Rhombus $ABCD$ has perimeter $148$, and one of its diagonals has length $24$. What is the area of $ABCD$?

Step 1: Understand the problem A rhombus is a type of quadrilateral where all four sides are of equal length. Given that the perimeter of the rhombus ABCD is 148, each side of the rhombus is 148/4 = 37 units.

Step 2: Use the properties of a rhombus In a rhombus, the diagonals are perpendicular bisectors of each other. This means they intersect at a right angle and divide each other into two equal parts. Given that one of the diagonals is 24 units, each half of this diagonal is 24/2 = 12 units.

Step 3: Use the Pythagorean theorem Since the diagonals of a rhombus form a right angle at their intersection, we can use the Pythagorean theorem to find the length of the other diagonal. Let's denote the other diagonal as d. Then, we have:

37^2 = 12^2 + (d/2)^2 1369 = 144 + d^2/4 1225 = d^2/4 d^2 = 4900 d = 70 units

Step 4: Find the area of the rhombus The area of a rhombus is given by the formula 1/2 * (product of the diagonals). Therefore, the area of the rhombus ABCD is:

Area = 1/2 * 24 * 70 = 840 square units.