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 xย cm as the length of each side of rhombus ABCD. Its perimeter isย 4x cm.
Thus, 4x = 40, which leads to x = $$x=\dfrac{40}{4}$$ = 10ย 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 toย $$\dfrac{BA}{2BA}=\frac{LM}{AC}$$ย , which further simplifies to 1/2 = 6/ACย , leading toย AC = 12 cm.
Hence, in triangle ABC, the sides are a = 10cm,ย b = 12cm, andย c = 10cm.
The semi-perimeter, s, is calculated as (10+12+10)/2 = 16
The area of triangle ABC is given by $$\sqrt{๐ (๐ โ๐)(๐ โ๐)(๐ โ๐)\ }$$, which equals$$\sqrt{16\times(16โ10)\times(16โ12)\times(16โ10)\ }$$โ, which further simplifies to $$\sqrt{16\times6\times4\times6\ }=48cm^2$$
As the rhombus ABCD consists of two congruent triangles ABC, its area equals 2รarea of triangle ABC=2ร48 cmยฒ, resulting in 96cmยฒ.