# Thread: urgent help with area of shapes

1. ## urgent help with area of shapes

ok, this might sound dumb but i am dyslexic, but basically i have been given the square footage of an area (30000feet) and knowing this i must make a shape out of it, not just a square or rectangle but a rhombus. it has a short front long back and the sides are angled. knowing the square footage i dont know how i would make a shape out of that would equal the area (if that makes sense).

I know how to find the area volume and perimeter of a shape knowing the side lengths but dont know how to do it the other way round.

cheers

2. Originally Posted by eve
ok, this might sound dumb but i am dyslexic, but basically i have been given the square footage of an area (30000feet) and knowing this i must make a shape out of it, not just a square or rectangle but a rhombus. it has a short front long back and the sides are angled. knowing the square footage i dont know how i would make a shape out of that would equal the area (if that makes sense).

I know how to find the area volume and perimeter of a shape knowing the side lengths but dont know how to do it the other way round.

cheers
You should know that:

(1) the area of a rhombus is equal to one half of the product of its diagonals.

(2) The diagonals of a rhombus are perpendicular.

(3) The sides of a rhombus are all the same length.

Let the diagonals be 2x and 2y. Let the side length be z.

From (1): $30,000 = \frac{1}{2} (2x)(2y) = 2xy$

$\Rightarrow 15,000 = xy$ .... (A)

From (2) and (3) (using Pythagoras' theorem): $x^2 + y^2 = z^2$ .... (B)

Let x have a convenient value: x = 100, say. Then from equation (A), y = 150.

Substitute x = 100 and y = 150 into equation (B) and solve for z. Note that z will not be a whole number.

(If you want z to be a whole number, then that's a much harder problem).

There are obviously many possible different answers (an infinite number, in fact) for your question unless further information is given.

3. ok algebra like that (or of any kind) is way over my head. i dont have any more info, basically i am trying to make a room (i am doing a sound course and have to create an acoustical room) and all we have been given is the square footage of the room, i know there are infinite different answers. is there any simpler way of doign it?

thank you

4. Originally Posted by eve
ok algebra like that (or of any kind) is way over my head. i dont have any more info, basically i am trying to make a room (i am doing a sound course and have to create an acoustical room) and all we have been given is the square footage of the room, i know there are infinite different answers. is there any simpler way of doign it?

thank you
Using the values I gave, a rhombus of side length 180.3 units will pretty much have the area you want. The angles at the corners need to be 67.4 degrees and 112.6 degrees (using trigonometry).

By the way ...... is the shape meant to be a rhombus - as you state - or a parallelogram .....? Note that a rhombus has all sides the same length ....

5. a parallelogram then. sorry, proves how much i dont know about maths, they are not equal only the walls at the sides that slant out wards are. sorry

6. Originally Posted by eve
a parallelogram then. sorry, proves how much i dont know about maths, they are not equal only the walls at the sides that slant out wards are. sorry
Are the opposite sides meant to be parallel? If not, it might be a trapezium ..... Without being sure about the shape there's not a lot I can do. Perhaps you can draw the shape, scan it and attach it to this thread. I have to go now, I'm sure someone else will come along in due course.

7. ok this is the front wall is parallel to the back wall but the back wall is much longer and the side walls meet the corners of the font and back corners. so the side walls are angled.

8. Originally Posted by eve
ok this is the front wall is parallel to the back wall but the back wall is much longer and the side walls meet the corners of the font and back corners. so the side walls are angled.
From your description I assume that you mean a trapezium. (See attachment).

If so: The area of a trapezium is calculated by:

$area_{trapezium} = \dfrac{front + back}2 \cdot width$

Without any specific informations about the variables at the right hand side of the formula I can't help you any more. Sorry

9. thank you so so much you have helped big time