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)`

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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.

Surface Area of Part of a Sphere

Introduction to Sphere:

Sphere is the one of the type of the geometrical figure.  Sphere is the three dimensional figure of the circle.  Surface area is the area covered by the sphere.  Radius of the sphere is the distance between centers to the circumference of the sphere.  Here we have to discuss about the surface area of sphere with example problems.

Brief Explanation Surface Area of Part of a Sphere

Sphere:

Sphere is a solid 3D shape figure.  It is look like a ball.  The picture of the sphere is look like in the following diagram

Here A is the area of the sphere and r is the radius of the sphere. And o is the center of the sphere.

Surface Area:

The surface area of the part of the sphere is mentioned by the following formulas,

Total surface area of the sphere =4 p r2 Square units
Curved surface area of the sphere=3 p r2 Square units
In the above formulas we have to substitute the radius of the sphere and the value of p is always equal to 3.14 or `(22)/(7)` . We can get the surface area of the sphere.  Curved surface area of the sphere is also called as lateral area of the sphere.


Example and Practice Problems

Find the Surface area of part of a sphere if the radius of the sphere is 105 cm.

Solution:

Given radius of the sphere r=105 cm.

Curved Surface Area:

A= 3 p r2 Square units

Here we have to substitute the values of r and p then we can get,

A= 3 x `(22)/(7)` x 105 Square centimeter

Divided by 7 we can get,

A=3 x 22 x15 Square centimeter

Multiplying this we can get,

A=990 Square centimeter

Total Surface Area:

A= 4 p r2 Square units

Here we have to substitute the values of r and p then we can get,

A= 4 x`(22)/(7)` x 105 Square centimeter

Divided by 7 we can get,

A=4 x 22 x15 Square centimeter

Multiplying this we can get,

A=1320 Square centimeter

Practice problems:

Find the surface area of the  sphere if the radius is 0.25 meter,        ans = 0.785 squrae meter
Find the surface area of the  sphere if the radius is 82 inches           ans = 84453.44 square inches.

Rectangular Pyramid Vertices

Introduction to rectangular pyramid vertices

The pyramid is the solid form objects through a polygon for a bottom.  Each faces joint on one point. The bottom of a rectangular pyramid is for all time a rectangle.  The rectangular pyramid includes the five vertices. It normally contains five sides.  Let us see about the rectangular pyramid vertices.

Rectangular Pyramid Vertices

A rectangular pyramid contains five vertices.

Rectangular pyramid is geometrical form within math. Usually pyramids are objects which contain a pyramid like formation through a triangular or else rectangular or else square or else pentagonal base etc. The bottom is which categorized the kind of pyramids. Pyramid through a rectangular bottom is identified as rectangular pyramid.

The rectangular pyramid contains the five vertices and five sides, eight edges.

In the above rectangular pyramid contains five vertices.

The volume of right rectangular pyramid = `1/3`  x base x height

Example

Given the length = 12 cm, width = 15 cm, height = 18 cm.

The volume of right rectangular pyramid = `1/3`  x base x height

= `1/3`  x (length x width) x height.

= `1/3`  x (12 x 15) x 18

= `1/3`  x 180 x 18

= `1/3`  x 3240

= 1080

Therefore the volume of right rectangular pyramid is 1080 cm3

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Examples for Rectangular Pyramid

Example 1

Compute the volume of rectangular pyramid if length = 8 cm, width = 10 cm, height = 14 cm.                             

Solution

Given the length = 8 cm, width = 10 cm, height = 14 cm.

The volume of right rectangular pyramid = `1/3`  x base x height

= `1/3`  x (length x width) x height.

= `1/3`  x (8 x 10) x 14

= `1/3` * 80 x 14

= `1/3` * 1120

= 373.3

Therefore the volume of right rectangular pyramid is 373.3 cm3

Example 2

Compute the volume of rectangular pyramid if length = 6 cm, width = 12 cm, height = 20 cm.                             

Solution

Given the length = 6 cm, width = 12 cm, height = 20 cm.

The volume of right rectangular pyramid = `1/3`  x base x height

= `1/3`  x (length x width) x height.

