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

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