Today's blog entry is how to find the area of an enclosed polygon, when you just know the vertices (corner points) of the polygon.
The area, with n vertices, is:
A ≈ 1/2 * | (x1 y0 - x0 y1) + (x2 y1 - x1 y2) + (x3 y2 - x2 y3) + ... + (x0 yn - xn y0) |
With | | representing the absolute value.
The sum ends with the last vertex of the set and the initial vertex. The calculation is a fairly simple one.
The following TI-84+ program handles this formula.
TI-84+ Program AREAVERT
: Input "X0:",X
: Input "Y0:",Y
: {X} → L1
: {Y} → L2
: 0 → T
: 1 → K
: Lbl Y
: 1+K→ K
: Input "NEXT X:", A
: Input "NEXT Y:", B
: augment(L1,{A})→ L1
: augment(L2,{B})→ L2
: T+A*L2(K-1)-B*L1(K-1)→ T
: Menu("MORE?","YES",Y,"NO",N)
: Lbl N
: T+X*L2(K)-Y*L1(K)→ T
: abs(T)/2→ T
: Disp "AREA=", T
: Pause
Notes:
* L1 is accessed by pressing [2nd], [ 1 ]
* L2 is accessed by pressing [2nd], [ 2 ]
* The program works best if you start with one vertex and cycle either clockwise or counterclockwise, and stay in one direction all the way through.
* The program will ask you if there are more vertices. You do not have to repeat the initial vertex in is program.
Example 1:
Find the area of a square with the vertices (0,0), (0,5), (5,5), and (5,0). This is a square with a side length of 5 units.
Area: 25.
Example 2:
Find the area of a polygon with vertices (-3,0), (3,0), (4,11), (2,15), and (-4,12).
Area: 95.5
Until next time,
Eddie
This blog is property of Edward Shore. © 2012
