Solving Geometry Homework

Geometry:

Geometry is the main branches of mathematics. The geometry different types of size, shape, relative position of figures, and the properties of space. Geometry is one of the oldest sciences. The word ‘Geometry’ means learn of properties of figures and shapes and the relationship between them. A system of geometry is called Euclidean geometry. A solid geometry classified to a set of problems or objects.

(Source-Wikipedia)


Solving homework 1:


To calculate the area of a cylinder given that its radius is 8 and its length or height is 6.

Solution:

The surface area of this cylinder is 2 `pi` RL+2 `pi` R2.

= 2*3.14*8*6+2*3.14*82=703.717

The surface are of a cylinder is 703.717

Solving homework 2:

To calculate the perimeter of a rectangle given that its width is 7 and its height is 6.

Solution:

The perimeter is the distance around the rectangle, or h+w+h+w or 2h+2w.

Perimeter = 2 * 6 + 2 * 7 = 26

The perimeter of rectangle is 26


Solving homework 3:


To calculate the area of a rectangle given that its width is 8 and its height is 7.

Solution:

The area enclosed by rectangle, is h × w

Area = 7 * 8 = 56

The area of a rectangle is 56

Solving homework 4:

To calculate the area of a right triangle given that its base is 10 and its height is 8.

Solution:

The area of a right triangle is 1/2bh

Area = ½ * 10 * 8 = 40

The area of a rectangle is 56

Solving homework 5:

To calculate the side of a square given that its area is 8.

Solution:

The area is the amount of space enclosed by the square is S × S or Area=S2.

Solve this equation for S to get that or S=Area1/2 Side = area ½=81/2=2.82843

The side of this square has a length of 2.82843

Solving homework 6:

To calculate the area of  rhombus whose diagonal lengths are  8 and 12.

Solution:

The area of the rhombus= 1/2  x length of the diagonal 1 x length of the diagonal 2 .

=1/2 x 8 x 12

The area of the rhombus = 48

Solving Geometry Perimeters

solving geometry perimeters:

Geometry is the study of all kinds of shapes and their properties.

Plane geometry  is the study of two dimensional shapes such as lines, circles, triangles etc.

Solid geometry is the study of three dimensional shapes like polygons, prisms, pyramids, sphere, cylinder, cone etc.

Perimeter is defined as the total distance around the outside of a 2D shape. Perimeter can be calculated by adding all the lengths along the periphery of a shape.

Let us see solving geometry perimeters in this article


Solving Geometry Perimeter


The Perimeter of rectangle P is the Addition of two times length l and two times width w.

Formula:           P = 2 * length + 2 * width

Example:

Find the Perimeter of rectangle for the length is 6 cm and Width is 5 cm.

Solution:

Perimeter of the rectangle          P =  2 * 6  + 2 * 5

P = 12 + 10

P = 22 cm. Answer.

Perimeter of square:

The Perimeter of any square P is the product of 4 and a side.

Formula:                P = 4 * side

Example:

Find the Perimeter of square for the side is 4 cm.

Solution:

Perimeter of the square A = 4 * 4

A = 16 cm. Answer.

Having problem with Surface Area of a Prism keep reading my upcoming posts, i will try to help you.

Solving Geometry Perimeter for parallelogram and triangle


Perimeter of Parallelogram:

The Perimeter of any Parallelogram is the addition of 2 times side a and 2 times side b.

Formula:           P = 2 * side a + 2 * side b

Example:

Find the Perimeter of Parallelogram for the side a is 7 and side b is 9 and height is 5.

Solution:

Perimeter of Parallelogram        P = 2 * 7 + 2 * 9

P = 14 + 18

P = 32 . Answer.

Perimeter of Triangle:

The Perimeter of Triangle P is the Addition of all the three sides.

Formula:           P = AB + BC + AC for triangle ABC

Example:

Find the Perimeter of triangle for the AB = 7 cm , BC = 5 cm and AC = 7 cm.

Solution:

Perimeter of the triangle            P =  7 + 5 + 7

P = 19 cm. Answer.

