+1490 The area of the square ABCD is 16 cm². Its area is divided into 4 equal squares. Line EH passes through the center of the large square.
Angle AHG is 90°. Find the area of the yelow-bordered region. 
+1490 On my PC the picture is clear enough. What part do you not understand? Let me know.
+2864 um the picture is blocked on my browser. You might have to screenshot it.
+118687 Yes thanks, we can see the pic now.
I would tackle this question by using coordinate geometry.
A((0,0)
B(0,4)
C((4,4)
D(4,0)
centre(2,2)
work out H (intersection of 2 lines)
then work out E
then look at areas.
+1490 AD = 4 cm
AG = 2 cm
Area of square ABCD = AD² = 16 cm²
AH = 1.789
Angle ABG = 26.565°
GH = sqrt(AG² - AH²) = 0.894
Area ABHFD = (AB * AG) - [(AH * GH) /2] = 7.2 cm²
BG = sqrt( 4² + 2² ) = 4.472
BH = BG - GH = 3.578
HL = cos(26.565°) * 3.578 = 3.2
BL = sqrt[(BH)² - (HL)²] = 1.6
Area of Δ BHL = (HL * BL) /2 = 2.56 cm²
Angle ABG = BHE = 45°
Angle EHL = ∠BHE - ∠BHL = 18.435°
EL = tan( 18.435°) * 3.2 = 1.067
Area of Δ EHL = ( HL * EL) /2 = 1.707 cm²
Area CFHE = 16 - ( 7.2 + 2.56 + 1.707 ) = 4.533 cm² 
