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# Geometry Question

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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? Jun 27, 2018

#1
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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    Jun 27, 2018
edited by CPhill  Jun 27, 2018
#2
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I put your answer into my online program, and unfortunately it was incorrect. I do however get one more chance to complete the problem.

Guest Jun 27, 2018
#3
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Sorry...I missed the last point.....let me delete my answer and start over....!!   CPhill  Jun 27, 2018
#4
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Thanks so much! Sorry, these type of questions are difficult for me.

Guest Jun 27, 2018