2014 AMC 10A Problem 13

The hexagon $DEFGHI$ is built from pieces we already know. Look at the picture: it is made of

• the middle triangle $ABC$,
• the three squares,
• three small "gap" triangles, one at each corner of $ABC$: triangle $AEF$ at $A$, triangle $BDI$ at $B$, and triangle $CGH$ at $C$.

Find the area of each kind of piece.

1. Middle triangle. An equilateral triangle with side $1$ has area $\frac{\sqrt3}{4}$.
2. Squares. Each has area $1$, so together they give $3$.
3. Angle of a gap triangle. Stand at $A$. The angles around $A$ add to $360^\circ$. Two of them are square corners, $90^\circ$ each, and one is the triangle's corner, $60^\circ$. So the gap angle is
$$\angle EAF = 360^\circ - 90^\circ - 90^\circ - 60^\circ = 120^\circ.$$
4. Area of a gap triangle. Its two sides $AE$ and $AF$ are sides of squares, so both are $1$. Drop a line from $A$ straight down to the middle of $EF$. It splits the gap triangle into two $30$-$60$-$90$ triangles, each with hypotenuse $1$. The side across from $30^\circ$ is half the hypotenuse, so the height is $\frac12$; the side across from $60^\circ$ is $\frac{\sqrt3}{2}$, so $EF = \sqrt3$. The area is $\frac12 \cdot \sqrt3 \cdot \frac12 = \frac{\sqrt3}{4}$.
5. Three gap triangles give $\frac{3\sqrt3}{4}$.

Add everything:
$$\frac{\sqrt3}{4} + 3 + \frac{3\sqrt3}{4} = 3 + \frac{4\sqrt3}{4} = 3 + \sqrt3.$$

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

When shapes are stuck onto the sides of a figure, the outer outline is made of the figure, the attached shapes, and the gaps between them. The gaps are found by angle bookkeeping: around any point the angles add to $360^\circ$. It is worth remembering that a triangle with two sides of $1$ and a $120^\circ$ angle between them has the same area as an equilateral triangle with side $1$.

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