The length of square ABCD is 4, M is the moving point on BC, n is the moving point on CD, am is perpendicular to Mn, BM =? ABCN has the largest area

The length of square ABCD is 4, M is the moving point on BC, n is the moving point on CD, am is perpendicular to Mn, BM =? ABCN has the largest area

The length of square ABCD is 4Mn, which is the two moving points of BCCD and always keeps am ⊥ Mn. When BM = what, the area of quadrilateral ABCN is the largest

Let BM = x and the area of trapezoid ABCN be y
The side length of square ABCD is 4, BM = X,
∴CM=4﹣x
Rt△ABM∽Rt△MCN
∴AB/MC=BM/CN
∴CN=(-x²+4x)/4
y=1/2*(AB+CN)*BC
   =-1/2x²+2x+8
   =-1/2(x-2)²+10
When x = 2,
Y has a maximum of 10
When BM = 2,
The quadrilateral ABCN has the largest area

The side length of square ABCD is 4, m and N are two moving points on BC and CD respectively, and always keep am ⊥ Mn?

Let BM = x, then MC = 4-x, ∵ - amn = 90 °, ∵ - AMB = 90 ° - NMC = ∠ MNC, ∵ △ ABM ∽ MCN, then abmc = bmcn, that is 44 − x = xcn, the solution is: CN = x (4 − x) 4, ∵ s quadrilateral ABCN = 12 × 4 × [4 + X (4 − x) 4] = - 12x2 + 2x + 8 = - 12 (X-2) 2 + 10, ∵ 0 ≤ x ≤ 4, ∵ s quadrilateral ABCN is the largest when x = 2, that is, when the length of BM is 2, the area of quadrilateral ABCN is the largest

The side length of square ABCD is 4, and point Mn is on line AB and ad respectively. If 2Mn square + cm square + CN square = 54, then am + a
What is the value range of am + an

Pythagorean theorem:
MN^2=AM^2+AN^2
CM^2=BC^2+BM^2
CN^2=CD^2+DN^2
BC=CD=4
Namely
2(AM^2+AN^2)+BM^2+DN^2=22
2(AM^2+AN^2)+(4-AM)^2+(4-AN)^2=22
3AM^2-8AM+3AN^2-8AN=-10
3(AM-4/3)^2+3(AN-4/3)^2=-10+32/3
(AM-4/3)^2+(AN-4/3)^2=2/9
That is to say, there is a right triangle with 62 / 9 square hypotenuse, and the right sides are am-4 / 3 and an-4 / 3
So am + an is between 8 / 3 + sqrt (2) / 3 and 8 / 3 + 2 / 3 = 10 / 3

In square ABCD, M is on ad, n is on CD, ∠ MBN = 45?: Mn = am + CN

The triangle BCN can be rotated 90 degrees clockwise around point B, so that BC and ab coincide. From point n to e, the triangle BCN and triangle Abe are congruent after rotation, so the angle EBA = angle NBC, CN = AE, BN = be. Then because the angle MBN = 45 degrees, the angle ABM + angle CBN = 45 degrees, that is, the angle ABM + angle EBA = angle MBN = 90 degrees, pass the (ASA) proof

As shown in the figure, the rectangle ABCD is folded along AE so that D falls on the f point on the edge of BC. If the angle BAF = 60 degrees, calculate the degree of angle AEF

As shown in the figure, the rectangle ABCD is folded along AE so that point d falls at point F on the edge of BC. If ∠ BAF = 60 °, then ∠ DAE=______ Degree

The results show that ∵ BAF = 60 ° and ∵ DAF = 30 ° and ∵ AF is obtained by ad folding, ∵ ade ≌ △ AFE and ∵ DAE = ∠ EAF = 12 ∠ DAF = 15 °. So the answer is 15

A rectangular piece of paper with width of 3 and length of 4, ABCD, is folded along the diagonal BD, point C falls at the position of C ', BC' intersects ad with point G, and the length of Ag is calculated
Write down the process

7/6
Let Ag be x, and the Pythagorean theorem is listed in △ ABG, where (3 + x) ^ 2 = 4 ^ 2 + x ^ 2, and the solution is 7 / 6

As shown in the figure, fold the rectangle ABCD along AE, so that point d just falls on a point F on the edge of BC. If ∠ BAF = 60 °, calculate ∠ DAE

The ∵ quadrilateral ABCD is a rectangle, ∵ bad = 90 °, ∵ BAF = 60 °, ∵ fad = 90 ° - 60 ° = 30 °, ∵ AEF is formed by ∵ AED folding, ∵ DAE = 12 ∵ fad = 12 × 30 ° = 15 °

In rectangle ABCD, ab = 6, BC = 8. If the rectangle is folded along the diagonal BD, then the area of overlap is————

If the intersection is O, then the overlap is Δ BDO; if Δ Ab0 ≌ Δ DCO, let do = x, then the square of X - (8-x) is 36, then x = 25 / 4; s Δ BDO = 1 / 2do × AB = 1 / 2 × 25 / 4 × 6 = 75 / 4 = 18.75