Solving Geometry Problems Online

Introduction to solving geometry problems online:

Geometry is a part of math which involves the study of shapes, lines, angles, dimensions, etc. it plays vital role real time application like elevation, projection. Learning geometry provides many foundational skills and helps to build the thinking skills of logic, deductive reasoning, and analytical reasoning. Flat shapes like lines, circles and triangles are called the Plane Geometry. Solid (3-dimensional) shapes like spheres and cubes are called Solid geometry. In this article we shall discuss about solving geometry problems online.

I like to share this Surface Area of a Hemisphere Formula with you all through my article.

Sample problems for online geometry solving:


Example 1:

The perimeter of a rectangle is 800 meters and its length L is 3 times its width W. Find W and L, and the area of the rectangle.

Solution:

Perimeter of rectangle=2L+2W,

2 L + 2 W = 800

We now rewrite the statement. Its length L is 3 times its width into a mathematical equation as follows:

L = 3 W

We have to substitute L =3W in the equation 2 L + 2 W = 800

2(3 W) + 2 W = 800

8 W = 800

W =100 meters

Use the equation L = 3 W to find L.

L = 3 W = 225 meters

Use the formula of the area.

Area = L x W = 225 * 100 = 22500 meters 2.

So, the area of the rectangle=22500 meters 2.

Example online geometry solving problem 2:

A perimeter of the triangle is 50cm. If 2 of its sides are equal and also the third side is 5cm more than the equal sides, find the length of the third side?

Solution:

Let x = length of the equal side.

Third side=5 more than the equal side=x+5

So, the three sides are x, x and x+5.

P = sum of the three sides

x+ x+(x+5) =50

Combine like terms

3 x + 5=50

3x = 50 – 5

3x = 45

x =15cm (equal sides)

Length of the third side=x+5=15+5=20cm

The length of third side is 20cm.

Example online geometry solving problem 3:

A circle has an area of 100pi square units. What is the length of the circle's diameter and circumference?

Solution:

Area of the circle

A = (pi)*r^2

100pi = (pi)*r^2

(100pi) / pi = [(pi)*r^2] / pi

100= r^2

10 = r

So, the radius=10units

Diameter=2(radius) =20 units

Circumference= (pi)*d

=20pi units (or)

Substitute the value of pi=3.14

=62.8units

Circumference=20pi units (or) 62.8 units.

Understanding Quadratic Equation Calculator is always challenging for me but thanks to all math help websites to help me out.

Practice problems for online geometry solving:


Problems:

A circle has an area of 80pi square units. What is the length of the circle's diameter and circumference?
Answer: 16 pi units.

A circle has an area of 60pi square units. What is the length of the circle's diameter and circumference?
Answer: 12 pi units.

Solving 5th Grade Geometry Problems

Introduction to 5th Grade geometry problems:

Geometric is construction of object form given measurement.5th grade geometry contains the topic of  Understanding what is meant by point, line, and plane Identifying acute, right, obtuse, and straight angles as well as complementary angles, supplementary angles, and vertical angles Identifying parallel lines and perpendicular lines. 5th grade geometric contains the content of triangle, square, circle, rectangle, used to finding the area and volume of the given figure.


Solving Example problems in 5th grade geometry:


5th grade geometry Problems in angle:

Example 1:

1. Which of the following can be the angles of a triangle?

(a)90°, 45°, 55° (b)30°, 70°, 80°(c) 45°, 80°, 50°

Solution:

(a)Sum of the three angles = 90° + 45° + 55° = 190° > 180°

Hence, these cannot be the angles of a triangle.

(b)Sum of the three angles = 30° + 70° + 80° = 180°

Hence, these can be the angles of a triangle.

(c)Sum of the three angles = 45° + 80° + 50° = 175° < 180°

Hence, these cannot be the angles of a triangle.

5th grade geometry Example Problems :

Example 2:

2. find the area of the triangle, its base =4cm and height =6cm?

Solution:

Area of triangle=1/2(Base * height)

=1/2(4*6)

=2*6

=12cm

Problems in square:

A square is a closed figure made up of four line segments

Its side lengths are equal.

