HP 32SII: area of a polygon by coordinates

[Carlos CM (Mexico)]

Program to calculate the area of a polygon by coordinates

Code:

[LBL] [A]
[CLVARS]
[INPUT] [N]
[1] [E] [3] [+/-]
[*]
[1]
[+]
[STO] [I]
[LBL] [B]
[INPUT] [X]
[INPUT] [Y]
[Rv]
[+]
[Rv]
[-]
[X<>Y]
[Rv]
[*]
[2]
[/]
[STO][-][A]
[RCL][Y]
[RCL][X]
[ISG][I]
[GTO][B]
[VIEW][A]
[RTN]

REGISTERS USED:

A: area of the polygon
N: total of points to enter
X: xi
Y: yi

LABELS USED:

A (20 bytes), B (28.5 bytes)

NOTES:

X<>Y is SWAP
Rv is ROLL DOWN


INSTRUCTIONS:

1. The polygon must be in the 1st Cuadrant of Cartesian System
2. Must begin/end with P1(0,0) coordinate
3. The order to enter the coordinates is clockwise to normal figures and counter clockwise if the figure has a HOLE.
4. Enter in order the coordinates, P1, P2...Pn-1,P1 in this way: [X coordinate] [R/S] [Y coordinate] [R/S]
6. At the end of the last point, P1(0,0) the program display the desired area.

EXAMPLE 1.
For example, in a square (3x3=9 squared units):

[XEQ] [A]
N? ***Total of points to enter
[5] [R/S] *** In this case 5 points
[0] [R/S] [0] [R/S] *** P1(0,0)
[0] [R/S] [3] [R/S] *** P2(0,3)
[3] [R/S] [3] [R/S] *** P3(3,3)
[3] [R/S] [0] [R/S] *** P4(3,0)
[0] [R/S] [0] [R/S] *** P1(0,0)
A=9 ***AREA OF THE POLYGON


EXAMPLE 2.
Example with a hole at center of the last square:

[XEQ][A]
N? ***Total of points to enter
[1][1] [R/S] *** In this case 11 points
[0] [R/S] [0] [R/S] *** P1 (0,0) (Begin with 0,0)
[0] [R/S] [3] [R/S] *** P2 (0,3)
[3] [R/S] [3] [R/S] *** P3 (3,3)
[3] [R/S] [0] [R/S] *** P4 (3,0)
[0] [R/S] [0] [R/S] *** P5 (0,0) (Close the polygon)
[1] [R/S] [1] [R/S] *** P6 (1,1) (here the order of the points in counter clockwise)
[2] [R/S] [1] [R/S] *** P7 (2,1)
[2] [R/S] [2] [R/S] *** P8 (2,2)
[1] [R/S] [2] [R/S] *** P9 (1,2)
[1] [R/S] [1] [R/S] *** P10(1,1) (close the polygon that generate the hole)
[0] [R/S] []0 [R/S] *** P11(0,0) (end with 0,0)
A=8 ***AREA OF THE POLYGON

RESULT=8 squared units


Thanks for your aditional comments.