Three Horizontal Lines

Introduction to three horizontal lines:

Three horizontal lines are nothing but three lines parallel to x – axis or three lines perpendicular to y – axis. Three horizontal lines mean their slopes will be zero. Because the slope of x – axis is 0. We know if there is any two lines are parallel their slope s will be equal. We will some example problems for graphing three horizontal lines. If the line is horizontal their y value is constant.

Examples for three Horizontal Lines:
If the line is parallel to x – axis we can say those lines are horizontal lines. The slopes of the horizontal lines are zero and the y value of the line is constant. So the equation of the horizontal lines are like y = some constant value.Having problem with geometric probability formula keep reading my upcoming posts, i will try to help you.

Example 1 for three horizontal lines:

Graph the following lines y = 1, y = 5 and y = -1.

Solution:

Here the line equations are y = 1, y = 5 and y = -1

The slope intercept form general equation is y = mx + c

If we compare the given equation with general form we can get the slope of the lines are 0.

If we graph these equations we will get the graph like the following.



More Examples for three Horizontal Lines:
Example 2 for three horizontal lines:

Graph the following lines y = 2, y = 3 and y = -2.

Solution:

Here the line equations are y = 0, y = 3 and y = -2

The slope intercept form general equation is y = mx + c

If we compare the given equation with general form we can get the slope of the lines are 0.

If we graph these equations we will get the graph like the following.



In this the line y = 0 is lies on the x axis.

These are some of the examples for three horizontal lines. From the above we can understand how to graph the three horizontal lines and slopes of the horizontal lines.

Line Segment Circle Intersection

Introduction:
A line can intersect a circle at two points or it can touch the circle at one point or never pass throught.

To find the point of intersection of line segment with a circle, we need to solve both equations. Thus we can get two points of intersection of the line with the circle.

Let us see few problems of this kind.

Example Problem on Line Segment Circle Intersection.

Ex 1: Find the point of intersection of the line 2x + y = 1 and x 2 + y 2 = 1.

Soln: Given: The line is 2x + y = 1 ……….. (1)

The circle is x 2 + y 2 = 1…………… (2)

(1) `=>` y = 1 – 2x

Therefore (2) = x 2 + (1 – 2x) 2 = 11 `=>` x 2 + 1 + 4x 2 – 4x = 11

`=>` 5x 2 – 4x – 10 = 0

Therefore x = `[4+- sqrt [(-4) ^2 ** 4(5) (-10)]] / [2 (5)]`

=` [4 +- sqrt [16 + 200]] / 10`

= `[4 +- sqrt 216] / 10`

=` [4 +- 6 sqrt 6] / 10` 

= `[2 +- 3 sqrt 6] / 5`

Therefore y = `1 ** [2 [(2 + 3 sqrt 6) / 5]]`

= `[5 ** 4 ** 6 sqrt 6] / 5`

= `[1 ** 6 sqrt 6] / 5`

When x = `[2 + 3 sqrt 6] / 5` , y = `[1 ** 6 sqrt 6] / 5`

When x = `[2 ** 3 sqrt 6] / 5` , y = `1 ** [2 [2 ** 3 sqrt 6] ]/ 5`

y =`[ 5** 4 + 6 sqrt 6 ]/ 5` = `[1 + 6 sqrt 6 ]/ 5`

Therefore the point of intersections are given by (`[2 + 3 sqrt 6] / 5` , `[1 ** 6 sqrt 6] / 5` ) and (`[2 ** 3 sqrt 6] / 5` , `[1 + 6 sqrt 6] / 5` )


Example Problem on Line Segment Circle Intersection.

Ex 2: Find the point of intersection of the following line and the circle. x – y = 1, x 2 + y 2 + 4x + 2y + 2 = 0.

Soln: Given: x– y = 1   =  x = y + 1 ……….. (1)

x2 + y 2 + 4x + 2y + 2 = 0 ………... (2)

By using (1) in (2), we get (y + 1) 2 + y 2 + 4 (y + 1) + 2y + 2 = 0

`=>` y 2 + 2y + 1 + y 2 + 4y + 4 + 2y + 2 = 0

2y 2 + 8y + 7 = 0

y =`[ -8 +- sqrt[ 8 ^2 ** 4 (2) (7)]] / [2 (2)]`

= `[-8 +- sqrt 64 ** 56] / 4`

=` [-8 +- 2 sqrt 2] / 4`   =`[-4 +- sqrt2]/2`

Therefore y = `[- 4 + sqrt 2 ]/ 2` , (1)  `=>` x =` [-4 + sqrt 2 ]/ 2` + 1

= `[- 2 + sqrt 2] / 2`

y =` [-4 ** sqrt 2] / 2` , (1) `=>` x =` [-4 ** sqrt 2] / 2` + 1 = `[-2 ** sqrt 2 ]/ 2`

Therefore the points are (`[-2 + sqrt 2] / 2` , `[-4 ** sqrt 2] / 2` )

Geometric Solids Patterns

Introduction to geometric solids patterns:

Geometry is a topic which deals with the shapes and sizes of any objects. The shapes are classified as solid and plane surface shapes. The geometric solids are the shapes that can be classified as cubes, cone, sphere, rectangular prism, hemisphere, etc. The geometric solid patterns can be explained with their properties. Here we see some of the solids pattern in geometry.

