Orthographic projection and isometric projection are two ways to show three-dimensional objects in a two-dimensional space, such as on a piece of paper or a computer screen. Each method gives a different perspective. Do some research and then compare and contrast the two methods for displaying three-dimensional shapes. Then try your hand at creating both types of projections for a simple geometric shape using paper, pencil, and a ruler. In your opinion, what are the pros and cons of each projection? What are the limitations? In which circumstances, environments, or occupations is one type of projection likely preferred over the other? Describe any special tools that might be needed to create the projection. Which projection is easiest for you to interpret visually? Why?

Orthographic Projection is used for making the projects but Isometric Projection is used to have better understanding of the object.

Orthographic drawings are typically two dimensional views of an object. For instance, if you were designing a table, you would draw a top view, side view and a bottom view. Should these three views not fully explain the design of the table other views would need to be drawn. When drawing an perspective view in an orthographic manner, you would utilize a 45 degree triangle for the lines that extend back or forward from the vertical lines. This type of perspective is not a true perspective because you can measure the true length of all the details shown. An isometric drawing is meant to depict a 3D image of an object in what appears to be a perspective view. However, similar to an orthographic perspective, all of the lines in an isometric drawing can be measured to their true length. What makes it different from an orthographic perspective is that its angled lines are drawn at 30 or 60 degrees or divisions of them. Drawing this by hand you would use a 30/60/90 triangle.

In either case, both types of perspectives can be accurately measured with a ruler in order to know the objects measurements.

Step-by-step explanation:

the answer is a

step-by-step explanation:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.