= `1/3`  x (6 x 12) x 20

= `1/3`  x 72 x 20

= `1/3`  x 1440

= 480

Therefore the volume of right rectangular pyramid is 480 cm3

Polar Equation Cartesian

Introduction on polar equation Cartesian :

Polar Equations:

The polar equation system is a two-dimensional equation system in which each point on a plane is determined by a distance from a fixed point and an angle from a fixed direction.

Cartesian Coordinates:

A Cartesian coordinate system specifies each point uniquely in a plane by a pair of numerical coordinates, which are the signed distances from the point to two fixed perpendicular directed lines, measured in the same unit of length.

Complex Form of Polar Equation Cartesian:
A complex number is of the form x + iy, where x, y belongs to R and 'i' is called imaginary unit (i = `sqrt(-1)` )

Let z = x+ iy,

Real part of z = Re(z) = x, and imaginary part of z = Im(z) = y.

The point (2, 8) written as 2 + 8i. Cartesian form are used to solve non linear shallow -water equations on the sphere.

Let, z1 = a + ib; z2 = c + id

z1 = z2; a = c; b = d.

Sinz = Sin (a+ib)

= Sina Cos hb + iCosa Sin hb

Cosz = Cos (a+ib)

= Cosa Cos hb  - iSina Sin hb

Sinz = Cosz

Sina Cos hb = Cosa Cos hb

Cosa Sin hb = Sina Sin hb

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Steps for Solving Polar Equation Cartesian:

The following steps are used to solve from Cartesian form to polar form,

Find the value of mod(z), | z | = `sqrt(x^2 + y^2)`

Find the `theta` value by using `tan theta = (y)/(x)`

Find q in radians.

By writing the equation in polar form, `z = r(cos theta + i sin theta)`.

Example for Solving Polar Equation Cartesian:

Question: Solve z = 1 + i in polar form.

Solution:  1) r = | z | = `sqrt(1^2 + 1^2) = sqrt(1 + 1) = sqrt2`

2) `tan theta = ` `1 / 1`  => `theta = ` 45°

3) `theta = (pi)/(4)`

4) The polar form is, z = r (`cos theta + i sin theta` )

=> z = `sqrt(2)(cos (pi /4) + sin(pi/4))`

The Polar form of z = 1 + i is, z = `sqrt(2)(cos (pi /4) + sin (pi/4))`

Semicircle Learning

Introduction of semicircle learning :

Semicircle is defined as half of a circle. That is, the angle is 180 degree arc of a circle. A triangle decorated in a semicircle is always called a right triangle.

If two curves or arcs are equal, then both the segments and sectors are similar. This each part of term is called as semicircle region.

Formulas of Semicircle Learning

A semicircle is the area enclosed by a diameter and an arc of the circle joining its two ends. The length of the resulting segment is called the geometric mean, which can be proved using the concept of Pythagorean Theorem.

Formulas:

Area of semicircle (A) =circle /2

A = (pr2)/2

Circumference of semicircle(C) = (2pr)/2

C = pr

A circumference of a semicircle is calculated for the circumference of circle divided by 2.we get,

C = 2pr ==> C/2 = pr

p = 3.14 ( approximately )

A perimeter of a semicircle is the sum of circumference and diameter of a semicircle. We get,

P = pr + 2r = r (p+2)

P = 5.14 r

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Examples of Semicircle Learning:

Semicircle learning Ex 1!:

Find the area of semicircle with radius of 12.5 cm.

Semicircle learning sol :

We can find the area of semicircle by using the following formula,

Area = (pr2)/2

Substitute the values of p and the radius into the above formula. Then we get,

= (3.14*(12.5)2)/2

Squaring the values of radius and multiplying with 3.14 then dividing by the value of 2.

= (3.14*156.25)/2

= (490.625)/2

Then we get the final answer.

=245.3 cm2

Answer: 245.3 cm2

Semicircle learning Ex 2:

Find the perimeter of semicircle with the radius 10 cm.

Semicircle learning sol :

We can find the perimeter of semicircle by using the following formula,

Perimeter = 5.14*r

Substitute the value of r into the above formula,

=5.14*10

=51.4 cm

Answer: 51.4 cm

Semicircle learning Ex 3:

Find the circumference of semicircle with the radius of 7.5 cm.

