Use a 54 cm long wire to form a rectangle. If the length is twice the width, what is the length and width of the rectangle? What's the area?

Use a 54 cm long wire to form a rectangle. If the length is twice the width, what is the length and width of the rectangle? What's the area?

So the length of the rectangle is 9 × 2 = 18 (CM), and the area is 18 × 9 = 162 (CM). A: the length of the rectangle is 18 cm, the width is 9 cm, and the area is 162 square cm

Use a 54 cm long wire to form a rectangle. If the length is twice the width, what is the length and width of the rectangle? What's the area?

So the length of the rectangle is 9 × 2 = 18 (CM), and the area is 18 × 9 = 162 (CM). A: the length of the rectangle is 18 cm, the width is 9 cm, and the area is 162 square cm

The length and width of a rectangle are increased by 5cm, the area is increased by 85CM, and the perimeter of the original rectangle is () cm

85-25=60
60/5=12
12*2=24cm

Given that the radius of the ball is r, make an inscribed cylinder in the ball. What is the value of the radius and height of the bottom of the cylinder, when its side area is the largest? What is the maximum side area?

As shown in the figure, if the height of the cylinder is h, the bottom radius is r, and the side area is s, then (H2) 2 + R2 = R2, that is, H = 2r2 − R2. ∵ s = 2 π RH = 4 π R · R2 − R2 = 4 π R2 · (R2 − R2) ≤ 4 π R2 + R2 − R22 = 2 π R2. If and only if R2 = r2-r2, take the equal sign, then the bottom radius of the inscribed cylinder is 22r, and the height is 2R

When the side length of a square increases by 3cm, its area increases by 39cm2. The side length of the square is ()
A. 5cmB. 6cmC. 8cmD. 10cm

Let the original side length of the square be x, then x2 + 39 = (x + 3) 2, the solution is x = 5, so a

When the side length of a square increases by 3cm, its area increases by 39cm2. The side length of the square is ()
A. 5cmB. 6cmC. 8cmD. 10cm

Let the original side length of the square be x, then x2 + 39 = (x + 3) 2, the solution is x = 5, so a

When the side length of a square increases by 3cm, its area increases by 39cm2. The side length of the square is ()
A. 5cmB. 6cmC. 8cmD. 10cm

Let the original side length of the square be x, then x2 + 39 = (x + 3) 2, the solution is x = 5, so a

When the side length of a square increases by 3cm, its area increases by 39cm2. The side length of the square is ()
A. 5cmB. 6cmC. 8cmD. 10cm

Let the original side length of the square be x, then x2 + 39 = (x + 3) 2, the solution is x = 5, so a

Question 1: if the side length of a square is increased by 3cm, its area will be increased by 39cm square. What is the side length of the square

Let the original side length be x cm
（x+3）²-x²=39
（x+3+x）（x+3-x）=39
2x+3=13
2x=10
x=5

The side length of a square tile is 6 decimeters. How big is the area of such a tile? Is it enough to lay a 12 square meter floor with 20 such tiles?
To write a column

6 decimeters = 0.6 meters
1. The area of this tile is: 0.6 × 0.6 = 0.36 (M2)
2. The total area of 20 such tiles is: 0.36 × 20 = 7.2 (M2)
3. It's only 7.2 square meters, so it's not enough