# 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 x be the width of the rectangle, then since the length of the rectangle is 7 cm larger than the width, then the length is 7 + x cm.The area of the rectangle is equal to the product of the length by the width: S = a * b, according to the condition of the problem, the value of the area is 120 cm square. Let’s make the equation: x * (x + 7) = 120, x ^ 2 + 7x – 120 = 0. Find the discriminant of the quadratic equation: D ^ 2 = b ^ 2 – 4ac = 72 – 4 · 1 · (-120) = 49 + 480 = 529. Since the discriminant is greater than zero, the quadratic equation has two real roots: x1 = (-7 – √529) / 2 · 1 = -15. x2 = (-7 + √529) / 2 1 = 8. The width of the rectangle is 8 cm, and the length is 8 + 7 = 15 cm.Let’s find the perimeter of the rectangle: P = 2 * a + 2 * b = 2 * 8 cm + 2 * 15 cm = 16 cm + 30 cm = 46 cm.