Find the area of the above figure. Round to the nearest whole number!!! Show all work!!! *especially how you find he other values!!! Solution lets denote the points X,Y,Z,W such that XY=20 YZ=55 ZW=65 angle WXY=100 degrees angle YZW=30 degrees join Y and W. from cosine rule, cos(30)=(YZ^2+ZW^2-YW^2)/(2*XY*YW) so sqrt(3)/2=(55^2+65^2-YW^2)/(2*55*65) YW=32.52 so now consider the triangle XYW. cos(100)=(XY^2+XW^2-YW^2)/(2*XY*XW) so (20^2+XW^2-32.52^2)=2*20*XW*cos(100) XW^2-657.55=-6.946*XW XW^2+6.946*XW-657.55=0 solving for XW, we get XW=22.4 so, area of XYW: XY=20 YW=32.52 XW=22.4 area=sqrt(p*(p-XY)*(p-XW)*(p-XW)) where p=(XY+XW+YW)/2 area=220.588 square unit area of YWZ: YZ=55 YW=32.52 ZW=65 area=sqrt(p*(p-YZ)*(p-YW)*(p-ZW)) where p=(XY+YW+YW)/2 area= 893.59 square unit total area=area of XYW +area of XYW=220.588+893.59=1114.2 square unit .

1. Find the area of the above figure. Round to the nearest whole number!!! Show all work!!!
*especially how you find he other values!!!
Solution
lets denote the points X,Y,Z,W such that XY=20 YZ=55 ZW=65 angle WXY=100 degrees angle
YZW=30 degrees join Y and W. from cosine rule, cos(30)=(YZ^2+ZW^2-YW^2)/(2*XY*YW)
so sqrt(3)/2=(55^2+65^2-YW^2)/(2*55*65) YW=32.52 so now consider the triangle XYW.
cos(100)=(XY^2+XW^2-YW^2)/(2*XY*XW) so (20^2+XW^2-32.52^2)=2*20*XW*cos(100)
XW^2-657.55=-6.946*XW XW^2+6.946*XW-657.55=0 solving for XW, we get XW=22.4 so,
area of XYW: XY=20 YW=32.52 XW=22.4 area=sqrt(p*(p-XY)*(p-XW)*(p-XW)) where
p=(XY+XW+YW)/2 area=220.588 square unit area of YWZ: YZ=55 YW=32.52 ZW=65
area=sqrt(p*(p-YZ)*(p-YW)*(p-ZW)) where p=(XY+YW+YW)/2 area= 893.59 square unit total
area=area of XYW +area of XYW=220.588+893.59=1114.2 square unit