# Parallelograms projection on a plain

• fawk3s
In summary, the problem is to find the area of a parallelogram projected onto a plain alpha, given that one of its sides is 5cm and forms an angle of 30 degrees with the plain, while another angle in the original parallelogram is 60 degrees. Using trigonometric functions, the attempt at a solution was to find the length of the projection of the 3cm side on the plain, and then use the sine function to calculate the area. However, the calculation resulted in an answer of 11.25cm^2, while the textbook solution was 10.6cm^2. After further analysis, it was discovered that the angle between the parallelogram and the plain
fawk3s

## Homework Statement

Parallelograms 5cm side is located on a plain alpha and its 3cm side forms an angle with the plain equal to 30 degrees. Find the area for the parallelograms projection on the plain alpha, if an angle in the original parallelogram is 60 degrees.

all kinds

## The Attempt at a Solution

I think I am visualizing something wrong. Because what I did gave me an answer or 11,25 cm2, but the answer is supposed to be 10,6cm2 according to the textbook.

Here's what I did:

Find the length of the 3cm side's projection on the plain: d=3cm*cos30=3*sqrt3/2 cm
Directly find the area of the projection with 2 sides of the projection and the angle between them: S=5*d*sin60=5*3*sqrt3*sqrt3/2*2=11,25 cm2

Where did I go wrong?

fawk3s

Last edited:
I got the same answer as you. *Shrug*

I just happened to figure it out. I have to go out at the moment, but incase you are interested in the solution dacruick, I can post it up later on.

fawk3s said:
I just happened to figure it out. I have to go out at the moment, but incase you are interested in the solution dacruick, I can post it up later on.

yes I'd like to see :)

[PLAIN]http://img267.imageshack.us/img267/1040/mathlol.png

The important thing to realize is that β is not equal to 60o.
Think of it this way: the side a of the parallelogram which is not on the plain alpha, is shifted by d relative to the side a which IS on the plain. As is its projection, no matter what the angle between the parallelogram and the plain.

d=b*cos60o=3*cos60o=1,5 cm

Since we know that the angle between b and the plain is 30o (which I accidently forgot to add in the drawing), we can find b's projection. Let it be b'.

b'=cos30o*b=3*sqrt(3)/2

We can now find β, because cosβ=d/b'. So β=54,74o (approx.)
Now we can find the area

S=a*b*sinβ=10,606601...

Last edited by a moderator:

## 1. What is a parallelogram projection on a plain?

A parallelogram projection on a plain is a method of representing a three-dimensional object on a two-dimensional surface using parallel lines. It is commonly used in technical and scientific drawings to accurately depict the shape and proportions of an object.

## 2. How is a parallelogram projection different from other types of projections?

A parallelogram projection differs from other types of projections in that it uses parallel lines to represent the object, while other projections may use converging lines or other techniques. This allows for a more precise and accurate representation of the object's shape and proportions.

## 3. Can any shape be projected onto a plain using a parallelogram projection?

Yes, any shape can be projected onto a plain using a parallelogram projection as long as it follows the rules of a parallelogram. This means that opposite sides are parallel and equal in length, and opposite angles are also equal.

## 4. How is a parallelogram projection useful in scientific research?

A parallelogram projection is useful in scientific research because it allows for accurate and detailed representation of three-dimensional objects, such as molecules, structures, and equipment. This can aid in analysis, understanding, and communication of scientific concepts and data.

## 5. Are there any limitations to using a parallelogram projection?

One limitation of using a parallelogram projection is that it can only represent objects with straight edges and flat surfaces. Curved or irregular shapes may not be accurately depicted. Additionally, the projection may introduce errors or distortions if not done carefully, so it is important to use proper techniques and measurements.

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