Semicircle learning sol :

We can find the circumference of semicircle by using the following formula,

Circumference C = (2pr)/2

Circumference C = pr

Substitute the value of p and the radius.

C = 3.14*7.5

C = 23.55 cm2

Answer: C = 23.55 cm2

Measuring Irregular Triangles

Introduction for measuring irregular triangles:

The irregular triangle is nothing but the triangle where the three sides are not equal and the angles present in it also different during its measurement. The only irregular triangle is the scalene triangle. Now we are going to see about the measuring of irregular triangle with some example problems.


About Measuring of Irregular Triangle:
Now we are going to see about the irregular triangles and its measurement. The irregular triangle is nothing but the scalene triangle where the sides are unequal in its length and the angles in it also unequal.

When the two sides and an angle are given and if we want to find the third side of an irregular triangle, then use the formula which is given below,

c2 = a2 + b2 - 2ac * cos ?

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Problems for Measuring Irregular Triangle:

Example 1:

The sides of the triangle are 5cm and 10 cm and the angle measuring is 40. Determine the third side of the irregular triangle.

Solution:

The third side of the triangle can be calculated by using the formula,

c2 = a2 + b2 - 2ac * cos ?

Now substitute the values in the formula we get,

c2 = 52 + 102  2(5)(10) * cos 40

Now square the values which are substituted in the formula as follows,

= 25 + 100  100 cos 40

c2 = 48.39

c = `sqrt 48.39`

c = 6.95

The value can be rounded as 7cm.

Example 2:

Find the third side of the triangle whose measurements of the triangle are 7cm, 8cm and angle measuring is about 50 degree.

Solution:

The third side of the triangle can be calculated by using the formula,

c2 = a2 + b2 - 2ac * cos ?

Now substitute the values in the formula we get,

c2 = 72 + 82  2(7)(8) * cos 40

= 49 + 64  112 cos 50

c2 = 41

c = ` sqrt 41`

c = 6.4

The value can be rounded as 6cm.

Non Coplanar Definition

Introduction to non-coplanar points:
The points which do not lie in the same plane or geometrical plane are called as non-coplanar points. Any 3 points can be enclosed by one plane or geometrical plane but four or more points cannot be enclosed by one. The points belong to the same plane are called as coplanar points. In this article we shall be discussing the non-coplanar points. Now we know what non-coplanar point is and we shall see some examples of the non-coplanar points and solve it for the same.

Example for Non Co-planar Points
1)      From the below shown figure the points are non coplanar points as they do not lie on the same plane it lies in different planes.

2)      We can see four planes with the help of four non co-planar points.

3)      Plane is the two dimensional geometrical object.

Non co-planar points

Non co-planar points

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Solved Examples for Non Co-planar Points:

Ex 1:  Check whether the following lines are co-planar or not.

3x+6y+9 = 0 and 4x+4y+11 = 0

Sol :  The given equations are  3x+6y+9 = 0 and 4x+4y+11 = 0

The slope intercept form can be given as y = mx+b

Where m indicates slope.

Comparing the above equation with the given equation, we get:

6y = -2x-9

Dividing by 6 on both sides we get:

We get,

--- (1)

The slope intercept form can be given as y = nx+b

Where n indicates slope.

Comparing the above equation with the given equation, we get:

4y = -4x-11

Dividing by 4 on both sides we get:

y = -1

We get n = -1--------- (2)

Equation (1) (2), that is m n

That is the slopes of the two equations are not equal and therefore the points lie on the two lines are non co-planar points.

Ex 2:   Check whether the following lines are co-planar or not.

x+5y+9 = 0 and 2x+10y+11 = 0

Sol :  The given equations are  x+5y+9 = 0 and 2x+10y+11 = 0

The slope intercept form can be given as y = mx+b

Where m indicates slope.

Comparing the above equation with the given equation, we get:

5y = -x-9

Dividing by 5 on both sides we get:

We get,

------- (1)

The slope intercept form can be given as y = nx+b

Where n indicates slope.

Comparing the above equation with the given equation, we get:

10y = -2x-11

Dividing by 10 on both sides we get:

We get

--------- (2)

Equation (1) =(2), that is m = n

That is the slopes of the two equations are equal and therefore the points lie on the two lines are co-planar points.

