# The difference between the diagonals of the rhombus is 14 cm. The area of the rhombus is 120 cm².

**The difference between the diagonals of the rhombus is 14 cm. The area of the rhombus is 120 cm². Find the perimeter of the rhombus**

Let the smaller diagonal of the rhombus be x, then the larger one is x + 14. The area of the rhombus is equal to half of the product of the diagonals, we compose the equation:

0.5 * x * (x + 14) = 120;

x * (x + 14) = 240;

x ^ 2 + 14x-240 = 0.

D = b ^ 2-4ac = 14 ^ 2-4 * (- 240) = 196 + 960 = 1156.

x1 = (- b-√D) / 2a = (- 14-34) / 2 = -48 / 2 = -24. The first root has a negative value, which means it does not satisfy the solution of the problem.

x2 = (- b + √D) / 2a = (- 14 + 34) / 2 = 20/2 = 10.

Therefore, one of the diagonals of the rhombus is 10 cm, the second 10 + 14 = 24 cm.

Consider a right-angled triangle in which the hypotenuse is the side of the rhombus, the legs are half of its diagonals. The sum of the squares of the legs is equal to the square of the hypotenuse, we can find the square of the side of the rhombus:

(10/2) ^ 2 + (24/2) ^ 2 = 25 + 144 = 169. The side of the rhombus is √169 = 13 cm. Therefore, the perimeter of the rhombus is 13 * 4 = 52 cm.