The length of the rectangle is 2 times its width. If the width of the rectangle is increased by 8 dm

The length of the rectangle is 2 times its width. If the width of the rectangle is increased by 8 dm, and the length is reduced by 10 dm, then the area of the rectangle will increase by 220 dm ^ 2. Find the area of the rectangle.

Let the width of the rectangle be x dm, then the length of the rectangle is 2x dm, and its area is equal to the product of its sides, i.e. x * 2x = 2x ^ 2 dm ^ 2. If the width of the rectangle is increased by 8 dm, then it will become equal to (x + 8) dm, and if the length is reduced by 10 dm, then it will be equal to (2x – 10) dm and the area will be equal to (x + 8) (2x – 10 ) dm ^ 2. By the condition of the problem, it is known that after this the area of ​​the rectangle will increase by (x + 8) (2x – 10) – 2x ^ 2 dm ^ 2 or by 220 dm ^ 2. Let’s make an equation and solve it.

(x + 8) (2x – 10) – 2x ^ 2 = 220;

2x ^ 2 – 10x + 16x – 80 – 2x ^ 2 = 220;

6x – 80 = 220;

6x = 220 + 80;

6x = 300;

x = 300: 6;

x = 50 (dm) – width;

2x = 50 * 2 = 100 (dm) – length.

Find the area of ​​the rectangle:

S = 50 * 100 = 5000 (dm ^ 2).

Answer. 5000 dm ^ 2.