View Full Version : Recreational math question

11-06-2001, 07:28 AM
I'm building a patio, hence the "recreational." Gotta calculate the size of the area I'll need to buy stone for, hence the "math." Normally I'd just guess but this time, if I'm wrong, it impacts the ol' wallet. Plus, I want them to deliver enough but I don't want a bunch left over either.

I've got a 5 sided area, each side is a different length and the angles vary (a polygon?).

What formula should I use to calculate the square footage?

For example: If my 5 lengths were 12, 14, 16, 18 and 20
and my 5 angles were 60, 65, 70, 75 and 90 degrees, what would my formula look like?

Many thanks in advance.

11-06-2001, 07:36 AM
I think you're going to need to break your irregular polygon up into a bunch of triangles. Then find the area for each triangle and add them up. The formula for the area of a triangle is one half base times height.

11-06-2001, 07:42 AM
Makes sense. Thanks, Perderabo.

11-06-2001, 07:42 AM
It also depends on the sort of stone you are going to use - not the area - but the amount extra that you need to order. If it is small cobbles or bricks, there won't be much wastage, but if it is large slabs, the irregular shape may mean a lot of wastage. The rule of thumb is 10% but if you're careful, you can get away with far less than that. Or try to get the providers to take back the unused stone.

11-06-2001, 07:46 AM
Good point, Pergau. It's a mixed lot that I'll buy by the pallet. I'll negotiate that in at purchase.

Again, many thanks!

11-06-2001, 07:50 AM
Originally posted by Perderabo
I think you're going to need to break your irregular polygon up into a bunch of triangles. Then find the area for each triangle and add them up. The formula for the area of a triangle is one half base times height.
Depending on the exact shape, it might be even easier to draw a rectangle around the outside of your polygon, and then subtract the missing triangles.

11-06-2001, 07:50 AM
The formula a*b*sin(C)/2 also gives the area of a triangle, and may be useful since you've already measured a lot of angles. ("a" and "b" are the lengths of two sides, and "C" is the angle between them.)

Other than that, what the other guys said.

11-06-2001, 07:55 AM
Since you have a 90 degree angle, why don't you first form a rectangle with the 2 lengths of that angle and then you will be dealing with less triangles. A scale drawing helps.

11-06-2001, 08:45 AM
Before you use a*b*sin(C)/2 to find the area of the component triangles, you should verify your angle measurements. Angles as listed sum to 360, should sum to 540 for pentagon.

11-06-2001, 08:53 AM
If dividing up the polygon into triangles gets too messy, make a scale drawing on graph paper. Enclose the polygon in a rectangle, and calculate the areas of the triangles between the polygon and the surrounding box.
subtract that number from the area of the box. Or count the squares in the polygon, whichever is easier.
These Stone Calculators (http://www.google.com/search?q=%22stone+calculator%22&hl=en&start=10&sa=N) could be useful once you get the area figured out.

11-06-2001, 09:04 AM
The name of the shape you're describing is "pentagon". One thing about pentagons - their angles have to sum to 540. So those angles you gave can't be right. As it turns out, you don't need all the angles - two will do, as long as you have the lengths of all five sides. So here's what you do. Measure the five sides. In order, call them A, B, C, D, and E. Measure the angle between A and B and call the cosine of it X. Measure the angle between D and E and call the cosine of it Y. Now we can go:

Divide the pentagon into three triangles, ABF, DEG, and CGF. You can get the lengths of sides F and G with the law of cosines:

F = sqrt(A2 + B2 - 2ABX)
G = sqrt(D2 + E2 - 2DEY)

Now, in my opinion, the best way to get the area of a general triangle is by the semiperimeter method. So define:

H = (A + B + F) / 2
J = (D + E + G) / 2
K = (C + G + F) / 2

So you can get the areas of each of the triangles:

sqrt(H(H - A)(H - B)(H - F))
sqrt(J(J - D)(J - E)(J - G))
sqrt(K(K - C)(K - G)(K - F))

Sum these up and you're done! As an example, let's use your numbers for side lengths, A=12, B=14, C=16, D=18, E=20. For the angles, let's say the angle between A and B is 66 and the angle between D and E is 99. Then X=cos(66)=0.4067 and Y=cos(99)=-0.1564. This is a really irregular shape, but I think it'll work. I get the following numbers:

F = 14.260
G = 28.924
H = 20.130
J = 33.462
K = 29.592

And the three areas are 76.739, 177.785, and 63.436. The sum of these three numbers is 317.96, which is your total square footage. I'm pretty bad at arithmetic, so I may have gotten some of the numbers wrong, but I think you'll get the gist. Good luck!

11-06-2001, 09:19 AM
If I may add to [bAchernar's[/b] post, forget about measuring the angles and using the law of cosines. I have found that measuring angles in the field quite difficult. Instead, directly measure distances F and G.

11-06-2001, 10:26 AM
My angles were hypotheticals. Thanks for catching that.

I'm going to measure everything tonite but wanted to know what to do with 'em once I had 'em. You guys are great. Thanks again.

What did I ever do before SDMB?

11-06-2001, 10:33 AM
Here's another method that might be of interest:

Pick's Theorem (http://www.cut-the-knot.com/ctk/Pick.html)

11-06-2001, 11:45 AM
That's a good point, NotMrKnowItAll. I definitely agree that measuring F and G directly would be much easier. As you can probably tell, I'm much better with numbers on paper than in real life.

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