# Irregular room area

Ragnorak
Posers and Puzzles 21 Feb '05 17:40
1. Ragnorak
21 Feb '05 17:40
Hey guys,

I know there are a lot of maths enthusiasts here, and I was hoping to use your skills for a practical purpose. I need to find the area of 2 irregularly shaped rooms.

1 is 410 x 275 x 401.5 x 293 (all cms)
2 is 564 x 244 x 619 x 415

I have made a rough guesstimate, but need something a bit more accurate. Is there enough info, or would I also need to find out the angles?

If anybody can point me in the right direction, great. If somebody can actually work it out, then even better.

Thanks,

D
2. The Plumber
Leak-Proof
21 Feb '05 17:56
Originally posted by Ragnorak
Hey guys,

I know there are a lot of maths enthusiasts here, and I was hoping to use your skills for a practical purpose. I need to find the area of 2 irregularly shaped rooms.

1 is 410 x 275 x 401.5 x 293 (all cms)
2 is 564 x 244 x 619 x 415

I have made a rough guesstimate, but need something a bit more accurate. Is there enough info, or would I ...[text shortened]... right direction, great. If somebody can actually work it out, then even better.

Thanks,

D
You need one of the angles.

Think of it this way: If you took 4 sticks each of the lengths you specified (or maybe shrink them all by factor of 10 so you can actually deal with them) and tie them end to end to end to end, you could pull two opposite corners and stretch the whole contraption, thus changing the area. However, if you were to fix one of the joints (glue, rather than tie), then the rest of the sticks would have only two possible combinations, either the interior angle formed by the two remaining sticks is less than 180 degrees or greater than 180 degrees. (I'm making the assumption that it would be less than 180, so there would only be one possible solution if you can identify any one of the four angles.)
3. The Plumber
Leak-Proof
21 Feb '05 18:05
Originally posted by The Plumber
You need one of the angles.

Think of it this way: If you took 4 sticks each of the lengths you specified (or maybe shrink them all by factor of 10 so you can actually deal with them) and tie them end to end to end to end, you could pull two opposite corners and stretch the whole contraption, thus changing the area. However, if you were to fix one o ...[text shortened]... 0, so there would only be one possible solution if you can identify any one of the four angles.)

You can avoid needing to identify an angle if you can identify the length of a diagonal. If we go back to our 4 sticks tied at the ends, and we take a fifth stick (a diagonal) and place it in its appropriate location, the four sticks will be fixed.
4. Ragnorak
21 Feb '05 18:31
Originally posted by The Plumber

You can avoid needing to identify an angle if you can identify the length of a diagonal. If we go back to our 4 sticks tied at the ends, and we take a fifth stick (a diagonal) and place it in its appropriate location, the four sticks will be fixed.
Excellent. That's exactly the type of pointer I needed. But how would I find the area of the irregular triangles?

Thanks,

D
5. 21 Feb '05 20:223 edits
Using the Heron's formula? http://mathworld.wolfram.com/HeronsFormula.html
6. Ragnorak
21 Feb '05 21:13
Originally posted by ilywrin
Using the Heron's formula? http://mathworld.wolfram.com/HeronsFormula.html
Brilliant. Thanks.

Just to make double sure, I just need formulas 1 and 2 from that page, right?

I used to be good at maths. Then I went to university and for some unexplainable reason, I lost my mathematical ability. And my short term memory. ðŸ˜›

; D
7. 22 Feb '05 16:541 edit
Yes, the first two. All of the rest is "curios but highly useless".
8. Ragnorak