"What is the area? You should be able to solve this"

08082018, 04:44 PM
(This post was last modified: 08082018 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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