A rhombus has a perimeter of 40 cm. The line joining the midpoints of two adjacent sides is 6 cm long. Find the area of the rhombus.

Let's designate x  cm as the length of each side of rhombus ABCD. Its perimeter is  4x cm.

Thus, 4x = 40, which leads to x = $$x=\dfrac{40}{4}$$ = 10  cm.

In triangle ABC, L is the midpoint of side AB, and M is the midpoint of side BC, resulting in LM measuring 6 cm.

Applying the similarity criterion in triangles BLM and BAC, we get $$\dfrac{BL}{BA}=\frac{LM}{AC}$$

This simplifies to $$\dfrac{BA}{2BA}=\frac{LM}{AC}$$ , which further simplifies to 1/2 = 6/AC , leading to  AC = 12 cm.

Hence, in triangle ABC, the sides are a = 10cm,  b = 12cm, and  c = 10cm.

The semi-perimeter, s, is calculated as (10+12+10)/2 = 16

The area of triangle ABC is given by $$\sqrt{𝑠(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)\ }$$, which equals$$\sqrt{16\times(16−10)\times(16−12)\times(16−10)\ }$$​, which further simplifies to $$\sqrt{16\times6\times4\times6\ }=48cm^2$$

As the rhombus ABCD consists of two congruent triangles ABC, its area equals 2×area of triangle ABC=2×48 cm², resulting in 96cm².