Find the area of a rectangle whose length is 4 times its width and the numerical value of the area

Find the area of a rectangle whose length is 4 times its width and the numerical value of the area is equal to the perimeter.

Let x denote the width of this rectangular quadrangle.

In the initial data for this task, it is reported that the length of this rectangular quadrangle is four times its width, which is 4x.

Also in the problem statement it is said that the sum of the lengths of all four sides of a given quadrangle is equal to its area, therefore, we can draw up the following equation:

x + x + 4x + 4x = x * 4x,

solving which, we get:

10x = 4x²;

4x² – 10x = 0;

4x * (x – 10/4) = 0;

x1 = 0;

x2 – 10/4 = 2.5.

Since the width of the rectangle cannot be 0, the value x = 0 is not suitable.

Find the length:

4x = 4 * 2.5 = 10.

Let’s find the area of ​​this geometric figure:

2.5 * 10 = 25.

Answer: The area of ​​the rectangle is 25.