9) A parallelogram with sides 1 and 2 has an angle of $60^{\circ}$. What is the length of its shorter diagonal?
(A) $\sqrt{2}$
(B) $\sqrt{3}$
(C) $2 \sqrt{2}$
(D) $2 \sqrt{3}$
(E) $\frac{\sqrt{10}}{2}$.
9) The answer is (B). Let $A B C D$ be the parallelogram and $A$ be the vertex of the $60^{\circ}$ angle. Since opposite sides are congruent, two sides measure 1 and two measure 2. Therefore, one of the sides measuring 1 has $A$ as an endpoint, and the same is true for one of the sides measuring 2.

![](https://cdn.mathpix.com/cropped/2024_04_17_8c48e706878e314feb6bg-2.jpg?height=274&width=602&top_left_y=2210&top_left_x=1321)
Assume the notation such that $A D$ measures 1 and $A B$ measures 2. Then the shorter diagonal is $B D$. The figure suggests that $A B D$ is half of an equilateral triangle, leading to the answer (B), and a simple reasoning confirms the conjecture. Let $D H$ be the height of $A B D$ relative to the base $A B$. Then the measures of $A H$ and $D H$ are immediately obtained, which are $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$, respectively. The measures of $H B$ and $B D$ then follow, which are $\frac{3}{2}$ and $\sqrt{3}$, respectively.

Second Solution. With the given data, $A B D$ is necessarily similar to half of an equilateral triangle (two sides in a 2:1 ratio and the included angle of $60^{\circ}$), leading to the conclusion.