = slant height,

• Area of Base(A) = `(5/2)axxs`
• Surface Area of Pyramid = `(5/2) a xx s + (5/2) s xx l = A + (5/2) s xx l`
•  Volume of Pyramid = `(5/6) axxbxxh`

Properties of pentagonal pyramid:

• A pentagonal pyramid has 6 faces. One base and 5 lateral faces
• A pentagonal pyramid has 10 edges
• A pentagonal pyramid has 6 corners

 

Examples on pentagonal pyramid:

1) Find  volume and surface area of a pentagonal pyramid with the given apothem length 7, side 8, height 9 and the slant height 10.
 Solution:   a = 7, s = b = 8, h = 9, l = 10

  Step 1:  Area of the base(A) = `(5/2) axx s`                                 
                                               = 2.5 * 7 * 8
                                               = 2.5 * 56
                                               = 140

  Step 2:  Surface Area of Pyramid = `A + (5/2) s xxl`
                                                       = 140 + ((5/2) * 8 * 10
                                                       = 140 + (2.5 * 80)
                                                       = 140 + 200
                                                       = 340

  Step 4: Volume of Pyramid =`(5/6) a xx b xxh`
                                              =  `(5/6)` * 7 * 8 * 9
                                              = (0.833) * 56*9
                                              = 0.8333 * 504
                                                = 419.98

2) Find the surface area a pentagonal pyramid with the given apothem length 2.5, side 5 and the slant height 9.
 Solution:   a = 2.5, s = b = 5, , l = 9

       Area of the base(A) = `(5/2)axxs`                         
                                          = 2.5 * 2.5 * 5
                                          = 6.25 * 5
                                          = 31.25
    Surface Area of Pyramid = `A + (5/2) s xxl`
                                               = 31.25 + ((5/2) *5 * 9
                                               = 31.25 + (2.5 * 45)
                                               = 31.25+ 112.5
                                               = 143.75

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Pictures of pentagonal pyramid

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Sin Angle Formula

Sin is a basic trigonometric function. In this article we are going to deal with sin angle formulas and how to use the sin angle formulas. Sin angle formulas used to calculate the sin values of the angles. Using the sin angle formulas we have to find the side lengths of the triangles normally from a right angle triangle we can say the sin function as
Sin A = `(opposite) / (hypotenuse)` . In this A is the angle of the opposite side.

Once you've gone through these, take a look at our Arc Formula for more refernce.

Sin angle formulas:

           There are five types of sin angle formulas are there.
Sum and angle formula
              Sin(X +Y) = Sin X Sin Y + Cos X Cos Y
              Sin (X – Y) = Sin X Sin Y – Cos X Cos Y
Double angle formula
             Sin 2A = 2 Sin A Cos A
Triple angle formula
             Sin 3A = 3 Sin A – 4Sin³A
Half angle formula
             Sin²(X / 2) = ((1 – Cos A) / 2)
Product and sum formula:
             Sin X Sin Y = Cos(X – y) – Cos (X + y) / 2
             Sin X Cos Y = Sin (X + Y) + Sin (X – Y) / 2

Example problems for sin angle formulas:

 Example 1:
      Find the value of 75^o
Solution:
         Sin 75^o = Sin (30^o+ 45^o)               
             We know the sin angle formula
                                        Sin(X +Y) = Sin X Sin Y + Cos X Cos Y
                                        Sin (30^o+ 45^o) = Sin 30^o Sin 45^o + Cos 30^o Cos 45^o
                                        Sin (30^o+ 45^o) = `(1 / 2)xx (1 / sqrt(2)) + (sqrt(3) / 2) xx (1 / sqrt(2))`
                                        Sin (30^o+ 45^o) = `(1 / (2 sqrt(2))) + (sqrt(3) / (2 sqrt(2)))`
                                        Sin (30^o+ 45^o) = `((1 + sqrt(3)) / (2 sqrt(2)))`

Example 2:
         Find the value of Sin 150^o using double angle formula. Where sin 75^o = 0.9659, Cos 75^o = 0.2588
Solution:
         Given sin 75^o = 0.9659, Cos 75^o = 0.2588
         Sin 150^o = Sin 2 `xx` 75^o
           We have the sine angle formula Sin 2A = 2 Sin A Cos A
                 Sin 150^o = 2 Sin 75⁰ Cos 75^o
                 Sin 150^o = 2 `xx` 0.9659 `xx` 0.2588

                 Sin 150^o = 0.4999