Angle of Rotational Symmetry

Introduction to angle of rotational symmetry:

Rotational symmetry is definite as angle, while we rotate as alternate a shape in its center point; you may notice that at a certain angle, the shape coincides with its not rotated itself. When this happens, the shape is said that it have rotational symmetry. A shape has rotational symmetry if it fits on to itself two or more times in one turn. The number of times the rotational symmetry is the shape fits on to itself in one turn. Is this topic Straight Angle Definition hard for you? Watch out for my coming posts.


Type of symmetry:

Symmetry has,

Line symmetry
2D rotational symmetry
3D rotational symmetry
A 2D shape has a line of symmetry if the lines separate the shape into two share equally – one being the mirror image of the other.

Rotation: whatever rotate the shape on it around, every rotation has depending on center point and an angle.

Translation:  Translation is to be in motion without rotating or reflecting, every translation has depending on distance and

a direction.

Reflection: reflection is seemed mirror image. Every reflection has a mirror line.

Glide Reflection: With the direction of the reflection line, glide reflection is the symmetry of its collection of reflection and  translation.

Angle of rotation:


When you can turn a figure around a center point by less than 270° and the figure appear that there is no changed, and then the figure has rotation symmetry. The middle of rotation, and then the smallest angle require turning and the point around which you rotate is called the angle of rotation.

For instance take any figure on the left can be turned by 160 degree the same way you would turn an hourglass and it look like the same. Her take 2nd figure one can be turned by 120 degree and other one is 72 degree. The 2nd figure comes from 72 degree that it has five points, just rotate it until it looks the same; we need to make 1 / 5 of a completed 360 degree. So 1 / 5 * 360 degree = 72 degree.

Geometry Book Answers

Introduction:

Geometry is nothing but the act of determining the dimensions or volume of an object. Buildings, cars, planes, maps are of great examples of geometry. The plane geometry, normally used in finding the area, perimeter, circumference of  Two-dimensional figures like triangle, circle, rectangle, rhombus, trapezoid, quadrilateral etc.  Let us solve some sample problems from plane geometry. Understanding Geometry Lines is always challenging for me but thanks to all math help websites to help me out.


Sample Geometry Problems:


Example 1:
The radius of a right circular cylinder is 7 cm and its height is 20 cm. Find its curved surface area and total surface area.

Solution:
Radius(r) = 7cm, Height(h) = 20 cm.


Curved surface area = (2`pi`r h )= 2 * 22 / 7 * 7 * 20 cm^2

= 880 cm^2

Area of the base and the top    = (2`pi`r2)

= 2 * (22 / 7) * 7 * 7 cm^2

= 308 cm^2

Total surface area = (2`pi`r h) + (2`pi`r2) = 880 cm^2 + 308 cm^2

= 1188 cm^2

Example 2:

Surface area of a right circular cylinder of height 35 cm is 121 cm^2. find the radius of the base.

Solution:

Curved surface area = (2`pi`r h) = 121 cm^2

2*(22/7)* r *35 = 121

r = (7*121)/(2*22*35)

= 11/20

= 0.55 cm

Hence, radius of the base = 0.55 cm.

Example 3:

Curved surface area of a right circular cylinder is 6.6 m^2. if the radius of the base of the cylinder is 1.4 m , find the height.

Solution:

Curved surface area = (2`pi`r h) = 6.6

= 2 * (22/7) * 1.4 * h

= h = (7*6.6) / (2*22*1.4)

= (7*66) / (2*22*14)

= 3/4 = 0.75

Required height of the cylinder = 0.75 m.

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Geometry Practice Exam Problems:


1. the circumference of the base of a right circular cylinder is 220cm. if the height of the cylinder is 2m, find the lateral surface area of the cylinder.

2. A closed circular cylinder has diameter 20cm and height 30 cm. find the total surface area of the cylinder.

3. the radius of the base of a closed right circular cylider is 35 cm and its height is 0.5m. find the total surface area of the cylinder.

Answers:

1. 4.4 m^2

2. 2512 cm^2

3. 18700 cm^2

Construction of Triangles

Introduction to construction of triangles:

The triangles can be constructed if the following requirements are given such as follows,

The measurement of three sides should be given (or)

The measurement of the two sides and the included angle should be given (or)

The measurement of a side and any two angles should be given.

Now we are going to see about the construction of triangles. Is this topic Scalene Triangles hard for you? Watch out for my coming posts.


Construction of triangles:


Construction of triangles if three sides are given:

Construct a triangle if three sides are given with x, y and z measurements.

Steps of construction:

First a line segment QR of x cm length should be drawn.

With Q as center and radius of y cm be drawn and it equals to PQ and draw an arc of a circle.

With R as center and radius of PR = z cm and draw an arc and it will intersects at the first arc of point P.

Now join the points of line segments PQ and PR.