Quadrilaterals Shapes Learning

The word Quadrilateral may look difficult but it is not so. It just means a four sided figure with four vertices (corners). Absolutely, for any figure the number of sides equals the number of angles. It has four angles too. Based on the sides and the angles, we name the figures. The word quadrilateral formed from the words quad (meaning "four") and lateral (meaning "of sides").

The word Quadrilateral doesn't rhyme with the triangles (which has one fewer side and angle) and the pentagons (which has one more side and angle). Hence, we can say Quadrangle by analogy with the Triangle. Similarly, we can also call as "Tetragon" by analogy with the Pentagon.The sides of a quadrilateral may be congruent or non-congruent, may be parallel or perpendicular. The sum of the angles in a quadrilateral is found out by assuming the right angles for each angle. So, the sum of the angles = 90 + 90 + 90 + 90 = 360 degrees.
Lets have a look on the quadrilaterals shapes.

Quadrilaterals Shapes
The quadrilaterals are mainly divided by two types "Regular" and "Irregular". If all the sides and the angles are congruent, it is a Regular Quadrilateral and if they are non-congruent, then they are Irregular Quadrilaterals.Only the Square belongs to Regular quadrilateral, whereas all the other quadrilaterals are Irregular.Now, you may think that Rhombus whose all sides are congruent, why is it not included in Regular quadrilaterals. Well, I agree that the Rhombus has all sides congruent, but the angles ? Are all angles congruent too ? No, the angles in a Rhombus is not congruent.

Making all Quadrilaterals from a Square
If all the sides are equal and all the angles same in a quadrilateral, then it is a Square. Right ?
Let me show you to make all types of quadrilaterals using just a single square.
Draw a square. If you drag one side far away, there becomes a Rectangle. If you drag any two adjacent vertices to any one direction, there becomes a Parallelogram. We can also call a  Parallelogram as a Squashed Rectangle. Pull in one side of a Parallelogram inside the figure (provided the horizontal lines should not get changed), it makes a Trapezoid. If you make all the sides slant in a square, there becomes Rhombus. If you drag one vertex of a Rhombus far away, there becomes Kite.

Conic sections

The conic sections are curves obtained by making sections, or cuts, at particular angles through a cone. First we will discuss about history of conic sections. Conics are amongst the oldest of the curves, and are the oldest math subject studied systematically and thoroughly. The conics had been discovered by Menaechmus, the tutor to the great Alexander. These conics were perceived in order to solve the three problems (a) trisecting an angle, (b) duplicating the cube, and (c) squaring the circle.

If we cut a cone at different angles, then we will obtain different types of conics. There are four different types of conics that we can obtain. That is circle, where the cone is cut at right angle to its axis, ellipse, where the cone is cut at an oblique angle, parabola, where the cone is cut parallel to the generator and finally hyperbola, where a double cone is cut at an angle steeper than the generator.

Let’s discuss about conic sections formulas and conic sections equations. First is circle, the standard formula of circle is  (X^2+y^2=r^2),  where centre is (0,0) and radius is r. second is ellipse, the standard formula of a ellipse is (x^2/a^2+y^2/b^2 =1 a = 1/2) length major axis,b = 1/2 length minor axis, third one is parabola, the standard formula of parabola is 4px=y^2, where p=distance from vertex to focus. And the last one is hyperbola, the standard formula of hyperbola is (x^2/a^2-y^2/b^2=1), where a = 1/2 length major axis, b = 1/2 length minor axis. The general equation for all conic section is (Ax^2+Bxy+Cy^2+Dx+Ey+F=0). And by using the quadratic formula

If (B^2-4AC) <0 circle="circle" curve.="curve." curve="curve" ellipse="ellipse" is="is" no="no" or="or" p="p" point="point" the="the" then="then">If (B^2-4AC) =0, then the curve is a parabola or two parallel lines or a single line or no curve.
If (B^2-4AC)>0, then the curve is hyperbola or two intersecting lines.

The procedure of graphing conic sections, in this part first we will focus on graphing circle. If we cut a circular cone with a plane which is perpendicular to the symmetric axis of cone, then a circle is formed. The intersection line is parallel to the plane generating circle of the cone. A circle means from its center all points are at equal distance.

