17) In the adjacent figure, triangle $A B C$ is equilateral with a side length of $1 \mathrm{~m}$ and $D E F G$ is a square. What is the length of side $D E$?
(A) $\frac{1}{3} \mathrm{~m}$
(B) $(2 \sqrt{3}-3) \mathrm{m}$,
(C) $\frac{1}{2} \mathrm{~m}$,
(D) $\frac{1+\sqrt{3}}{4} \mathrm{~m}$,
(E) $(\sqrt{3}-1) \mathrm{m}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_880093126f4ae9b83064g-2.jpg?height=283&width=314&top_left_y=832&top_left_x=1070)
17. The answer is $(\mathbf{B})$.

Let $C H$ be the height of triangle $A B C$ relative to the base $A B$ and let $x=\overline{D E}$. Observe that triangles $A H C$ and $G K C$ are similar; therefore, the proportion holds:

$$
\overline{A H}: \overline{C H}=\overline{G K}: \overline{C K}
$$

On the other hand, $\overline{A H}=1 \mathrm{~m}, \overline{C H}=\sqrt{3} / 2 \mathrm{~m}, \overline{G K}=x / 2 \mathrm{~m}$ and $\overline{C K}=\overline{C H}-x=\sqrt{3} / 2 \mathrm{~m}-x$. It follows then:

$$
\frac{1}{2}: \frac{\sqrt{3}}{2}=\frac{x}{2}: \frac{\sqrt{3}}{2}-x, \quad \text { i.e. } \quad \sqrt{3} x=\sqrt{3}-2 x .
$$

Solving for $x$,

$$
x=\frac{\sqrt{3}}{2+\sqrt{3}} \mathrm{~m}=\frac{\sqrt{3}(2-\sqrt{3})}{4-3} \mathrm{~m}=(2 \sqrt{3}-3) \mathrm{m} .
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_3d973c0a184ce9fc7311g-4.jpg?height=454&width=480&top_left_y=1418&top_left_x=1413)