There is a rectangular sports ground. If its length and width are increased by 6 meters, the area will be increased by 1236 meters. What is the perimeter of the original sports ground If the equation, please say the practice. Kneel ah!

There is a rectangular sports ground. If its length and width are increased by 6 meters, the area will be increased by 1236 meters. What is the perimeter of the original sports ground If the equation, please say the practice. Kneel ah!


Let the original length be a and the width be B, then (a + 6) (B + 6) - AB = 1236, that is ab + 6 (a + b) + 36 AB = 1236
6 (a + b) = 1200, a + B = 200, perimeter = 400



There is a rectangular sports ground. If its length and width are increased by meters, the area will increase by 1236 square meters. What is the perimeter of the original sports ground
6m


If the length is x and the width is y, then
(X+6)*(Y+6)-XY=1236
The solution is x + y = 206
Therefore, the previous perimeter was 206 * 2 = 412m



If the length and width of a rectangular stadium increase by 6 meters, the area will increase by 1236 square meters. What is the perimeter of the original rectangle?


Suppose: the original rectangle is x in length and Y in width
xy+1236=(x+6)(y+6)
6x+6y=1200
2X + 2Y = 400m
So the circumference of the rectangle was 400 meters



There is a rectangle. Its length and width are increased by six meters and its area is increased by 1236 square meters. How many meters is the perimeter of the original rectangle?


Let the length be x and the width be y
There are (x + 6) * (y + 6) - x * y = 1236
The equivalence of the above formula and
XY+6X+6Y+36-XY=1236
So 6 (x + y) = 1200
With perimeter = 2 (x + y) = 400
It's done



Cut a 20 cm long iron wire into two sections, and make a square with the length of each section as the perimeter. To make the sum of the two squares equal to 17 square centimeters, what are the lengths of the two sections?


Let the length of one square be xcm, then the 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 20 cm long wire into two sections, and make a square with the circumference of each section
To make the sum of the areas of the two squares equal to 17 square centimeters, what are the lengths of the two sections of the wire


Let one of the squares be X
Then the side length of the other square is (20-4x) / 4 = 5-x
x²+(5-x)²=17
x²-5x+4=0
The solution is x = 1 or x = 4
That is, the side length of two squares
So the length of the wire cut into two sections is 4cm and 16cm respectively



There are () kinds of spelling a: 2 B: 3 C: 4
There are () ways to make a rectangle with 12 squares of 3cm on each side
A:2 B:3 C:4


B,1*12,2*6,3*4



There are () different ways to make a rectangle with 48 squares whose sides are 2 cm long. The minimum perimeter of the rectangle is () cm


There are (5) different ways to make a rectangle with 48 squares whose sides are 2 cm long. The minimum perimeter of the rectangle is (56) cm
The length is 16 cm, the width is 12 cm, and the circumference is 56 cm



Eight cubes with the edge length of 3cm can be put together into several large cuboids or cubes with different shapes. Draw a picture, spell a spell, and calculate the sum of their edge lengths respectively. Which figure has the smallest sum of edge lengths?


Three species: 1 * 1 * 8, 1 * 2 * 4, 2 * 2 * 2
1 * 1 * 8: 4 * 8 * 3 + (4 + 4) * 3 = 120 cm
1 * 2 * 4: 4 * 4 * 3 + (6 + 6) * 3 = 84cm
2 * 2 * 2: 12 * 2 * 3 = 72cm
2 * 2 * 2 is the smallest



Use 12 cuboids with 1 cm long edges to make cuboids of different shapes
A. 4B. 8C. 12D. 3


There are four kinds of spelling: a