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 $b$.
Let shorter diagonal equal $2x$ and longer one equal $4x-4$.
Let the angle between longer diagonal and base equal $\alpha$. By definition,
$$\sin\alpha=\frac{\frac{24}{5}}{2x-2}$$ 
By cosine rule,   $$\cos\alpha=\frac{5^2+(2x-2)^2-x^2}{2·5·(2x-2)}$$
If you calculate $\sin^2\alpha+\cos^2\alpha=1$, you can get the 4th power equation,
$$9x^4-48x^3-162x^2+336x+2745=0$$
The integer solution is only $5$ by GDC. By cosine rule,
$$b^2=5^2+8^2-2·5·8·\cos(π-\alpha)=25+64-80\cdot\left(-\frac{4}{5}\right)=153$$  $$b=\sqrt{153}=3\sqrt{17}$$ 
So that the perimeter is
$$2a+2b=10+6\sqrt{17}$$

Another real number solution is very complicated.
Another $x$ is approximately $5.68$