Find the area of a rhombus whose perimeter is 80 m and one of whose diagonals is 24 m

To find the area of a rhombus, we need to know the lengths of both diagonals. However, in this case, we only know the length of one diagonal and the perimeter.

The perimeter of a rhombus is given by 4a, where a is the length of one side. So, if the perimeter is 80 m, then each side of the rhombus is 80/4 = 20 m.

Now, we know that in a rhombus, the diagonals are perpendicular bisectors of each other. This means they cut each other at right angles and each half of one diagonal is equal to half of the other.

Since we know one diagonal is 24 m, we can use the Pythagorean theorem to find the other diagonal. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

Let's denote the unknown diagonal as d. We can set up the equation as follows:

(20^2) = (24/2)^2 + (d/2)^2 400 = 144 + (d/2)^2 256 = (d/2)^2 d/2 = √256 d/2 = 16 d = 32 m

Now that we have the lengths of both diagonals, we can find the area of the rhombus. The area A of a rhombus is given by the formula A = 1/2 * d1 * d2, where d1 and d2 are the lengths of the diagonals.

So, A = 1/2 * 24 m * 32 m = 384 m^2.

Therefore, the area of the rhombus is 384 square meters.