9) In the adjacent figure, the segment $D E$ is parallel to $A B$. Knowing that the area of $D E C$ is equal to $\frac{3}{4}$ of that of $A B C$ and that $A C$ measures $1 \mathrm{~m}$, what is the length of $D C$?
(A) $\frac{2-\sqrt{3}}{2} \mathrm{~m}$,
(B) $(2-\sqrt{3}) \mathrm{m}$
(C) $\frac{\sqrt{3}}{3} \mathrm{~m}$,
(D) $\frac{3}{4} \mathrm{~m}$,
(E) $\frac{\sqrt{3}}{2} \mathrm{~m}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_880093126f4ae9b83064g-1.jpg?height=332&width=298&top_left_y=692&top_left_x=2417)
9. The answer is $(\mathbf{E})$.

The triangles $A B C$ and $A^{\prime} B^{\prime} C$ are similar. We know that the ratio of the areas of two similar figures is equal to the square of the ratio of two corresponding sides. Consequently,

$$
\left(\frac{\overline{A^{\prime} C}}{\overline{A C}}\right)^{2}=\frac{\operatorname{area}\left(A^{\prime} B^{\prime} C\right)}{\text { area }(A B C)}=\frac{3}{4}, \quad \text { i.e. } \quad \overline{A^{\prime} C}=\frac{\sqrt{3}}{2} \overline{A C}=\frac{\sqrt{3}}{2} \mathrm{~m}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_3d973c0a184ce9fc7311g-2.jpg?height=411&width=300&top_left_y=1965&top_left_x=1660)