# The area of the rectangle is 120 cm2; its length is 7 cm more than the width of the rectangle.

The area of the rectangle is 120 cm2; its length is 7 cm more than the width of the rectangle. Find the perimeter of the rectangle.

Let the width of the rectangle be x cm. The length of the rectangle is 7 cm longer than the width, which means that it is equal to (x + 7) cm. The area of ​​the rectangle is equal to the product of its length and width x (x + 7) cm² or 120 cm². Let’s make an equation and solve it.

x (x + 7) = 120;

x² + 7x – 120 = 0;

D = b² – 4ac;

D = 7² – 4 * 1 * (-120) = 49 + 480 = 529;

x = (-b ± √D) / (2a);

x = (-7 ± √529) / (2 * 1) = (-7 ± 23) / 2;

x1 = (-7 + 23) / 2 = 16/2 = 8 (cm) – the width of the rectangle;

x2 = (-7 – 23) / 2 = -30/2 = -15 – the length of the segment cannot be negative.

x + 7 = 8 + 7 = 15 (cm) – the length of the rectangle.

Find the perimeter of the rectangle. The perimeter of a rectangle is equal to the sum of the lengths of its four sides. P = 2 (a + b).

P = 2 (8 + 15) = 2 “23 = 46 (cm). 