Classifying Solids Patterns:

Solids patterns can be classified into many shapes. We see some of them with properties.

Sphere:

It is a 3-dimensional solid shape pattern which does not have a base point. It is rounded in shape, as it is spherical. The parameters used in the solid sphere are calculated using the following formulas,

•    Volume of a sphere = 4/3 ?r3

•    The surface area of a sphere = 4?r2

It is rounded in shape, as well as spherical.

Cube:

The cube is also one of the three-dimensional solid shapes. It is made up of six equal sides with 12 edges and 8 vertices. Some parameters like volume of cube and surface area are calculated using the given formulas.

Volume of a cube can be found as a3

•    Area of a cube is given as 6a2

Rectangular prism:

The rectangular prism is a 3-dimensional solid shape which is having six numbers of sides.

The formulas used to find the parameters of a rectangular prism are given below:

•    The surface area of rectangular prism = 2(lb + bh + hl)

•    Volume of the rectangular prism can be given as lbh

Where,

l is the length, b is the breadth and h is the height of the rectangular prism.

These are some basic geometric solid patterns.

Problems to Geometric Solids Patterns:

We can solve some example problems for geometric solids patterns.

Example1:

Find the volume of the rectangular prism if its length, breadth and height are given as 6 cm, 4 cm and 2 cm.

Solution:

Formula to find volume of rectangular prism is lbh.

On substituting the given values in formula, we get

lbh = `6 xx 4 xx 2`

= `24 xx 2`

= 48

Hence the volume of the rectangular prism is found to be 48 cubic cm.

Example 2:

Find the volume of a cube if the side is measured as 3 cm.

Solution:

Given that,

Side length = 3 cm

We know the formula to find volume of cube as,

Volume of the cube is a3 = 27

Hence the volume of the cube is given as 27 cm3.

Thus these are some examples to geometric solids patterns

The Measure of Acute Triangles

Introduction to the measure of acute triangles :

Definition to the measure of acute triangles:A triangle which has all the three angles is less than 90°. This can also be tell in the words such as the angles which are smaller than the right angle triangles can be called as acute angles.The method of finding acute angles triangles which can be done by measure of two angles which are given and if measure of a side and any two angles are given.

Prolems for the Measure of Acute Triangles:

Ex1:The measurement of one of the acute angle of a triangle is 52°. Find the measure of other acute angles of the triangle?

Solution: The following steps to be taken for the measure of angles are

Step1: The addition of the measures of the two acute angles must be 90°.

Step2: If one acute angle of a right triangle is 52°, then the measure of the other acute angle is 90° - 52° =38 °

Ex 2:The measurement of an acute angle of a triangle which is given as 30° 53'. Find the other measure of  acute angles of the right angled triangle?

Solution:The addition of the measure of the two acute angles must be 90°.

If one acute angle of a right triangle is 30° 53', then the measure of the other acute angle is givan as,

90° - 30° 53' = 89° 60' - 30° 53' = 59° 07'.

Ex 3: Find the measure of the angle X° from the given diagram. The other two angles which are 50° and 60°.Find the third angle of an acute angle triangle?

Solution:The addition of three angles in the triangles should be equal to 180°

The sum of the measurement of the three angles is

50° + 60° + X° = 180°.

The third angle which can be measure by the following step

X° + 110° = 180°

The X° which can be performed as follows,

X° = 180° - 110°

X° = 70°.


Practice Problem for Measure of Acute Triangles:

Find the measure of the angle X° of an acute angle triangles with the given angles  40°and 80°?

Solution:  The measure of the third acute angle X° = 60°.

Number of Quadrilaterals

Introduction to number of quadrilaterals:

A quadrilateral is a polygon with four sides or edges and four vertices or corners
Quadrilaterals are either simple (not self interesting).simple quadrilaterals are either convex or concave.
Plane is a shape which consist of four sides, and consequently four angles.

