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# Geometric Properties of Motion

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Introduction for geometry properties of motion:

The geometrical motion properties of an entity is must not depend on other entity. These motion entities are may be two-dimensional or three dimensional entity with dissimilar geometrical properties used for explaining. Each entity has dissimilar geometrical properties. The geometrical properties are depends on the nature of the entities. In geometry properties of motion are extremely useful for explaining the convoluted geometrical difficulties.

Concept for introduction for geometry properties of motion:

In geometry properties of motion the transformational or motion geometry occupies the examination and realizing of what properties of outlines and solids modify as they are shifted, and what properties do not modify. Geometry properties of motion involves three types of rigid transformations or motions:

1) Translations (slides),

2) Rotations (turns),

3) Reflections (flips).

Translations:

Translations shift a figure an unchanging distance in a certain route. Translation arrows symbolize the remoteness and path for moving the figure.

Rotations:

A rotation is a conversion that rotates a figure approximately one point (or, to use the arithmetical term, ‚¬"rotates a figure about a point‚¬) known as the center of rotation.

Rotations:

The series of problems in the reflection strand necessary to reflect figures, get lines of reflection, and place the position of entities before they were reflected. Reflections over performed over a line of reflection.

Example problems for geometry properties of motion:

1) Find the area of parallelogram, which base is 12 and height is 6.

Solution:

Given that length of base (b) = 12 cm, length (i) = 6 cm

Area of the parallelogram = b €" l

= 12cm €" 6 cm

= 72 cm2

2) Diameter of a circular garden is 8.4m, find the area of the diameter.

Solution:

Diameter, d = 8.4m.

= 4.2 m

Area of the circle =p*R*R

= 3.14 * 4.2 * 4.2

= 55.3896

3) Find the volume of cylinder Which Diameter of the base of a right circular cylinder is 12 cm, height is 30 cm?

Solution:

Diameter of the base is 12 cm,

V = Area of the base €" Height= pR2H

=3.14*6*6*30

=3391.2cm2

Introduction to problem solving table:

Make a Table is a problem-solving strategy which students can use to solve mathematical word problems by writing the information in a more organized format. Here is an example of a problem which is to be solved by making a table. The example problem is shown below. We have to tabulate the data for the given problem.

Problem solving table - Example:

How many hours will be the van traveling at 65 miles per hour takes to catch up with a van traveling at 55 miles per hour if the slower van starts one hour before the faster van?

Problem solving table - Solution

Make a table to organize the data. For this problem, to create a row for the slower van, a row for the faster van, and a column for each hour. Find the distance will be travel through each hour by look at the distances planned in each column. The distance of the faster van was more than the distance of the slower van in hour seven. The faster van traveled six hours to catch up to the slower van.

Table

Hour 1 2 3 4 5 6 7
Slower Van 55 110 165 220 275 330 385
Fastre Van 0 65 130 195 260 325 390

Check

Read the problem again to be sure the question was solved.

Did you find the number of hours it took for the faster van to catch up? Yes, it took 6 hours.
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