The perimeter of the rectangle is 36m. If its length is increased by 1m, and its width is increased by 2m

The perimeter of the rectangle is 36m. If its length is increased by 1m, and its width is increased by 2m, then its area will increase by 30m. Determine the area of the original rectangle.

Let’s say that the length of this rectangle is x m and the width is y m.

Then its perimeter will be equal to:

36 = 2 * (x + y),

x + y = 18.

And the area is S = x * y.

If the length is increased by 1 m, and the width by 2 m, then the area of ​​the resulting rectangle will be equal to:

S = (x + 1) * (y + 2).

By the condition of the problem:

(x + 1) * (y + 2) = x * y + 30.

From the first equation we get that y = 18 – x.

Let’s plug this value into our equation:

(x + 1) * (18 – x + 2) = x * (18 – x) + 30,

(x + 1) * (20 – x) = 18 * x – x² + 30,

20 * x – x² + 20 – x = 18 * x – x² + 30,

x = 30 – 20,

x = 10 (m) – the length of the rectangle.

y = 18 – 10 = 8 (m) – the width of the rectangle.

The area of ​​the original rectangle is:

10 * 8 = 80 (m²).