August 23rd, 2017, 07:42 AM  #1 
Senior Member Joined: Jan 2015 From: London Posts: 106 Thanks: 2  Area of irregular Polygon
Hi here is the image of a polygon that I need to find the area of, my dad set me this challenge and I am not back at school yet, for anyone thinking this is homework. Also, there are right angles at the top right and bottom right I was thinking splitting it up into smaller triangles but after the first one didn't really know where to go from there. Last edited by jamesbrown; August 23rd, 2017 at 07:46 AM. 
August 23rd, 2017, 08:22 AM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,064 Thanks: 722 
Change the lengths (clockwise from top) to: 39,53,37,36,24,12 (easier to work with; change them back later). Flip diagram sideways so that "53" is on yaxis and "37" is on xaxis: so diagram fully in 1st quadrant. Go to town with that: that's all you get from me, since you did not label your diagram!! Last edited by greg1313; August 23rd, 2017 at 08:25 AM. 
August 23rd, 2017, 08:46 AM  #3  
Senior Member Joined: Jan 2015 From: London Posts: 106 Thanks: 2  Quote:
 
August 23rd, 2017, 08:57 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,064 Thanks: 722 
I meant labelling the vertices; like AB being the top line (3.86); then going clockwise, ending up with polygon ABCDEF. 
August 23rd, 2017, 09:05 AM  #5  
Senior Member Joined: Jan 2015 From: London Posts: 106 Thanks: 2  Quote:
Edit: still don't understand like if I'm plotting the points, I need 2 values to plot. Like I get the top line cause it's straight but when trying to do BC I wouldn't know where to plot it because I don't know the height nor the length? I only know the hypotenuse if you understand? Last edited by jamesbrown; August 23rd, 2017 at 09:11 AM.  
August 23rd, 2017, 01:15 PM  #6 
Senior Member Joined: Jan 2015 From: London Posts: 106 Thanks: 2 
All good my dad added an extra measurement in and was able to calculate it!


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area, irregular, polygon 
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