Thus, PQR is a required triangle. I have recently faced lot of problem while learning Geometry Definition, But thank to online resources of math which helped me to learn myself easily on net.

Other constructions of triangles:

Construction of triangles if two sides and angle are given:

Construct a triangle if two sides and an angle are given.

Steps of Construction:

First we have to draw a ray of QX of some length.

With the help of protractor measure the given angle and draw the line to meet Q.

The ray QY which may cut line segment QR of x cm.

The ray QY which may cut the line segment QP of y cm.

Now we can join the two points P and R.

Thus, PQR is the required triangle.

Construction of triangles if two angles and Side are given:

Construct a triangle if two angles and a side are given.


Steps of Construction:

First we should draw the line segment of QR of given length.

With the help of the protractor measure the given angle at RQX

Then, draw QRY for the given angle such that XY lie on the same side of the PQ.

Then, label the point where it intersects at QX and QY as P.

Thus, the PQR is the required triangle.

Base Pentagon Prism

Introduction:-
In geometry, the pentagonal prism is a prism with a pentagonal base.

Pentagonal prism is a type of heptahedron.

If faces are all regular, the pentagonal prism is a semi regular polyhedron and the third in an infinite set of prisms formed by square sides and two regular polygon caps.

A pentagonal prism has                                                                                                                                                                         Source:- Wikipedia.

7 faces
10 vertices
15 edges.
The pentagonal prism looks like .


Formulas For Pentagonal Prism:-


`Base area = 5/ 2 * a * s`
`Base perimeter = 5s`
`Prisms Surface area = 5as + 5sh.`
`Volume of Prism = (5/2)ash`
`here`
`a = apothem Leng th,`
`s = side,`

`h = height.`

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Problems on Prism:-


Problem 1:-

Find the base area  and base perimeter of the pentagonal prism if the apothem length is 4 and the side is 2.

Solution:-

Given

The apothem, length is 4

The side is 2.

The formula that is used to calculate the area of the base is `5/ 2 a* s.`

By plugging in the values of  a and s in the formula we get

Area of the base = `5/ 2` * 4 * 2.

By crossing 2 and 2 in the above equation we get

Area of the base = 5 * 4 = 20.square units.

The formula that is used to find the base perimeter is  `5s` .

By plugging in the given values we get

Base perimeter = 5 * s = 5 * 2 = 10.

Problem 2:-

Find the base area  and base perimeter of the pentagonal prism if the apothem length is 6 and the side is 4.

Solution:-

Given

The apothem, length is 6

The side is 4.

The formula that is used to calculate the area of the base is` 5/ 2 a* s.`

By plugging in the values of  a and s in the formula we get

Area of the base = `5/ 2 * 6 * 4.`

By crossing 2 and 4 in the above equation we get

Area of the base = 5 * 12 = 60.square units.

The formula that is used to find the base perimeter is  `5s.`

By plugging in the given values we get

Base perimeter = 5 * s = 5 * 4 = 20.

Problem 3:-

Find the base area and base perimeter of the pentagonal prism if the apothem length is 5 and the side is .

Solution:-

Given

The apothem, length is 5

The side is 7.

The formula that is used to calculate the area of the base is `5/ 2 a* s.`

By plugging in the values of  a and s in the formula we get

Area of the base = `5/ 2 * 5 * 7.`

Area of the base = `175/ 2` square units.

The formula that is used to find the base perimeter is  5s.

By plugging in the given values we get

Base perimeter = 5 * s = 5 * 7 = 35.

angles at a point

Introduction (Angles at point):

In geometry an angle is the figure produced by two ray’s distribution a common endpoint, called the vertex of angle. The degree of the angle is the quantity of revolution that separates the two waves, and deliberate by considering the length of circular curve is out when one ray is rotate regarding the vertex to correspond with the other. The angle along with a line and a curve or along with two intersecting curve.


Positive and negative angles at a point:

In mathematical script is that angles specified a sign are positive angles if considered anticlockwise and negative angles ? is efficiently the same to a positive angle of one full rotation less ?. if considered clockwise, from a known line. If no line is specified, that can be understood to be the x-axis in the Cartesian plane. In many geometrical situations a pessimistic angle of ?? is efficiently the same to a positive angle of one full rotation less ?.

Example, a clockwise rotation of 45° (angle of ?45°) is efficiently the same to an anticlockwise rotation of 360° ? 45° (angle of 315°).

Types of Angles:

Right angle
Acute angle
obtuse angle
reflex angle
Vertical opposite angles
Co-responding angles and Alternative angles
Interior angle
Identifying angles:

Angles may be recognized by the labels involved to the three points to identify them. Example, the angle by vertex A with this by the rays AB and AC.

Potentially, an angle denoted,  ?BAC may refer to any of four angles: the clockwise angle from B to C, the anticlockwise angle from B to C, the clockwise angle as of C to B, or the anticlockwise angle as of C to B, wherever the way in that the angle is deliberate determines its sign.