Now graphing of ellipse, in ellipse long axis is major axis and short axis is minor axis. Intersection point of two axis called vertices. The vertices along horizontal and vertical axis form points. This point along with center will provide method to graph ellipse in standard form.

Introduction to geometry word problems

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 the body of the practical knowledge is concerning with the lengths, areas, and volumes. (Source:wikipedia)

In this article we shall discuss about geometry word problems.

Geometry Word Problems Involving Area:

A rectangle is 4 times as extended as it is wide. If the measurement lengthwise is more than 4 inches and the width is less than 1 inch, the area will be 60 square inches. What were the dimensions of the original rectangle?

Solution:

Let x = original width of rectangle

Area of the rectangle is

A = l w

Plug in the values from the question

60 = (4x + 4) (x –1)

Use distributive property to remove brackets

60 = 4x2 – 4x + 4x – 4

Put in quadratic formula 4x2 – 4 – 60 = 0

4x2 – 64 = 0

This quadratic written as a difference of two squares

(2x) 2 – (8)2 = 0

Factorize Difference of two squares are

(2x) 2 – (8)2 = 0

(2x – 8)(2x + 8) = 0

Therefore the values for x is

Since x is a dimension, it would be positive.

So, we take x = 4

The question wants the dimensions of the original rectangle.

The width of the given rectangle is 4.

The length is 4 times the width = 4 × 4 = 16

Answer: The dimensions of the rectangle are 4 and 16.

Geometry Word Problems Involving Perimeter:

 A triangle has a perimeter of 50. If 2 of its sides are equivalent and the third side is five more than the equivalent sides, what is the length of the third side?

Solution:

Let x = length of the equal side

The formula for perimeter of rectangle

P = sum of the three sides

Plug in the values from the question

50 = x + x + x+ 5

Combine like terms

50 = 3x + 5

Isolate variable x

3x = 50 – 5

3x = 45

x =15

The question requires length of the third side.

The length of third side is = 15 + 5 =20

Answer: The length of third side is 20

Practice Geometry Word Problems:

A triangle has a perimeter of 80. If 2 of its sides are equivalent and the third side is five more than the equivalent sides, what is the length of the third side?
                   Answer: The length of third side is 30

A triangle has a perimeter of 95. If 2 of its sides are equivalent and the third side is five more than the equivalent sides, what is the length of the third side?
                   Answer: The length of third side is 35

Various types of conics sections

Definition of conic sections: We can define conic section as follows: Consider a double cone. If we have a plane that cuts this double cone, the cross section thus obtained at the intersection of the plane and the cone is called a conic section.

Depending on the angle at which the plane cuts the cone and the position of the plane, we can have mainly 4 different conics: circle, ellipse, parabola and hyperbola.

Conic s have been studied for over 2000 years. Greek mathematician Apollonius studied them intensely. He wrote a book ‘the conic’ that remained a standard work on the topic for eighteen centuries. In the sixteenth century, Galileo declared that the trajectory of a projectile was a parabola. The reflectors in head lights of a car, the speakers in a sound system and the mirrors in a telescope are all in the shape of a parabola.

Parabolic mirrors are used to harness solar energy. In the seventeenth century, Kepler declared that planets revolved around the sun in elliptical orbits. It is because of our knowledge of ellipses that precise predictions of time and place of solar and lunar eclipses is possible. The path of motion of a comet in the solar system is also in the shape of a parabola, ellipse or a hyperbola. Knowledge of conics is extremely useful in such terrestrial sciences as architecture and bridge building.

Thus, the study of conics has been proved very useful in space sciences, mechanics, optics, engineering, architecture and other fields.

How to graph conic sections?
To be able to graph conics, let us look at the following concept. Suppose line l is a fixed vertical line and another line m intersects l in the point V and makes an angle of measure a(0 < a < pi/2) with it. If m is made to rotate around V in such a way that a remains constant, then the surface generated is called a double cone. The point V is the vertex and the line m is a generator of the double cone. The line l is the axis of the double cone.  Then the plane cutting this double cone defines the four conics as follows: (see picture below)

Solving Solid Geometry

In solid geometry we study three dimensional geometry (3-D geometry).
For examples: Cube, cuboid, cylinder, cone, sphere, Pyramids, Prisms etc. Dimensions are the terms as length, width, height, thickness etc. A three dimensional figure must have length, width and height.

