@@MathandEngineering Let G be the intersection of CF and AE. Draw a perpendicular to DF, from G and label the point on DF as H.Let GH = b. Draw another perpendicular to AF from G and label the point on AF as I. Let GI = GF = a. Triangles GHF and CDF are similar. Therefore, a/b = 60/30, a = 2b. Solve for x : 2x + x = 60, so x = 20. Then, HE = 40 - a = 40 - 2b. Notice that triangles GHE and AFE are similar, so HE / GH = FE / AF. (40 - 2b)/b = 40/60. 60(40 - 2b) = 40b, so 2400 = 160b, b = 15. Therefore, a = 2b = 30. Finally, blue area = area AFD - area AGF - area FGD = (60)(60)/2 - (30)(60)/2 - (15)(60)/2 = 450 m^2
We can solve it with trigonometry only without any construction lines. Let angle GAF=a ,angle GFA=angle FCD=b then angle AGF=180-(a+b) tan a=2/3 hence sin a=2/sqrt(13) and cos a=3/sqrt(13) tan b=2 hence sin b=2/sqrt(5) and cos b=1/sqrt(5) From above we can calculate sin(a+b)=sin a * cos b+cos a*sin b=8/sqrt(65) sin(45-a)=sin 45*cos a-cos 45*sin a=sqrt2)/(2*sqrt(13)) From the law of sine AG/sin b=AF/sin(180-(a+b) where sin(180-(a+b))=sin(a+b) we,get AG=sqrt(13)/4*AF The area is 1/2*AD*AG*sin(45-a)=1/2*sqrt2)*AF*sqrt(13)/4*AF*sqrt(2)/(2*sqrt(13))=450 where AF=60.
Let CF intersect DA at K, CF intersect AE at G. Draw a perpendicular from G to DF intersect DF at N. As CD parallel to AF then CD/AF=CK/FK=KD/KA=1/2. As CK/FK=DE/EF=1/2, we can easily prove that CD parallel to KE then KE/CD=EF/FD=2/3 or KE=2/3*CD=2/3*30=20. Additionally, as CD parallel to KE and angle CDF=90 then angle KEF=90. As a result KE parallel to GN. We now have GN/KE=GF/FK, but GF/GK=AF/KE=60/20=3 then GF/FK=3/4 or GN/KE=3/4 or GN=3/4*KE=3/4*20=15. The blue area= area of ADG=area of ADF-(area AFE+area GED)=1/2*AF*FD-(1/2*AF*EF+1/2*GN*ED)=1/2*60*60-(1/2*60*40+1/2*15*20)=450.
