5. $A B$ and $C D$ are two segments, both 4 units long, sharing the midpoint $M$ and such that $B \widehat{M} D=60^{\circ}$. Let $X$ be the set of all and only the points that are at most 1 unit away from at least one of the two segments. What is the area of $X$?
(A) $8-\frac{4}{3} \sqrt{3}$
(B) $16-\frac{8}{3} \sqrt{3}$
(C) $16-\frac{4}{3} \sqrt{3}+\pi$
(D) $16-\frac{8}{3} \sqrt{3}+2 \pi$
(E) $8+2 \pi$.
5. The answer is (D). Let $X_{1}$ (respectively $X_{2}$) be the set of points with distance $\leqslant 1$ from only the segment $A B$ (respectively $C D$). Clearly, $X_{1}$ and $X_{2}$ are congruent and $X$ is the union of the two. Then $\operatorname{Area}(X)=2 \operatorname{Area}\left(X_{1}\right)-\operatorname{Area}\left(X_{1} \cap X_{2}\right)$.

$X_{1}$ consists of a rectangle with base 4 and height 2, joined to two terminal semicircles with diameters on the shorter sides. The area of $X_{1}$ is therefore $8+\pi$.

The intersection $X_{1} \cap X_{2}$ is a parallelogram with both heights of length 2 and an angle of $60^{\circ}$, so it is the union of two equilateral triangles with side $\frac{4 \sqrt{3}}{3}$. Its area is $\frac{8 \sqrt{3}}{3}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_cf63dde6cb9f14f4f823g-08.jpg?height=417&width=573&top_left_y=2139&top_left_x=1324)

In conclusion, $\operatorname{Area}(X)=16-\frac{8 \sqrt{3}}{3}+2 \pi$.