The vertices of a convex pentagon are $(-1, -1), (-3, 4), (1, 7), (6, 5)$ and $(3, -1)$. What is the area of the pentagon?

Guest Jun 27, 2018

#1**+1 **

Here's one way to do this.....construct triangles BFA , BGC, HDC and EFD

And the area of the pentagon =

Area of rectangle FGHI - area of triangle BFA - area of triangle BGC - area of triangle HDC - area of triangle EFD =

FG * GH - (1/2)AF * BF - (1/2) CG * BG - (1/2) CH * DH - (1/2) EI * DI =

8*9 - (1/2) 2 * 5 - (1/2) 4 *3 - (1/2) 5 * 2 - (1/2) 3 * 6 =

72 - 5 - 6 - 5 - 9 =

72 - ( 5 + 6 + 5 + 9) =

72 - ( 25)

47 units^2

CPhill Jun 27, 2018