Find the area ANLU in the figure below

Question

Two congruent circles, whose radii measure $R=3$. Calculate the area $ANLU$.Consider that $D$ and $T$ are points of tangency.(S:$3(9+\sqrt3)$

$\triangle OBE: OB^2=6^2-3^2 \implies OB = 3\sqrt3\\ \therefore NB=3+3\sqrt3$

$S_{ANLU} = \left(\dfrac{AN+3+3\sqrt3}{2}\right).6=9+9\sqrt3+3AN$

I didn't find $AN$

Answer 1

You've made a good start. Below is your diagram with a couple of lines and a line length added:

As shown, a horizontal line is drawn through $D$, meeting $LN$ at $G$ and $UB$ at $H$. Also, the vertical line $DJ$ is drawn, meeting $NB$ at $J$. Finally, $\lvert LG\vert = m$, so

$$\lvert GN\rvert = \lvert DJ\rvert = 6 - m \tag{1}\label{eq1A}$$

Next, we have

$$90^{\circ} = \measuredangle LDG + \measuredangle GDO = \measuredangle GDO + \measuredangle ODJ \;\;\to\;\; \measuredangle LDG = \measuredangle ODJ \tag{2}\label{eq2A}$$

Thus, $\triangle ODJ \sim \triangle LDG$. Since $\frac{\lvert OD\rvert}{\lvert LD\rvert} = \frac{3}{6} = \frac{1}{2}$, each side of $\triangle ODJ$ has a length one-half that of the corresponding side of $\triangle LDG$. This means

$$\lvert OJ\rvert = \frac{1}{2}\lvert LG\rvert = \frac{m}{2} \tag{3}\label{eq3A}$$

Also,

$$\lvert GD\rvert = \lvert NJ\rvert = 3 + \frac{m}{2} \tag{4}\label{eq4A}$$

Using this, $\triangle ODJ \sim \triangle LDG$ and \eqref{eq1A}, we then also have

$$\begin{equation}\begin{aligned} \lvert DJ\rvert & = \frac{1}{2}\lvert GD\vert \\ 6 - m & = \frac{1}{2}\left(3 + \frac{m}{2}\right) \\ 12 - 2m & = 3 + \frac{m}{2} \\ -\frac{5m}{2} & = -9 \\ m & = \frac{18}{5} \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

Thus, using \eqref{eq4A}, we next get

$$\lvert DH\rvert = \lvert GH\rvert - \lvert GD\rvert = (3 + 3\sqrt{3}) - (3 + \frac{m}{2}) = 3\sqrt{3} - \frac{9}{5} \tag{6}\label{eq6A}$$

Since $\triangle ABU \sim \triangle DHU$, then

$$\frac{\lvert AB\rvert}{\lvert UB\rvert} = \frac{\lvert DH\rvert}{\lvert UH\rvert} \;\;\to\;\; \frac{\lvert AB\rvert}{6} = \frac{3\sqrt{3} - \frac{9}{5}}{\frac{18}{5}} \;\;\to\;\; \lvert AB\rvert = \frac{5}{3}\left(3\sqrt{3} - \frac{9}{5}\right) = 5\sqrt{3} - 3 \tag{7}\label{eq7A}$$

Next, we have

$$\lvert AN\rvert = \lvert NB\rvert - \lvert AB\rvert = (3 + 3\sqrt{3}) - (5\sqrt{3} - 3) = 6 - 2\sqrt{3} \tag{8}\label{eq8A}$$

Substituting this into your last equation gives

$$S_{ANLU} = 9 + 9\sqrt{3} + 3(6 - 2\sqrt{3}) = 27 + 3\sqrt{3} = 3(9 + \sqrt{3}) \tag{9}\label{eq9A}$$

Answer 2

You know $\theta:=\angle OLD$ since $\triangle OLD$ has a right angle at $D$, and $OD=3$, $DL=6$.

Find point $G$ on segment $LN$ so that segment $\overline {GD}$ is horizontal. You can solve right triangle $\triangle GLD$ since $\angle GLD=2\theta$ and $DL=6$. This gives you the lengths $LG$ and $GD$. Now obtain $AN$, given that $LU=NB=3+3\sqrt 3$ and $LN=6$.