Draw a rectangle at will, and then increase the length and width of the rectangle by 1 / 2 respectively? What do you find?

Now the area of the rectangle is the same as that of the original rectangle
（1+1/2）×（1+1/2）
＝3/2×3/2
＝9/4

There are 48 small square pieces with an area of 1 square centimeter. Now we need to use them all to make a big rectangle. How many different kinds of shapes can we make
rectangle?

It can be made into a rectangle with a width of 48 and a width of 1
It can be made into a rectangle with a width of 24 and a width of 2
It can be made into a rectangle of 16 by 3
It can be made into a rectangle 12 wide and 4 wide
It can be spelled into a rectangle with width of 8 and width of 6
Five in all

There are several pieces of rectangular paper of 5cm and 4cm in width. How much is the minimum square area? How many pieces of rectangular paper are needed?
Come on
beg
It's urgent

Use this paper to make a square to find the minimum square area (4 * 5) * (4 * 5) = 400 square centimeters
Need this kind of rectangular paper 400 (4 * 5) = 20 sheets

How many square cards with side length a can make a new square? Try three answers and show the area of the new square in two ways
What can you find in the same representation?

4
nine
twenty-five
Square number

What is the side length of a square piece of paper with an area of 80 square cm;

The side length of a square piece of paper is 8.94 cm

Cut a 20cm long iron wire into two sections, and make a circle with the length of each section of iron wire as the perimeter, then the minimum value of the sum of the areas of the two circles is

20 / 2 = 10 cm, 10 / 2 = 5 cm, 5 * 5 * 3.14 * 2 = 157 square cm is the minimum

Cut a 20cm long iron wire into two sections, and make a square with the length of each section as the perimeter. (1) to make the sum of the two squares equal to 17cm2, what are the lengths of the two sections? (2) Can the sum of the areas of two squares be equal to 12 cm2? If yes, find out the length of two pieces of wire; if not, please explain the reason

(1) Let the side length of one square be xcm, then the side length of the other square be (5-x) cm. According to the equation, we can get x2 + (5-x) 2 = 17, and we can get x2-5x + 4 = 0, (x-4) (x-1) = 0. Solving the equation, we can get X1 = 1, X2 = 4, 1 × 4 = 4cm, 20-4 = 16cm; or 4 × 4 = 16cm, 20-16 = 4cm

Cut a 20cm long iron wire into two sections, and make a square with the length of each section of iron wire as the perimeter, then the minimum value of the sum of the areas of the two squares is______ cm2．

Let the length of one section of iron wire be x and the length of the other section be (20-x), then the side lengths are 14x and 14 (20-x) respectively, then s = 116x2 + 116 (20-x) (20-x) = 18 (X-10) 2 + 12.5. By the function, when x = 10cm, s is the smallest, which is 12.5cm2

Cut a 20cm long iron wire into two sections, and make a square with the length of each section as the perimeter. (1) to make the sum of the two squares equal to 17cm2, what are the lengths of the two sections? (2) Can the sum of the areas of two squares be equal to 12 cm2? If yes, find out the length of two pieces of wire; if not, please explain the reason

(1) Let the side length of one square be xcm, then the side length of the other square be (5-x) cm. According to the equation, we can get x2 + (5-x) 2 = 17, and we can get: x2-5x + 4 = 0, (x-4) (x-1) = 0. Solving the equation, we can get X1 = 1, X2 = 4, 1 × 4 = 4cm, 20-4 = 16cm; or 4 × 4 = 16cm, 20-16 = 4cm Reason: let the sum of the areas of two squares be y, then y = x2 + (5-x) 2 = 2 (X-52) 2 + 252, ∵ a = 2 ＞ 0, ∵ when x = 52, the minimum value of y = 12.5 ＞ 12, ∵ the sum of the areas of two squares can not be equal to 12cm2; (in addition, from (1) we can know that x2 + (5-x) 2 = 12, after simplification, we get 2x2-10x + 13 = 0, ∵ △ = (- 10) 2-4 × 2 × 13 = - 4 ＜ 0, ∵ the equation has no relation So the sum of the areas of two squares can't be equal to 12cm2

Cut a 20cm long iron wire into two sections, and make a square with the length of each section of iron wire as the perimeter, then the minimum value of the sum of the areas of the two squares is______ cm2．

Draw a rectangle at will, and then increase the length and width of the rectangle by 1 / 2 respectively? What do you find?