7) 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_84f5f6e83b5066604abfg-1.jpg?height=317&width=283&top_left_y=371&top_left_x=2432)
7. The answer is $(\mathbf{E})$.

The triangles $A B C$ and $D E C$ 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,

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

![](https://cdn.mathpix.com/cropped/2024_04_17_4701f996ee50400fb9c5g-2.jpg?height=403&width=317&top_left_y=1523&top_left_x=1646)