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"What is the area? You should be able to solve this"
08-08-2018, 04:44 PM (This post was last modified: 08-08-2018 04:45 PM by Voldemar.)
Post: #20
RE: "What is the area? You should be able to solve this"
Area of quadrangle by coordinates:
S = ((x1*y2 - y1*x2) + (x2*y3 - y2*x3) + (x3*y4 - y2*x4) + (x4*y1 - y4*x1))/2
   
We have four quadrangles:
16 = ((0*y - a*x) + (x*2a - y*a) + (a*2a - 2a*0) + (0*a - 2a*0))/2
20 = ((0*0 - 0*a) + (a*y - 0*x) + (x*a - y*0) + (0*0 - a*0))/2
32 = ((a*0 - 0*2a) + (2a*a - 0*2a) + (2a*y - a*x) + (x*0 - y*a))/2
S = ((x*a - y*2a) + (2a*2a - a*2a) + (2a*2a - 2a*a) + (a*y - 2a*x))/2
After simplification:
16 = (a*x - a*y +2a^2)/2
20 = (a*x + a*y)/2
32 = (a*y - a*x + 2a^2)/2
S = (-a*x - a*y + 4a^2)/2
Take Prime:
   
   
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RE: "What is the area? You should be able to solve this" - Voldemar - 08-08-2018 04:44 PM



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