E. F is the edge BC of square ABCD with side length 4, the point on CD, CE = 1, CF = 4 / 3, the extension line of line Fe intersecting AB is at g, passing through a moving point h on line FG, making HM perpendicular to Ag and m, let HM be x, and the area of rectangle amhn be y 1. Find the function expression 2 when x =? Amhn, what is the largest area When x = 3, the maximum value of Y is 12 How to - -,

E. F is the edge BC of square ABCD with side length 4, the point on CD, CE = 1, CF = 4 / 3, the extension line of line Fe intersecting AB is at g, passing through a moving point h on line FG, making HM perpendicular to Ag and m, let HM be x, and the area of rectangle amhn be y 1. Find the function expression 2 when x =? Amhn, what is the largest area When x = 3, the maximum value of Y is 12 How to - -,

Let the intersection of HN and BC be o
Then Bo = x, EO = 3-x
∵△EOH∽△ECF
∴OH/OE=4/3
∴OH=4/3(3-x)
∴MH=4+4/3(3-x)=8-4/3x
∴y=x(8-4/3x)
∴y=-4/3x²+8x
(2)
y=-4/3(x-3)²+12
When x = 3, y is the maximum, and the maximum is 12

The side length of square ABCD is 2, the points E and F are BC and CD respectively, and CE = CF, the area of △ AEF is equal to 1, so the length of EF can be obtained
emergency

Let CE = CF = X
be
BE=DF=2-x
therefore
S△AEF=SABCD-S△ABE-S△ADF-S△EFC
2*2-(1/2)*2*(2-x)-(1/2)*2*(2-x)-(1/2)*x*x＝1
4-(2-x)-(2-x)-(1/2)x^2=1
(1/2)x^2-2x+1=0
x^2-4x+4=2
(x-2)^2=2
x=2-√2
EF^2=EC^2+FC^2
=(2-√2)^2+(2-√2)^2
=2(2-√2)^2
EF=√2(2-√2)
=2√2-2

The length ab of rectangle ABCD is 4cm and the width BC is 2cm. Now the rectangle rotates around AD to get a cylinder. Try to find the volume of the cylinder

The radius of the cylinder r = AB = 4cm; the height of the cylinder H = BC = 2cm
Volume of cylinder = area of cylinder bottom * height = Πxr & # 178; XH = 3.14x4 & # 178; x2 = 101cm & # 179;