The side of the rhombus is 65 and the diagonal is 120. Find the area of the rhombus.

Let a rhombus ABCD be given.
The diagonals of the rhombus (AC, BD) intersect at right angles and halve at the intersection point (K)
Consider the triangle ABK, because the angle K of the straight line can be used by the Pythagorean theorem.
Let AK = a, KB = b, AB = c
AK = AC / 2 = 120/2 = 60
Then we get:
c ^ 2 = a ^ 2 + b ^ 2
65 ^ 2 = 60 ^ 2 + b ^ 2
65 ^ 2 – 60 ^ 2 = b ^ 2
625 = b ^ 2
b = 25

b = BK, BK = 25
BK = KD, so BD = 25 + 25 = 50

Since we know both diagonals of the rhombus, we find its area by the formula:
S = 1/2 * AC * BD = 1/2 * 50 * 120 = 6000/2 = 3000