6th Grade Geometry Problems

Introduction:

Sixth grade geometric contains the basic of geometricals .It includes the topic of geometric in Points, Lines ,Line segment, Triangles, Types of triangles, circles, Angles, Types of Angles, Quadrilaterals

Geometric Definitions:

Point: A point   determines the location of particular area.

Line:   A line through two points A and B is written as AB.. It extends

Indefinitely in both directions. So it contains countless number of points. Two points are enough to fix a line

Types of lines:

Intersecting lines
Parallel lines
Perpendicular lines

Triangles in Geometry:


Triangles:

A triangle is a three-sided polygon. In fact, it is the polygon with the least number of sides

Types of Triangle:

Equilateral Triangle
When all the three sides of a triangle are equal to each other, it is called an Equilateral triangle. Each angle measures to 60 degrees. It is a type of regular polygon.

Isosceles Triangle
When two sides of a triangle are equal it is called an Isosceles triangle. It also have two equal angles.

Scalene Triangle
When no two sides of a triangle are equal the triangle is called Scalene triangle. It has three unequal sides.

Area of triangle: 1/2(Base*Height)

Perimeter of Triangle: (Sum of three sides)

Example problem:

1.Find the area of triangle base is 4cm,height is 2cm

Solution:

Area=1/2(4*2)

=8/2

=4cm2

2.Find perimeter of Triangle side lengths are 5cm,5cm,8cm

Solution:

Perimeter=(A+B+C)(Sum of three side lengths)

A=5, B=5, C=8

=(A+B+C)

=5+5+8

=18cm


Angles and Circle in Geometry


Angle:

Right Triangle
. Right angle is equal to 90 degrees. It obeys Pythagoras theorem.

Acute angle
. Acute angle is an angle which is less than 90 degrees.

Obtuse angle
An Obtuse angle is an angle which is greater than 90 degrees but less than 180 degrees.

Acute angle:
Acute and Obtuse triangles are also called as Oblique triangles because they don’t have any angle measuring 90 degrees.

Quadrilateral:
A four sided polygon is a quadrilateral. It has sides and 4 angles

Circle:

Are of circle=Pi*r*r

Circumference of Triangle=2*Pi*r

Diameter=2*Radius

Example:

Find the area  and circumference of the circle when the radius is 4cm?

Solution:

1.     Area=Pi*r*r (r=4) (Pi=3.14 constant)

=3.14*4*4

=50.24cm2

2. Circumference =2*pi*r

=2*3.14*4

=25.12cm

Answers to Geometry Homework

Introduction to answers to geometry homework:

Learning geometry has traditionally been regarded as important in the secondary schools, at least partly because it has been the primary means of teaching the art of reasoning.

Geometry is a theoretical subject, but easy to understand, and it has many real practical applications. Eventually, geometry has evolved into a skillfully arranged and sensibly organized body of knowledge. I like to share this Triangular Prism Net with you all through my article.


Part 1 -answers to geometry homework:


Geometry homework example 1:

If the perimeter of a cube is 52.5 ft, find its surface area.

Geometry homework solution:

Perimeter of a cube P=12a

52.5=12a

a=52.5/12

a=4.375 ft.

So, the value of a=4.375 ft.

Surface area of a cube SA=6a2

=6(4.375) 2

=6(19.14) ft2

=114.84 ft2

Answer of example 1: Surface area of cube   = 114.84 ft2

Geometry homework example 2:

A barrier of length 15 m was to be built across an open ground. The height (h) of the wall is 5 m and thickness of the barrier is 32 cm. If this barrier is to be built up with bricks whose dimensions are 25 cm × 18 cm × 12 cm, how many bricks would be required?

Geometry homework solution:

1 m=100 cm

Here, Length = 15 m = 1500 cm

Thickness = 32 cm

Height = 5 m = 500 cm

Therefore, Volume of the barrier = length × thickness × height

= 1500 × 32 × 500 cm3

Now, each brick is a cuboid with length = 25 cm, breadth = 18 cm and height = 12 cm

So, volume of each brick = length × breadth × height

= 25 × 18 × 12 cm3

So, number of bricks required =volume of the barrier divided by volume of each brick.

Substituting the values,then we get the final answer.

= (1500 × 32 × 500)/ (25 ×18 × 12)

= (24000000)/5400

=4444.44

Answer: The barrier requires 6416 bricks.

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answers to geometry homework:


Geometry homework example 3:

A line passes through (–3, 4) with a slope of -1/5. If another point on this line has coordinates (x, 2), find x.

Geometry homework solution:

Slope m= (y2-y1)/(x2-x1)

-1/5= (2-4)/ (x-(-3))

-1/5= (-2)/(x+3)

We can take cross multiplication.

-1(x+3) =5(-2)

-x-3= -10

-x=-10+3

-x=-7

In both sides cancel for the negative sign and then we get the final answer.

x =7

Answer: The x value is 7

Positive and Negative Angles

Introduction to Positive and Negative Angles:

An ANGLE is strong-minded by rotating a ray about it's endpoint. The initial location of the ray is the INITIAL SIDE of the angle, and the ending position of the ray is its TERMINAL SIDE. The endpoint of ray called the VERTEX.

The unit circle of the angle is said to be in STANDARD POSITION because its vertex is the origin, and its initial side lies on the x-axis. This is also called positive angle, meaning it's created by a COUNTERCLOCKWISE rotation.
The unit circle of angle is in standard position, but it's called a NEGATIVE ANGLE, since it is created by a CLOCKWISE rotation. I like to share this Inscribed Angles with you all through my article.


Rules of Positive and Negative angles:

Positive Angle:

An angle formed by anti-clockwise rotation is a positive angle. In the figure initial side is OX. When these side is rotated by an angle θ in counter clockwise direction then angle is generated is called positive angle.

Negative Angle:

An angle generated by clockwise rotation is a positive angle. In these diagram let the initial side is OX. When these side is rotated by an angle θ in clockwise direction then angle called as negative angle.

Rule I:

Sign of an angle is always positive when measured in anti-clockwise direction.

Rule II:

Sign of an angle is always negative when measured in clockwise direction. Understanding Volume of Right Prism is always challenging for me but thanks to all math help websites to help me out.


Example of Positive and Negative Angles:

Positive and Negative Angles:

Positive Angles start from 0 degrees and turn around counterclockwise.

Negative Angles start from 0 degrees and turn around clockwise.

You can translate your negative angle to its equivalent positive angle by adding 360 degrees to it until it turns positive.

Once it is positive, you can pleasure it the same as you would any other positive angle in the quadrant that it is in.

Example 1:

Angle is -135 degrees.

sum  360 degrees to it until it turns positive.

It turn positive then first time we add 360 degrees to it.

The equivalent is positive angle is 225 degrees.

It is in the quadrant of 3.

Example 2:

Is 300 not same as -300?

Solution:

The answer to this question is NO. Here why the angle have  two attributes attached to it: Degree of rotation (or magnitude of rotation) and Direction of rotation (clockwise or anticlockwise). While those wo angles have same degree of rotation, direction of rotation is just opposite as signified by there opposite signs. Therefore those two angles are different.

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.