11.9. Since the area of an equilateral triangle with side $a$ is $a^{2} \sqrt{3} / 4$, the side of an equilateral triangle with area 1 is $2 / \sqrt[4]{3}$, and its height is $\sqrt[4]{3}$. We will prove that an equilateral triangle with area 1 cannot be cut from a strip of width less than $\sqrt[4]{3}$. Let the equilateral triangle $A B C$ lie inside a strip of width less than $\sqrt[4]{3}$. Let the projection of vertex $B$ onto the boundary of the strip lie between the projections of vertices $A$ and $C$. Then the line drawn through point $B$ perpendicular to the boundary of the strip intersects segment $A C$ at some point $M$. The height of triangle $A B C$ does not exceed $B M$, and $B M$ is no greater than the width of the strip, so the height of triangle $A B C$ is less than $\sqrt[4]{3}$, i.e., its area is less than 1.

It remains to prove that any triangle with area 1 can be cut from a strip of width $\sqrt[4]{3}$. We will prove that any triangle with area 1 has a height not exceeding $\sqrt[4]{3}$. For this, it is sufficient to prove that it has a side not less than $2 / \sqrt[4]{3}$. Suppose all sides of triangle $A B C$ are less than $2 / \sqrt[4]{3}$. Let $\alpha$ be the smallest angle of this triangle. Then $\alpha \leqslant 60^{\circ}$ and $S_{A B C}=(A B \cdot A C \sin \alpha) / 2<(2 / \sqrt[4]{3})^{2}(\sqrt{3} / 4)=1$. A contradiction is obtained. A triangle that has a height not exceeding $\sqrt[4]{3}$ can be placed in a strip of width $\sqrt[4]{3}$ by placing the side to which this height is dropped on the side of the strip.
11.9*. What is the minimum width that an infinite strip of paper must have so that any triangle with an area of 1 can be cut out of it?