Posted by **PAT** on Wednesday, April 6, 2011 at 5:13am.

Find the area of a pentagon ABCDE with verticles A(0,4),B(3,2),C(3,-1), D(-3,-1) and E (-3,2)

- geometry -
**MathMate**, Wednesday, April 6, 2011 at 7:47am
If the pentagon's boundary does not cross itself, I would use the surveyor's formula:

1. lay out the coordinates in order, repeat the first at the end.

A(0,4)

B(3,2)

C(3,-1)

D(-3,-1)

E (-3,2)

A(0,4)

2. Cross multiply between each successive pair and sum.

For A and B, we get 0*2-3*4=-12

For B and C, we get 3*-1 - 3*2 = -9

Continue until the last pair.

The sum of all 5 products is twice the area. If the vertices have been named clockwise, the sum will be negative.

3. Divide the sum by 2. If the sum is negative, write down only the positive value.

For the example:

vertex sum

A -12

B -9

C -6

D -9

E -12

Sum -48

Area = 48/2=24

Check my calculations.

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