Geometry Theorem List

Geometry is the study of shapes and configurations. It attempt to understand and classify spaces in various mathematical contexts. For a space with lot of symmetries the study naturally focuses on properties which are invariant (remaining the same) under the symmetries. Other type of geometry-In general, any mathematical construction which has a notion of curvature falls under the study of geometry.

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Examples of Geometry Theorem List


Differential geometry: which is the  natural extension of calculus and linear algebra and is known simplest form of vector calculus
Algebraic geometry: This studies objects defined by polynomial equations. This is vital to recent solution is  many difficult problem in number theory, such as the finiteness of solutions to the polynomial equations considered in Fermat's Last Theorem.
Semi-Riemannian geometry: which Einstein is used to study the four dimensional geometry of space and time.
Simplistic geometry: which originated with the study of the evolution of simple mechanical systems, but now pervades all aspects of theoretical physics.



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Lines of Geometry Theorem List


Lines  A line is one of the basic terms in geometry. We may think of a line is a "straight" line that we might draw with a ruler on a piece of paper, except that in geometry, a line extends forever in both directions. We write the name of  line  is passing through two different points A and B as "line AB" or as the two-headed arrow over AB signifying a line passing through points A and B. Points-A point is one of the basic terms in geometry. We may think of a point as a "dot" on a piece of paper. We identify this point with a number or letter. A point has no length or width. It just specifies an exact location. Intersection The term intersect is used when lines, rays, line segments or figures meet, that is, they share a common point. The point they share is called the point of intersection. We say that these figures intersect.

Solving Geometry Examples

Introduction of solving geometry examples:-

In geometry, we deal with the problems in triangles, circle, and square are solved using certain formulas is called Solving geometry examples. Here the formulas are very important to solve any examples problems. From this we can find area, volume, and perimeter etc.

To solving the geometry examples are,

Area of triangle formula =½(bh)

Area of square formula = a²

Area of circle formula = πr²

Area of rectangle formula = l*b


Examples for solving Geometry Problems


Example 1:

Find the area of square given that a= 35?

Solution:

Area of square = a²

= 35²  (calculate the area square )

= 1225

Area of square =1225

Example 2:

Find the area of triangle given that base = 21, height = 18?

Solution:

Area of triangle = ½(bh)

= ½(21*18) ( multiply the values)

= ½(378)

= 189

Area of triangle = 189

Example 3:

Find the area circle given that diameter = 26?

Solution:

Area of circle = πr²

But the radius is not given here; we have found the radius from diameter.

Radius = diameter /2

Radius = 26/2 ( dived the values)

Radius = 13

Area of circle = πr²

Area of circle = π*13²

=3.14*169 ( multiply the values)

Area of circle = 530.66sq.m.

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Examples for solving Geometry Problems


Example 4:

Find the length of the rectangle given that area = 155 and base = 23?

Solution:

Area of rectangle = l*b

155 = l*23

l= 155/23

l = 6.7

l = 7

Example 5:

Find the dimension of 3rd side. When the perimeter is fifty and the two sides are same. The equal sides have five greater the 3rd side?

Solution:

Let us assume s be the unknown length of the triangle, then

We know that perimeter of triangle is

P = a+b+c

50 = s + s + s+ 5

50 = 3s + 5

Then solving  the value of variable s, we get
3s = 50 – 5
3s = 45
s =15

The value of third side = 15 + 5 =20

The value of the third side is 20

Solve Online Geometry Problems

Introduction to solve geometry problems online:

Geometry is one of the important topics in math which includes the study of all kinds of shapes and their properties. Points is one of the basics of geometry. There is no length, height or width.  Points have  four main properties that is exact location, dot, node and ordered pair. Geometry deals with plane geometry, solid geometry. Let us see about solve online geometry word problems.


Solve the geometry word problems online


Q 1: A triangle has a perimeter of 78. If 2 of its sides are equal and the third side is 6 more than the equal sides, what is the length of the third side?

Sol:

Step 1: Let us take Y is the length of the equal sides of triangle

So, the third side of the triangle is Y+6

Step 2:  Perimeter is derived by sum of three sides on triangle.

Step 3: To Plug in the values of above problems.

78= Y+Y+(Y + 6)

Then Combine the similar terms
78 = 3y + 6

3y = 78 – 6
3y = 72
y =24

That is the third side is 6 more than the equal sides.

So, the length of third side = 24 + 6 =30

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More examples to solve geometry word problems online


Q 2 : The perimeter of a rectangle is 1000 meters and its length L is 6 times its width W. Find W and L, and the area of the rectangle.

Sol:

Step 1: Using formula for perimeter of the rectangle.

2 L + 2 W = 1000 ----------- (1)

Step 2: Here the length (L) of rectangle is 6 times more than its width (W)

L = 6 W --------------------- (2)

Step 3: To plug L value in the equation (1).

2(6 W) + 2 W = 1000

Step 4: 12W+W=1000
14 W = 1000

Step 5: Now Find the value of W.

W = 1000 / 14

W = 71.42 meters

Step 6: To plug W value in to equation (2)
L = 6 W

= 6 * 71.42 meters

= 428.57 meters

Step 7:  Area of a rectangle = L * W

L * W = 428.57 * 71.42

= 30608.57 meters 2.

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.

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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.

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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.

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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

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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