$ABCD$ is a square of side length 1. Points $E, F, G, H $ are such that $ AE : EB = 1:1$, $ BF: FC = 1 : 2 $, $ CG : GD = 1 : 3 $, $ DH : HA = 1 : 4 $.
What is the area of the quadrilateral $ EFGH $?
Note that if AE : EB = 1:1, then EB = 1/2
And if BF : BC = 1 :2 , then BF = 1/3
So the area of right triangle EBF =(1/2) (1/2)(1/3) = 1/12
Similarly FC =2/3
And if GC : GD =1:3 then GC = 1/4
So the area of right triangle FGC = (1/2)(2/3)(1/4) = 1/12
And GD = 3/4
And if HD : HA =1 : 4 then HD = 1/5
So the area of right trinagle HDG =(1/2) (3/4)(1/5) = 3/40
And finally HA =4/5 and AE =1/2
So the area of right triangle HAE =(1/2) (1/2) (4/5) = 1/5
So....the blue area is area of the square - 1/12 -1/12 - 3/40 - 1/5 =
1 - 1/12 -1/12 - 3/40 - 1/5 = 67/120