2014 AMC 10A Problem 13

The hexagon $DEFGHI$ is made of seven pieces: the equilateral triangle, the three squares, and three "gap" triangles at the vertices $A$, $B$, $C$ (for example $\triangle AEF$ at $A$).

• Triangle $ABC$: $\frac{\sqrt3}{4}\cdot 1^2 = \frac{\sqrt3}{4}$.
• Three unit squares: $3 \cdot 1 = 3$.
• Gap triangle at $A$: its sides $AE$ and $AF$ are square sides of length $1$. The angles around $A$ are two right angles from the squares and $60^\circ$ from the triangle, so $\angle EAF = 360^\circ - 90^\circ - 90^\circ - 60^\circ = 120^\circ$. An isosceles triangle with legs $1$ and apex angle $120^\circ$ has area $\frac12 \cdot 1 \cdot 1 \cdot \sin 120^\circ = \frac{\sqrt3}{4}$. (Equivalently, dropping the altitude gives two $30$-$60$-$90$ halves with height $\frac12$ and base $\sqrt3$.)

The three gap triangles contribute $3 \cdot \frac{\sqrt3}{4} = \frac{3\sqrt3}{4}$.

Total area:
$$\frac{\sqrt3}{4} + 3 + \frac{3\sqrt3}{4} = 3 + \sqrt3.$$

The answer is $\boxed{\textbf{(C)}\ 3+\sqrt3}$.

A polygon built by attaching shapes to a core figure decomposes into the core, the attachments, and the gaps between attachments. The gaps are determined by angle bookkeeping at each vertex: the full $360^\circ$ minus the angles already used. A $120^\circ$ isosceles triangle with unit legs has the same area as a unit equilateral triangle, a fact that recurs whenever squares are built on the sides of an equilateral triangle.

• Forgetting the three gap triangles between the squares, which gives 3 + root 3 / 4, or treating them as right triangles.
• Using $\sin 120^\circ = \frac12$ instead of $\frac{\sqrt3}{2}$, which shrinks each gap triangle to $\frac14$.