Let’s denote the altitude of the rhombus as h.

The formula for the area of a rhombus is A = (d1 * d2) / 2, where d1 and d2 are the lengths of the diagonals.

Since the diagonals of a rhombus are perpendicular bisectors of each other, we can use the Pythagorean theorem to find the lengths of the diagonals.

Let’s denote the lengths of the diagonals as d1 and d2.

We know that the perimeter of the rhombus is 28 m, so each side of the rhombus is 28 / 4 = 7 m.

Since the diagonals bisect each other at right angles, we can divide the rhombus into four congruent right triangles.

The hypotenuse of each right triangle is a side of the rhombus, which is 7 m.

Let’s denote the legs of each right triangle as a and b.

Using the Pythagorean theorem, we have:

a^2 + b^2 = 7^2 a^2 + b^2 = 49

Since the diagonals of a rhombus bisect each other, the lengths of the diagonals are twice the lengths of the legs of each right triangle.

So, d1 = 2a and d2 = 2b.

The area of the rhombus is given as 56 m², so we have:

(2a * 2b) / 2 = 56 4ab = 112 ab = 28

Now, let’s substitute the value of ab into the equation a^2 + b^2 = 49:

a^2 + (28/a)^2 = 49 a^2 + 784/a^2 = 49 a^4 + 784 = 49a^2 a^4 - 49a^2 + 784 = 0

This is a quadratic equation in terms of a^2. Let’s solve it using the quadratic formula:

a^2 = (-(-49) ± sqrt((-49)^2 - 4 * 1 * 784)) / (2 * 1) a^2 = (49 ± sqrt(2401 - 3136)) / 2 a^2 = (49 ± sqrt(-733)) / 2

Since the area of the rhombus is positive, we can ignore the negative square root.

a^2 = (49 + sqrt(-733)) / 2

Since the square root of a negative number is not a real number, there is no real solution for a^2.

Therefore, there is no real solution for the altitude of the rhombus.