Calculate perimeter of rhomboid I am trying to solve the following problem but I got stuck In a rhomboid with an area of $48 \space cm^2$, the major diagonal is $4$ cm shorter than the double of the minor diagonal. Calculate the exact value of the perimeter knowing that the shorter sides are of $5 \space cm$. So, if I call the shorter sides $a$, the longer sides $b$, and $D$ and $d$ the major and minor diagonals respectively, we have the following: $a=5$, $D=2d-4$, $b=?$, $perimeter=?$ In order to find the perimeter, we have to calculate the value of $b$. The area of the rhomboid is the sum of the areas of the two triangles obtained by dividing the figure by its shorter diagonal. We have $$\text{area of rhomboid}=\dfrac{2(a.h)}{2},$$$$48=5.h,$$$$h=\dfrac{48}{5}$$ So now I have the height of the rhomboid, but I still don't know how to arrive at the value of $b$. I would really appreciate some help, thanks in advance.

Hint: You can use the forumula $A = \frac{1}{2}ab\sin C$ to get the value of the diagonals. What relationship do the central angles have? (supplementary)

Once those are found, law of cosines easily finds $b$.