"What is the area? You should be able to solve this" - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: General Forum (/forum-4.html) +--- Thread: "What is the area? You should be able to solve this" (/thread-11181.html) Pages: 1 2 RE: "What is the area? You should be able to solve this" - roadrunner - 08-08-2018 05:03 PM On the hp prime I used the geometry app: Code: ```"GA:= point(0.,0.); // c(FF000000) v(1)  GB:= point(xyc,0.); // c(FF000000) v(1)  GC:= point(xyc,xyc); // c(FF000000) v(1)  GD:= point(0.,xyc); // c(FF000000) v(1)  GE:= segment(GA,GB); // c(FF000000) v(1) l(0)  GG:= segment(GB,GC); // c(FF000000) v(1) l(0)  GH:= segment(GC,GD); // c(FF000000) v(1) l(0)  GI:= segment(GD,GA); // c(FF000000) v(1) l(0)  GJ:= midpoint(GE); // c(FF000000) v(1)  GK:= midpoint(GG); // c(FF000000) v(1)  GL:= midpoint(GH); // c(FF000000) v(1)  GM:= midpoint(GI); // c(FF000000) v(1)  GN:= point(xn,yn); // c(FF000000) v(1)  GO:= segment(GN,GM); // c(FF000000) v(1) l(0)  GP:= segment(GN,GJ); // c(FF000000) v(1) l(0)  GQ:= segment(GN,GK); // c(FF000000) v(1) l(0)  GR:= segment(GN,GL); // c(FF000000) v(1) l(0)  "``` define: a:=(x1,y1,x2,y2,x3,y3,x4,y4)->(ABS(x1*y2+x2*y3+x3*y4+x4*y1-x2*y1-x3*y2-x4*y3-x1*y4))/2 // shoelace formula for 4 points xyc -- x and y coordinates of point C xn -- x coordinate of point N yn -- y coordinate of point N initially, {xyc,xn,yn}:={10,3,6} // just a guess [attachment=6200] use fsolve with 3 equations for areas 1, 2, and 3; and 3 unknowns {xyc,xn,yn}: {xyc,xn,yn}:=fsolve({(a(0,(xyc/2),xn,yn,(xyc/2),xyc,0,xyc)) = 16,(a(0,0,(xyc/2),0,xn,yn,0,(xyc/2))) = 20,(a((xyc/2),0,xyc,0,xyc,(xyc/2),xn,yn)) = 32},{xyc,xn,yn},{10,3,6}) [attachment=6202] area 4 = 28 -road