Area of square =(side * side)

Perimeter of square =4*side

Having problem with Volume of a Pyramid keep reading my upcoming posts, i will try to help you.

5th grade geometry Example Problems :


Example 3:

3.Find the area and perimeter of  square which side length is 5?

Solution

Side of the square=5m

The perimeter of the floor is given by P= 4×side

=4×5m = 20m

Thus, the perimeter is 20 m.

Area =side * side

=5*5

=25cm2

Problems in circle:

Example 4:

Length of the diameter = 2 × length of the radius

Find the diameter of the circles whose radii are:

(a)2 cm(b)5 cm (c)4.5 cm

Solution:

We have Diameter = 2 × radius

= 2×2cm

= 4cm

Diameter = 2 × radius

= 2×5cm

=10cm

Diameter = 2 × radius

= 2 × 4.5 cm

= 9 cm

Problems in rectangle:

Example 5:

Area of rectangle: Length * Width)

Perimeter of rectangle=2(length+ width

Find the area of rectangle which length is 4 and breath is 7?

Area of rectangle: Length * breath

=4*7

=28cm2

Introduction to Co-ordinate Geometry

Introduction to co-ordinate geometry:
Co-ordinate geometry is a branch of Mathematics that studies about points, lines and geometrical figures using co-ordinate systems. In geometry, we study the same using geometrical constructions and actual measurement but in co-ordinate geometry it is predominantly using co-ordinates of points.

Some of the topics covered in co-ordinate geometry are  finding distance between two points, slope of line, equations of lines, circles and geometrical figures etc

Let us solve some of co-ordinate geometry problems to get a feel of the subject

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Solve Points Problems of Co-ordinate geometry:


Problem 1:

Find the slope of the line for these two points. ( 5 , 6 ) and ( -3 , 9 )

Solution:

The slope formula is given by

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

Given two points: (x1, y1) =  (5, 6) and (x2, y2) = (-3, 9)

Apply these two points into that formula for finding slope.

m = `(9-6)/(-3-5)`

m = `3/-8`

m = `- 3/8`

The slope of these two points is `-3/8` .

Problem 2:

Solve the distance for the given two points (5, 4) and (4,6)

Solution:

The distance formula is given by

d = `sqrt((x2-x1)^2+(y2-y1)^2)`

d = `sqrt((4-5)^2+(6-4)^2)`

d = `sqrt((-1)^2+(2)^2)`

d = `sqrt(1+ 4)`

d = `sqrt(5)`

The distance for these points is 2.23.


Solve Lines problem of Co-ordinate geometry:


Problem:

Solve the equation of a line between these two points (3, 8) and (-6, 4).

Solution:

Line equation form is y = mx + b

Solve m, slope between these two points.

The slope formula is given by

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

Given two points: (x1, y1) =  (3, 8) and (x2, y2) = (-6, 4)

Apply these two points into that formula for finding slope.

m = `(4-8)/(-6-8)`

m = `-4/-14`

m = `4/14` ---------------Simplify it.

m = `2/7`

Solve b, y intercept

For one point (x1, y1) = (3,8), the line equation becomes

y1 = mx1 + b

8  = `2/7` (3) + b

8  =  `6/7`   + b

b = 8 – `6/7`

b = `50/7`

Substitute m and b into line equation, we get

y = mx + b

y = `2/7` x + `50/7`

y = `1/7` (2x+50)

Multiply by 7 both on sides,

7y = 2x + 50

2x – 7y +50 = 0.

So the equation of line between these two points is 2x – 7y + 50 = 0.


Solve Circle Problem of Co-ordinate geometry:


Problem:

Find the center and radius of (x – 5)^2 + (y – 7)^2 = 25 circle.

Solution:

The circle equation form is (x-h)^2 + (y-k)^2 = R2.

Here the center is (h, k) and Radius is R.

The given equation looks the same as circle equation form.

(x-5)^2 + (y-7)^2 = 52

From the given equation, the center is (5, 7) and Radius is 5.

