Geometry: triangulation for measuring a lot/yard

[roadrunner]

Interesting problem. I used the brute force method, example:

Measured distances:
S side = 8
W side = 7
E side = 9
NW to SE = 12
NE to SW = 11

Here’s what I did:

Go to cas define variables a and b to be x and y coordinates of the NW corner and variables c and d to be the coordinates of the NE corner. Start out assuming a square so:

b:=c:=d:=8 and a:=0

set up your 4 points:

Instruction:="GA:= point(0.,0.); // c(FF000000) v(1)
GB:= point(8.,0.); // c(FF000000) v(1)
GC:= point(a,b); // c(FF000000) v(1)
GD:= point(c,d); // c(FF000000) v(1)
"

Use fsolve to get a, b, c, and d:

fsolve({(distance(GA,point(a,b))) = 9,(distance(point(a,b),GB)) = 12},{a,b},{3,12})
fsolve({(distance(point(c,d),GB)) = 7,(distance(point(c,d),GA)) = 11},{c,d},{8,8})

set a, b, c, d to the answers from above

a:={0.0625,8.99978298349}[1]
b:={0.0625,8.99978298349}[2]
c:={8.5,6.98212002188}[1]
d:={8.5,6.98212002188}[2]

you should get something like this:

   

I'll admit, it's not as elegant as Casio, but it gets the job done. If you're going to be doing it a lot you can always make it into a program.

-road