So the area of triangle ABF =(3/ 10) the area of square ABCD.
=(3/ 10)x30=9 cm2,
The area of triangle DFC =(2/ 10) the area of square ABCD.
=(2/ 10)x30=6 cm2,
Because AE=( 1/2)EF,
So the area of triangle BEF =(2/3) the area of triangle ABF,
Because the area of triangle ABF =( 1/2) the area of square ABCD.
=( 1/2)x30
= 15 cm2,
So the area of the triangle BEF =(2/3)x 15.
= 10 cm2,
So the area of quadrilateral EFBC = the face of triangle BEF+the area of triangle DFC.
= 10+6
= 16 cm2。
2。 The intersection d is DG//BC, and the intersection AE is G point.
Because f is the midpoint of the CD,
So triangle GDF is equal to triangle ECF(A, s, a).
So DG=EC,
Because BE=2EC,
So BE=2DG, BE/DG=2,
Because DG//BC,
So AB/AD=BE/DG=2, AB=2AD, and d is the midpoint of AB.
So the area of triangle ACD =( 1/2) the area of triangle ABC.
=( 1/2)x 1
=1/2cm2,
Because f is that midpoint of the CD,
So the area of triangle ADF =( 1/2) the area of triangle ACD.
=( 1/2)x( 1/2)
=1/4cm2,
Because BE=2EC,
So the area of triangle ACE =( 1/3) the area of triangle ABC.
=( 1/3)x 1
=1/3cm2,
So the shadow area = the area of triangle ABC-the area of triangle ACE-the area of triangle ADF.
= 1-( 1/3)-( 1/4)
=5/ 12 cm2。
3。 Link DE
Because AE=EF=FB,
So the area of triangle ADF =( 1/3) the area of rectangle ABCD.
=( 1/3)x(6x2)
=4,
Triangle area ADE =(2/3) Triangle area ADF
=(2/3)x4
=8/3,
Area of triangular EDF =( 1/3) area of triangular ADF.
=4/3,
Because AB//DC, AE=EF=FB.
So FO/OD=EF/DC= 1/3.
So the area of triangle DOE =( 1/4) the area of triangle EDF.
=( 1/4)x((4/3)
= 1/3,
So the area of shaded part = the area of triangle ade+the area of triangle DOE.
=8/3+ 1/3
=3。
The Digital Mystery of Hutulo's Works
The expression of Hutuluo's calligraphy is the numerical relationship, which inevitably reflects the ancient mathematical l