In trapezoid $A B C D$, angle $B A D$ is right, angle $A B C$ equals $\operatorname{arctg} 2$ and $A B=A D$. Square $K L M N$ is positioned in space such that its center coincides with the midpoint of segment $A B$. Point $A$ lies on side $L K$ and $A L<A K$, point $M$ is equidistant from points $A$ and $D$. The distance from point $L$ to the nearest point of trapezoid $A B C D$ is $\frac{3}{2}$, and the distance from point $N$ to the nearest point of trapezoid $A B C D$ is $\sqrt{6}$. Find the area of trapezoid $A B C D$ and the distance from point $M$ to the plane $A B C D$.
Let $O$ be the midpoint of the base $A B$ of trapezoid $A B C D$. Since point $O$ is the center of square $KLMN$ and point $A$ lies on side $K L$ of this square, point $B$, symmetric to point $A$ with respect to $O$, lies on the opposite side $M N$ of square $K L M N$, and $A L = B N = \sqrt{6}$. Denote $A B = A D = a, \angle B A K = \alpha$. Then, if $E$ is the projection of point $C$ onto the base $A B$ of trapezoid $A B C D$, we have

$$
B E = C E \operatorname{ctg} \angle A B C = A D \operatorname{ctg} \operatorname{arctg} 2 = a \cdot \frac{1}{2} = \frac{a}{2}
$$

Let $F$ be the midpoint of $K L$. Point $F$ lies between points $A$ and $K$, since $A L 45^{\circ}$ (as the external angle of triangle $\left.A L O\right)$, then

$$
\alpha + 45^{\circ} = 180^{\circ} - \arcsin \frac{2}{\sqrt{5}}, \alpha = 180^{\circ} - 45^{\circ} - \arcsin \frac{2}{\sqrt{5}} = 135^{\circ} - \arcsin \frac{2}{\sqrt{5}}
$$

Then

$$
\sin \left(\alpha - 45^{\circ}\right) = \sin \left(90^{\circ} - \arcsin \frac{2}{\sqrt{5}}\right) = \cos \left(\arcsin \frac{2}{\sqrt{5}}\right) = \frac{1}{\sqrt{5}}
$$

$$
\sin \alpha = \sin \left(135^{\circ} - \arcsin \frac{2}{\sqrt{5}}\right) =
$$

$$
= \sin 135^{\circ} \cos \left(\arcsin \frac{2}{\sqrt{5}}\right) - \cos 135^{\circ} \sin \left(\arcsin \frac{2}{\sqrt{5}}\right) = \frac{\sqrt{2}}{2} \left(\sqrt{5} + \frac{2}{\sqrt{5}}\right) = \frac{3 \sqrt{2}}{2 \sqrt{5}}
$$

$$
\cos \beta = \frac{\sqrt{15}}{2 \sqrt{6} \sin n} = \frac{\sqrt{15}}{2 \sqrt{6} \cdot \frac{\sqrt{2}}{2 \sqrt{3}}} = \frac{5}{6},
$$

$$
\operatorname{tg} \beta = \sqrt{\frac{1}{\cos ^{2} \beta} - 1} = \sqrt{\frac{36}{25} - 1} = \frac{\sqrt{11}}{5}
$$

Applying the Law of Sines to triangle $A O L$, we get

$$
\frac{A O}{\sin 45^{\circ}} = \frac{A L}{\sin \left(\alpha - 45^{\circ}\right)}, \frac{\frac{a}{2}}{\frac{\sqrt{2}}{2}} = \frac{\sqrt{6}}{\frac{1}{\sqrt{5}}}
$$

Therefore,

$$
a = \frac{\frac{\sqrt{2} - \sqrt{6}}{1}}{\frac{1}{\sqrt{5}}} = 2 \sqrt{15}
$$

Then, if $S$ is the area of trapezoid $A B C D$, we have

$$
S = \frac{1}{2}(A B + C D) \cdot A D = \frac{1}{2}\left(a + \frac{1}{2} a\right) a = \frac{3}{4} a^2 = \frac{3}{4} \cdot 60 = 45
$$

Finally, from the right triangle $M P M'$, we find the distance from point $M$ to the plane $A B C D$:

$$
M M' = M' P \operatorname{tg} \beta = \frac{2}{2} \cdot \frac{\sqrt{11}}{5} = \sqrt{15} \cdot \frac{\sqrt{11}}{5} = \frac{\sqrt{165}}{5}.
$$

## Answer

$S = 45, d = \frac{\sqrt{165}}{5}$.