Solve Geometry Placement Test

Introduction to solve geometry placement test:

The division of math which deals with the measurement of lengths, angles, areas, perimeters and volumes of plane and solid figures is called geometry.we all knew placement is our dream in college life.In placement test, geometry plays an important role.Here solved geometry placement test papers with solutions given for your practice. Sample geometry placement test paper were given solve this test on your own without the help of a calculator, book, notes, or other people.


Solve geometry placement test:


Example 1:

A room inner space of diameter 150 cm has a wall around it. If the length of the outer edge of the wall is 60 cm, then find the width of the wall.

Solution:

Diameter of the room = 150 cm

Radius = 150 / 2 = 75 cm

Let width of wall = x cm then total radius = (75 + x) cm

Outer edge of the wall = 2  pi  (75 + x) = 44/7  (75 + x) cm

But outer edge of wall = 660 cm

44/7 (75 x) = 660

75 + x = 660  7 / 44

= 105

X = 105 - 75

X = 30 cm

Example 2:

Find number of times will the wheel of a car rotate in a journey of 76 km if the diameter of the wheel is 36 cm?

Solution:

Diameter of the wheel = 36 cm

The Circumference of the wheel of diameter  =  D = 22 / 7  36 = 113 cm

Length of the journey = 76 km = 76*1000*100 cm

Number of times the wheel will rotate in covering the above journey

= `7600000 / 113`

= 67,256.63 .

Please express your views of this topic Volume of Triangular Prism by commenting on blog

Solve geometry placement test:Examples


Example 3.

Find the area of a rectangle of length = 9 cm,breadth = 6 cm.

Solution:

Length and breadth is given here,

we need to find the area ,

Area = l × b sq.units

=` 9 * 6` sq.cm

= 54 sq.cm

Example 4.

Find the perimeter of a rectangle of length = 6 cm,breadth = 5 cm.

Perimeter = 2 (l + b) units

Perimeter  =  2 (6 + 5) units

= 22 cm.

Example 5:

A piece of thin wire which is circular, converted into a square of side 7cm. find the radius of circular wire.

Solution:

Side of a square = 7 cm

Its perimeter = `4* side` = `4 * 7` cm = 28 cm

Circumference of the circular wire = 28 cm. we know that c =`2*pi*r`

28 =` 2*(22/7)*r`

r = `(28*7)` /` (2* 22)`

= `196 / 44`

r = 4.454 cm

Transformations Geometry

Introduction for transformations geometry:

The transformation Geometry is a copy of a geometric figure, where the copy holds some certain property. The original shape is called the pre-image the new picture is called the image of the transformation. A transformation is single in which the pre-image and the figure equally has the exact same dimension and shape. I like to share this Definition of Parabola with you all through my article.


Basic Transformation Geometry:


The two types of transformation geometry is given by

Rigid transformations

Non-rigid transformations.

This page will covenant with three rigid transformations known as translations, reflections and rotations.

About geometry transformations:

The main geometry transformations in the mathematics are given as,

Translations

Reflections

Rotations

Scaling

Shear

Translations:

The mainly basic transformation is the translation. The definition of a translation is the pre-image and then it can be moved to the equal distance in the same direction to form the image .The transformation is would be

T(x, y) = (x+7, y+4).

Reflections:

The reflection is a "flip" of an aim over a line.

The two very common reflections is given by

horizontal reflection

vertical reflection.

The line of reflection will be both red points, blue points, and green points. The line of reflection which is directly in the center of both points. Having problem with Surface Area of a Circle keep reading my upcoming posts, i will try to help you.


Other types of Transformations:


Rotations:

The transformations which are performed by spinning the object just about a point of the center rotation .You can able to change your object at some of the degree measure, but 90° and 180° are very important degrees.

Rotation 180° around the origin: T(x, y) = (-x, -y)

Scaling:

The scaling is a linear transformation which diminishes the objects and the scale factor is same for direction is called scaling. The resultant image of the uniform scaling is similar to the original transformations

Shear:

The Shear which transforms effectively to rotate one axis and that the axes are no longer at right angle. A rectangle becomes a parallelogram, and a round becomes an ellipse. Constant lines parallel to the axes continue the same length, others do not. As a plot of the plane, it deception in the class of equilateral mappings.