Cube : A three dimensional shape having equal length(a), width(a) and height (a).
Cuboid: A three dimensional shape having different length(l), width(w) and height(h).
Cylinder: A three dimensional shape having two circular faces of radius(r) at two ends and a curved surface of height(h).
Cone: A three dimensional shape having a circular face at one end and a curved surface of height(h).
Sphere: A three dimensional shape of radius r. For example: a ball.

Formulas for Solving Solid Geometry
(1) Cube :           Lateral Surface Area ( Area of four sides i.e. front, back, left, right ) = 4a2
                           Total Surface Area( Area of all six faces) = 6a2
                            Volume = a x a x a = a3
(2) Cuboid:        Lateral Surface Area ( Area of four sides i.e. front, back, left, right ) = 2h(l+w)
                            Total Surface Area( Area of all six faces) = 2( lw + wh + hl )
                             Volume = lwh
(3) Cylinder:      Curved surface area = 2Ï€rh
                            Total surface area (including two circles on both ends) = 2Ï€r(r+h)
                            Volume = 2Ï€r2h
(4) Cone :          Curved surface area = Ï€rl where l is the slant height of the cone l = `sqrt(h^2 + r^2)`
                            Total surface area (including a circles on the base) = Ï€r(r+l)
                            Volume = 1/3 Ï€r2h
(5) Sphere :      Surface area = 4Ï€r2
                            Volume = 4/3 Ï€r3

How to Solve Problems for Solid Geometry
Step 1) Make a figure of solid given in the problem.
Step 2) Write the dimensions of the solid e.g. length, width, height, radius etc.
Step 3) Apply the formula for particular solid geometry figure.
Step 4) Write the unit of the particular physical quantity e.g. square meters, cubic centimeters etc.

Circles and semi circles

Important definitions related to circles:

1. Circle: A circle is a simple closed curve all of whose points are at a constant distance from a fixed point in the same plane. The fixed point is called the centre of the circle.

2. Circumference of a circle: The distance right around the circle is called its circumference. It is the perimeter of the circle. The traditional method to measure this perimeter of a circle was using a thread or a rope long the circumference. However this method is not too practical for very large circular fields or pieces of land. Therefore for all practical purposes, the following formula was derived by mathematicians for circumference of a circle.
 C = pi*d. Where, C = circumference of the circle, d = diameter of the circle and pi = ratio of the circumference of a circle to its diameter. The value of the Greek letter pi (read as pi) was experimentally calculated by mathematicians. It is an irrational number. A decimal of nonrecurring type. It is a constant. Indian mathematician Ramanujan gave two approximations for the value of pi in the year 1914. Generally all mathematicians have accepted the value of this constant as 3.141 592 653 589 793....

3. Semi circle: A diameter divides a circle into two equal parts which are called semi circles. The length of the curved portion of a semi circle is equal to half the circumference of the circle. The total perimeter of a semi circle is equal to sum of half the circumference and diameter. So putting that mathematically, perimeter of a semi circle = P,
P = C/2 + d, where C = circumference of a circle with diameter d.
From any point on the semi circle if we draw two lines that meet both the ends of the diameter, the angle so formed is called angle in a semi circle. This angle in a semi circle is always a right angle.

4. Unit of a circle (or unit circle): A circle which has a unit radius is called a unit circle. In other words a circle with radius = 1 and diameter = 2 is called a unit circle.

5. Intersecting circles: If there are two circles in a plane then any of the following three possibilities are there:
(a) The circles do not touch or intersect each other at all.
(b) The circles touch each other in exactly one point.
(c)  The circles intersect each other in exactly two points.

Beginners Guide to Geometry of Circle

A branch of Mathematics, Geometry is a study of the size, shape and position of two and three dimensional figures. Geometry of a Circle is a study of a circle, its parts and its properties. A math circle is an important and special figure and as such its parts have special names. Circle in Geometry is a planar figure in which all points are equidistant from a fixed point. This fixed point is called the centre of the circle. A segment with one endpoint at the centre of the circle and the other endpoint on the curve of the circle is a radius; the plural of radius is radii.