Description about Number of Quadrilaterals :
Quadrilaterals are classified into the following types

They are:

Trapezium

Parallelogram

Rhombus

Rectangle

Square

Kite

Number of Quadrilaterals with Example:
Square :

The square is one of the best example for quadrilaterals. It is defined by sides are equal, and its sum of  interior angles of all right angles should be (90°). and its opposite sides will be parallel.
we can say square is a specific case of regular polygin, in this case there are  4 number of sides. All the facts and properties described for regular polygons apply to a square.
Rectangle:

The rectangle is somwhat similar to the square, and this is one of the most generally known quadrilaterals. It is defined by having all four interior angles 90° (right angles).
Parallelogram

A quadrilateral has both pairs of opposite sides parallel and equal in length.
Parallelogram is defined by opposite sides are parallel and it should be congruent. It is the the root way  of some other quadrilaterals, which are consisting by adding some restrictions of various techniques:
A rectangle is a parallelogram but all the angles fixed at 90°
A rhombus is a parallelogram but with all sides equal in length
A square is a parallelogram but with all sides equal in length and all angles fixed at 90°.
Trapezoid

A quadrilateral should have  at least one pair of parallel sides
Rhombus:

In rhombus, the number of sides are four and its all equal.
Rhombus is looking like a special type of parallelogram. Iin a parallelogram, each sets of opposite sides are equal in length. but in rhombus, all four sides are the same length.so the properties are also same.
Rhombus is somewhat light similar ro square but the angles will not be 90o
Kite

A quadrilateral which contains  two distinct pairs of equal adjacent sides.
kite is a part of the quadrilateral relations, and while easy to understand visually, is a little tricky to define in precise mathematical terms. It has two pairs of equal sides. Each pair must be adjacent sides (sharing a common vertex) and each pair must be distinct. That is, the pairs cannot have a side in common.

Base of a Polygon

Introduction to base of polygon:

Let us discuss the base of polygon. The base of the polygon is declaring the lowest part. The base is commonly known as bottom line of the shape. Solid objects are placed on a plane by the bottom line of the surface. The straight side shape is called  the base of a polygon. Next we see the polygon base. For example in triangle we take one of the sides as base (from three sides). Similarly in square we take one of the sides of a base (from four sides).

Base of the Polygon

The two dimensional polygons are any side can be declare a base. There is the polygon “sits” bottom side is normally known as the base. The triangle is also known as the polygon. The three side polygon called as the triangles. We can say any side of triangle is base of the triangle. The all type of triangle contains three bases. The following diagram is representing the base of the polygons.

`Base = (2A)/(h)`           

The formula for the base of the triangle polygon is b = 2A / h. The b is representing the base of   the triangle. A is represent the area of the triangle. The h is representing the height of the triangle. Next we see the base of rectangle.

The rectangle is another part of the polygons. The rectangle polygons is represented the four sides. The base of the rectangle is declaring the bottom side of the diagram. The general formula for base of rectangle is b = A / h. the b is represents the base of the rectangle polygon. The A is represents the area of the rectangle polygon. The h is represents the height of the polygons.

`Base = (A)/(h)`

I am planning to write more post on How to Construct a Triangle, Properties of Quadrilaterals. Keep checking my blog.

Other Polygon Base

The next polygon declares the parallelogram. The side of the parallelogram is four. The base of the parallelogram is specifying the bottom side of the diagram. The general formula for base of parallelogram is b = A / h. the b is represents the base of the parallelogram polygon. The A is represents the area of the parallelogram polygon. The h is represents the height of the parallelogram polygons.

`Base = (A)/(h)`

These Form the Bases of a Prism

Introduction :

The prisms are the shapes that exist in the three dimensions. All the prisms are formed by the two bases and the bases of the prism formed by the faces of the prism. Regular polygons form the bases of a prism. In the following article we will discuss more about the online Volume of Right Prism help in detail.

More about the Topic these Form the Bases of a Prism

As we described before the prisms are the shapes formed by the two bases the upper base and the lower bases. All bases of the prism are the regular polygons and these form the bases of the prism. In the right regular prisms the bases of the prism are equal. The area of the bases is calculated as the product of the perimeter fo the base and the height of the prism. And the Formula for the area of the bases of the prism and the area of the total prism including the faces of the prism with the base made of n sides and side length S are,

Area of a base of the prism = `[n*S^2*cot (pi/n)]/4`

Total surface area = `[n*S^2*cot (pi/n)]/2 + S*H*n`



Example Problems on these Form the Bases of a Prism:

1. Calculate area of the base and the total area of the hexagonal prism with the height 10cm and side length of the base 5cm.

Solution:

Area of a base of the prism `= [n*S^2*cot (pi/n)]/4`

`= [6*5^2*cot (pi/5)]/4`

`= 43 cm^2`

Total surface area `= [n*S^2 cot (pi/n)]/2 + S*H*n `

`= [6*5^2*cot (pi /6)]/2 + 6*5*10`

`= (48*1.732) + 300`

`= 129.93 + 300`

`= 429.93 cm^2`

Practice problems on these form the bases of a prism:

1. Calculate area of the base and the total area of the pentagonal prism with the height 10 cm and side length of the base 6cm.

Answer: Total surface area = 423.9 cm2 and Area of a base= 61.94 cm2.

2. Calculate area of the base and the total area of the octagonal prism with the height 9cm and side length of the base 3cm.

Answer: Total surface area = 302.9 cm2 and Area of a base = 43.46 cm2.