# The perimeter of the rhombus is 68, the area of the rhombus is 240. Find the diagonals of the rhombus.

Since the lengths of the sides of the rhombus are equal, then AB = BC = CD = AD = Ravsd / 4 = 68/4 = 17 cm.

Let the length of the smaller diagonal be 2 * X cm, and the length of the larger diagonal is 2 * Y cm.

Then OB = 2 * X / 2 = X cm, AO = 2 * Y / 2 = Y cm.

The area of the rhombus is equal to: Savsd = AC * BD / 2 = 2 * Y * 2 * X / 2.

2 * X * Y = 240.

X * Y = 120. (1)

In a right-angled triangle AOB, according to the Pythagorean theorem:

AB ^ 2 = AO ^ 2 + OB ^ 2.

289 = Y ^ 2 + X ^ 2. (2).

Let’s solve the system of equations 1 and 2.

Y = 120 / H.

289 = (120 / X) ^ 2 + X ^ 2.

289 * X ^ 2 = 14,400 + X ^ 4.

X ^ 4 – 289 * X ^ 2 + 14,400 = 0.

Let X ^ 2 = Z, then Z ^ 2 – 289 * Z + 14400 = 0.

Let’s solve the quadratic equation.

Z1 = 64, Z2 = 225.

X1 = 8 cm, X2 = 15 cm.

Y1 = 120/8 = 15 cm.

Y2 = 120/15 = 8 cm

Then, if BD = 2 * 8 = 16 cm, AC = 2 * 15 = 30 cm.

If BD = 2 * 15 = 30 cm, AC = 2 * 8 = 16 cm.

Answer: The diagonals of the rhombus are 16 cm and 30 cm.