20 10 detailed explanation of 22 math problems in Ezhou senior high school entrance examination

Solution: (1) is MN⊥AE, the length of the rectangle is x decimeter, and the width is y decimeter;

∫△EMN∽△EFA

∴AF AE =MN NE and ∵ Mn = AD-X =12-X.

NE=y-(AB-AE)=y-6

AF=4，AE=2

4 ^ 2 = 12-x y-6，

S=xy=70

Solution: x= 10, y=7.

Answer: The length and width of the rectangular MGCH are 10 decimeter and 7 decimeter respectively.

(2) Let the length of EM be a, △EMN∽△EFA, EM EF =MN AF =NE AE,

EF=2 5，MN=2 5 5 a，NE= 5 5 a，

MH=AD-MN= 12-2 5 5 a，MG=BE+EN=AB-AE+EN=6+ 5 5 a

∴S=MH×MG

=( 12-2 5 5 a)×( 6+ 5 5 a)

=-2 5 a2+72

Therefore, when a=0, the area is the largest, that is, point M and point E coincide.

At this time, the circumference L=2MH+2MG=36 decimeters.

Answer: When EM is 0, the area of the rectangular MGCH is the largest, and the circumference of the rectangle is 36 decimeters.