ABCD is a rectangle and there are four equilateral triangles. Area of $\Delta ASD$ equals to area of  $\Delta BQC$ and area of $\Delta DRC$ equals to area of $\Delta APB$. The perimeter of the rectangle is $12 cm$. Also the sum of the area of the four triangles is $10\sqrt{3 }c{m}^2$ , then the total area of the figure thus formed is

$ABCD$ is a rectangle

$\begin{array}{l}\text{area}\left(\Delta {\rm A}SD\right)=\text{area}\left(\Delta BQC\right)\\ \text{area}\left(\Delta DRC\right)=\text{area}\left(\Delta APB\right)\end{array}$

Perimeter of $\text{ABCD}=12 cm$

Sum of the $4$ triangles $=10\sqrt{3}c{m}^2$

Sum of 4 triangles $=10\sqrt{3}c{m}^2$

$\begin{array}{l}2\left(\frac{\sqrt{3}}{4}{x}^2\right)+2\left(\frac{\sqrt{3}}{4}{y}^2\right)=10\sqrt{3}\\ \frac{\sqrt{3}}{2}(x^2+y^2)=10\sqrt{3}\\ x^2+y^2=\frac{10\sqrt{3}\times 2}{\sqrt{3}}\\ x^2+y^2=20\to \left(1\right)\end{array}$

Perimeter of rectangle $=12 cm$

$\begin{array}{l}2(x+y)=12\\ x+y=6\to \left(2\right)\end{array}$

From (1)  and (2)

$\begin{array}{l}{(x+y)}^2=x^2+y^2+2xy\\ 36=20+2xy\\ 16=2xy\\ xy=8\end{array}$

Total area $=2\left[\frac{\sqrt{3}}{4}{x}^2\right]+2\left[\frac{\sqrt{3}}{4}{y}^2\right]+xy$

$\begin{array}{l}=\frac{\sqrt{3}}{2}[x^2+y^2]+xy\\ =\frac{\sqrt{3}}{2}\left[20\right]+8\\ =10\sqrt{3}+8\\ =2(4+5\sqrt{3})c{m}^2\end{array}$