It is known that the length and width of rectangular A are 2 and 1, respectively

1) Let the length of the rectangle a be a and the width be b

a+b=2(1+2)=6

a*b=2(1*2)=4

Solve b = 3 + 2 5, a = 3 - 5, or a = 3 + 2 5, b = 3 - 52) Let the length of the rectangle c be c and the width be d

c+d=3/2

c*d=1c, d has no solution.

1 Given that the length and width of rectangle a are 2 and 1, respectively, is there another rectangle b whose circumference and area are twice the circumference and area of rectangle a? For the above problems, Xiao Ming solved them from the perspective of "graphics" and using function images. The process of Xiao Ming's argumentation began like this:

If x and y are used to represent the length and width of the rectangle, then the rectangle b satisfies x y = y = 4. Please follow Xiao Ming's argumentation ideas to complete the following argumentation process. x y=6 xy=4 x(6-x)=4 -x square 6x-4=0 x=3 root number 3 or 3-root number 3 so there is.

2 Given that the length and width of rectangle a are 2 and 1, respectively, is there another rectangle c whose circumference and area are half the circumference and area of rectangle a? Xiao Ming thinks this question is yes, do you agree with Xiao Ming's point of view? Why?

Same as above x y=3 2 xy=1 x(2 3-x)=1 -x square 2x 3-1=0 According to the discriminant formula of the root, x has no solution. So it doesn't exist. References:

Solution: (1) Point (x,y) can be regarded as the coordinates of the image of the primary function y=-x+6 in the first quadrant, and the point (x,y) can be regarded as the coordinates of the image of the inverse proportional function y= in the first quadrant, and the point (x,y) that satisfies the requirements of the problem can be regarded as the coordinates of the intersection of the image of the primary function y=-x+6 and the image of the inverse proportional function y= in the first quadrant

Two images are drawn separately (Fig. 1), and it can be seen from the diagram that such an intersection exists, i.e., a rectangle b that satisfies the requirements.

1 Given that the length and width of rectangle a are 2 and 1, respectively, is there another rectangle b whose circumference and area are twice the circumference and area of rectangle a? For the above problems, Xiao Ming solved them from the perspective of "graphics" and using function images. The process of Xiao Ming's argumentation began like this:

If you use x and y to represent the length and width of the rectangle, respectively. Then the rectangle b satisfies x+y=. Please follow Xiao Ming's argumentation ideas to complete the following argumentation process.

x+y=6 xy=4

x(6-x)=4 -x-square + 6x-4=0x=3 + root number 3 or 3-root number 3

So there is. 2 Given that the length and width of rectangle a are 2 and 1, respectively, is there another rectangle c whose circumference and area are half the circumference and area of rectangle a? Xiao Ming thinks this question is yes, do you agree with Xiao Ming's point of view? Why?

Same as above. x+y=3/2 xy=1

x(2 3-x)=1 -x square + 2x 3-1=0 According to the discriminant formula of the root, x has no solution.

So it doesn't exist.

"Pink Plum Language": Hello.

There is no such thing as a blessing. Good bye.

Yes. The length of rectangle b is about and the width is about.

According to the conditions given in the question, the area of the rectangle is A2 + 2AB + B2. Suppose the width is a and the length is b.

We know that the area of a rectangle is equal to the width multiplied by the length, so we get the following equation:

A * B = A 2 + 2AB + B 2 gives AB to the left of the equation :

ab = a^2 + ab

ab - ab = a 2 + ab - ab gets: 0 = a 2

Therefore, according to the above equation, we can know that a = 0, i.e. the width of the rectangle is 0.

Note: There may be an error in the rectangular area-type checker given in the question, because usually the width and length of the rectangle are positive, and the above calculation results in a width of 0, which is not in accordance with the norm. If there are other requirements or conditions for the question, please provide more detailed information.

(a22a) a=a+2, chaotic mountain.

The other key is long a+2, so the answer is: a+2

The area of the unpainted part is the area of the blank rectangle with the side length of A-C, B-C, and the area of the unpainted part of the rectangle is Yes (Spine Bishen A-C) (b-c).

Therefore, C

The image of x is the coordinates of the point in the first quadrant, and the point (x,y) that satisfies the requirements of the problem can be regarded as the image of the primary function y=-x+6 and the inverse proportional function y=4

The coordinates of the intersection of the image of x in the first quadrant

Two images are drawn separately (as shown in the figure on the right), and it can be seen from the diagram that such an intersection exists, that is, a rectangle B that satisfies the requirements

2) Disagree with Xiao Ming's point of view

If x and y are used to represent the length and width of the rectangle, then the rectangle c satisfies x+y=322

The image is inversely proportional to y=1

The coordinates of the intersection of the image of x in the first quadrant

As you can see from the diagram (on the right), such an intersection does not exist, i.e., the rectangle c that satisfies the requirements does not exist

So I don't agree with Xiao Ming's point of view

The rectangle has four sides, 2 long and 2 wide.

2（a+3)+2(3a-1)=12a=1

The coordinates of 1 rotation are (3,2), and in 2009 it is also (3,2), because the center is centered on the vertex in the lower right corner, then every 4 times is 1 period.

Let the length and width of rectangle b be a and b, respectively

then there is a+b=12 ab=4

a b can be seen as the two roots of the equation x squared + 12x + 4 = 0, see if is greater than or equal to 0, =144-16 is greater than 0, so there is such a rectangle.