A segment whose endpoints lie on the circle is called the chord. A chord that passes through the centre of the circle is called the diameter of the circle. Let us learn more about Geometry circle, there are special lines and line segments in a circle; like secant, tangent and point of tangency.  Any line that contains a chord is called a secant. A tangent of a circle is a line in the plane of the circle that intersects the circle in exactly one point. The point where the tangent intersects a circle is called the point of tangency. Let us now learn a bit about tangents of a circle, a line that is tangent to two circles in the same plane is a common tangent. A common tangent that intersects the segment joining the centers of two circles is an external common tangent.

Now that we have a brief introduction to a circle and its parts, let us learn about the geometry circle formula. The major formulas in circles are as given below:
• Diameter is twice the radius. d=2r
• Circumference of a circle is the distance around the outer edge. It is like the perimeter of a circle. It is calculated using the formula 2 pi r, where r is the radius and pi is taken as 3.14
• Area of a circle is given by pi r2, r is the radius and pi value is taken as 3.14
• An arc is a part of circle. Length of an arc can be calculated using the formula, (a/360)x 2 pi r
• A sector is a portion of a circle bounded by two radii and the arc joining the radii. Area of a sector in degrees is, (sector angle/360) pi r2 and in radians it is (sector angle/2) r2

Know more about the solid geometry, Math Homework Help. This article gives basic information about geometry circle. Next article will cover more Geometryconcept and its advantages,problems and many more. Please share your comments.

Different kinds of graph

Lets learn kinds of graph below. We have learn t what is a bar graph a few days back.

Look below for the kinds of graph

Kind 1: Pictograph.

Kind 2: Bar graph.

Kind 3: Line graph.

Kind 4: Scatter plot.

These are some important graphs in mathematics, we will learn

Absolute values - Inequation and complex number

The absolute value of an integer is the numerical value of  the integer regardless of its sign .the absolute value of any integer say , a is denoted by |a|.On the number line the absolute value of an integer is regarded as the distance from  a point irrespective of its sign. The absolute value of a integer is always positive .
Some Absolute value examples are |-5| = 15 , |13|= 13.

Complex number

Absolute value inequality
To understand absolute value inequality , we will take few examples.

Example1 : |3x| ≤ 6
To  solve absolute inequality , here  we will use the absolute inequality results
|x|≤ a  =>  -a ≤  x ≤  a

=>  |3x|≤ 6
=>  -6 ≤  3x ≤ 6
=> Divide both sides by 3, we have
=>  -2 ≤  x ≤  2

Absolute value of complex number.

If a , b are two real number , then a number  a+ ib is  called as complex number.

Real and imaginary part of complex number : if z = a+ib  is a complex number , then a is called the real part of z and b is known as the imaginary part  of z . The real part of z is denoted by Re(z) and imaginary part is denoted by Im(Z).

Complex Number

The plane in which we represent a  complex number  geometrically  is known  as complex plane or argand plane or the Gaussian plane The point Z plotted on the argand plane is called the argand diagram.The length of the line segmemt 0z is called the absolute  of  complex number z and is denoted by |z|.thus |z|=√x²+y ²

= √(Re(z))²+(Im(z))²  In the above given figure z = 3+ 3i , so absolute of z , |z| = √3²+3 ²

=√18 = 3√2

Absolute value equations and inequalities

Now let us solve absolute value equations

Example1: Solve absolute value equation|x+ 5|= 4

Solution : For solving absolute value equations we will consider two cases

=> x + 5 = 4 or x + 5  = -4

=> x= -1 or x =-9 ans

Example 2: Solve the absolute inequality |x-2| ≥ 5

Solution : For solving absolute inequality , we will use the result

|x-a|≥ r => x ≤ a-r  or x ≥ a+r

=> |x-2| ≥ 5 => x ≤ 2-5  or x ≥ 2+5

=> X ≤ 3  or x ≥ 7

=> X ∈ ( -∞ , -3] or x ∈ [ 7 ,∞)

The solution set  of absolute inequality  is ( -∞ , -3] ∪ [ 7 ,∞)