Examples for angles at a point:


Example 1:

Find the value of x.

Solution:

x + 80° + 2x + x = 180° (contiguous angles on a straight line)

4x = 180° - 80°
= 100°

x = 100°
4
The answer of x = 25°

Example 2:

Find the value of x.

Solution:

48° + 90° + 120° + x = 360° ( Angles at a point )
x = 360° - (48° + 90° + 120° )
= 360° - 258°
The answer of x= 102°

Solve Allied Angles Axiom

Introduction to solve allied angles axiom:

The allied angles are nothing but the co-interior angles where they are transversely cuts the two parallel lines and the allied angles will be formed. The supplementary angles present in the geometry figures are also called as one of the types of allied angles. The allied angles total measurements are about 180 degrees. The tutors will describe the concepts to students with some example problems. Now we see how to solve allied angles axiom with the help of the tutor.

About How to Solve Allied Angles Axiom:

Now we see about how to solve for the allied angles axiom and its concepts with the help of the tutor. The allied angles are nothing but the angles where they are formed when two parallel lines are cut by a transverse line and the interior angles forms are called as the allied angles. The allied angles measurements are be about 180 degrees.

From the above figure it is given that as there are two parallel lines such as L1 and L2. These two lines can be cut by the traversal line T and the allied angles forms in the figure are given as a, b, c, d.

`angle a` +`angle d` = 180 degrees.

`angle b`  +`angle c` = 180 degrees.

The angle measures are given and we can determine the other angle with the total measurement of the angles are about 180 degrees.

Now we see some of the problems on allied angles axiom with the help of the tutor. Understanding Alternate Interior Angle is always challenging for me but thanks to all math help websites to help me out.

Problems to Solve Allied Angles Axiom:
Example:

Calculate the allied angles from the given figure?

Solution:

Now we see how to solve the allied angles measurement as follows,

The allied angles measurements are about 180 degrees.

From the given diagram, S1and S2 are the two parallel lines and cut by the transverse line T.

The allied angle are thus formed between the parallel lines.

For calculating the allied angle we have to do as follows,

Y + 70 = 180

Y = 180 - 70

Y = 110

Thus, the allied angles for 70 is about 110 degrees.

11th Grade Geometry

Introduction of 11th grade geometry :-

In grade (11) means eleventh grade is a year of education in all over the world. The eleventh grade is the final year of the secondary school. Students are usually 16 - 17 years old. Geometry is concerned with size, shape, relative figures etc. In 11th grade geometry lessons study about the angle, circle , quadrilateral etc.

Example Problems for 11th Grade Geometry :-

Problem 1:-

Determine the equation of the straight line passing through the points (1, 2) and (3, − 4).

Solution:

The equation of a straight line passing through two points is

`(y - y1)/ (y1 - y2)` = `(x - x1)/( x1 - x2)`


Here (x1, y1) = (1, 2) and (x2, y2) = (3, − 4).

Substituting the above, the required line is

`(y - 2)/(2 + 4)` = `(x - 1)/(1 - 3)`

`(y - 2)/6` = `(x - 1)/(- 2)`

`(y-2)/3` = `(x-1)/(-1)`

y − 2 = − 3 (x − 1)
y − 2 = − 3x + 3

3x + y = 5 is the required equation of the straight line.Is this topic AAA Postulate hard for you? Watch out for my coming posts.

Problem 2:-

Find the equation of the straight line passing through the point (1,2) and making intercepts on the co-ordinate axes which are in the ratio 2 : 3.

Solution:-

The intercept form is

`x/a +y/b` = 1 … (1)

The intercepts are in the ratio 2 : 3  a = 2k, b = 3k.

(1) becomes

`x/(2k) +y/(3k)` = 1     i.e. 3x + 2y = 6k

Since (1, 2) lies on the above straight line, 3 + 4 = 6k i.e. 6k = 7

Hence the required equation of the straight line is 3x + 2y = 7


Problem 3:-

Find the distance between the parallel lines 2x + 3y − 6=0 and 2x + 3y + 7 = 0.

Solution:-

The distance between the parallel lines is

`|(c_1 - c_2)/sqrt(a^2 + b^2)|` .

Here `c_1` = − 6, `c_2` = 7, a = 2, b = 3

The required distance is

`|(- 6 -7)/sqrt(2^2 + 3^2)|` = `| (-13)/sqrt(13)|`

= `sqrt(13)` units.


Practice Problems for 11th Grade Geometry :-

Problem 1:-

Find the equation of the straight line, if the perpendicular from the origin makes an angle of 120° with x-axis and the length of theperpendicular from the origin is 6 units.

Answer: The required equation of the straight line is x − `sqrt(3)` y + 12 = 0


Problem 2:-

Find the points on y-axis whose perpendicular distance from the straight line 4x − 3y − 12 = 0 is 3.

Answer: The required points are (0, 1) and (0, − 9).