# 2014 AMC 10A Problems/Problem 13

## Problem

Equilateral $\triangle ABC$ has side length $1$, and squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle. What is the area of hexagon $DEFGHI$? $[asy] import graph; size(6cm); pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); pair B = (0,0); pair C = (1,0); pair A = rotate(60,B)*C; pair E = rotate(270,A)*B; pair D = rotate(270,E)*A; pair F = rotate(90,A)*C; pair G = rotate(90,F)*A; pair I = rotate(270,B)*C; pair H = rotate(270,I)*B; draw(A--B--C--cycle); draw(A--E--D--B); draw(A--F--G--C); draw(B--I--H--C); draw(E--F); draw(D--I); draw(I--H); draw(H--G); label("A",A,N); label("B",B,SW); label("C",C,SE); label("D",D,W); label("E",E,W); label("F",F,E); label("G",G,E); label("H",H,SE); label("I",I,SW); [/asy]$ $\textbf{(A)}\ \dfrac{12+3\sqrt3}4\qquad\textbf{(B)}\ \dfrac92\qquad\textbf{(C)}\ 3+\sqrt3\qquad\textbf{(D)}\ \dfrac{6+3\sqrt3}2\qquad\textbf{(E)}\ 6$

## Solution 1

The area of the equilateral triangle is $\dfrac{\sqrt{3}}{4}$. The area of the three squares is $3\times 1=3$.

Since $\angle C=360$, $\angle GCH=360-90-90-60=120$.

Dropping an altitude from $C$ to $GH$ allows to create a $30-60-90$ triangle since $\triangle GCH$ is isosceles. This means that the height of $\triangle GCH$ is $\dfrac{1}{2}$ and half the length of $GH$ is $\dfrac{\sqrt{3}}{2}$. Therefore, the area of each isosceles triangle is $\dfrac{1}{2}\times\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{4}$. Multiplying by $3$ yields $\dfrac{3\sqrt{3}}{4}$ for all three isosceles triangles.

Therefore, the total area is $3+\dfrac{\sqrt{3}}{4}+\dfrac{3\sqrt{3}}{4}=3+\dfrac{4\sqrt{3}}{4}=3+\sqrt{3}\implies\boxed{\textbf{(C)}\ 3+\sqrt3}$.

## Solution 2

As seen in the previous solution, segment $GH$ is $\sqrt{3}$. Think of the picture as one large equilateral triangle, $\triangle{JKL}$ with the sides of $2\sqrt{3}+1$, by extending $EF$, $GH$, and $DI$ to points $J$, $K$, and $L$, respectively. This makes the area of $\triangle{JKL}$ $\dfrac{\sqrt{3}}{4}(2\sqrt{3}+1)^2=\dfrac{12+13\sqrt{3}}{4}$. $[asy] import graph; size(10cm); pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); pair B = (0,0); pair C = (1,0); pair A = rotate(60,B)*C; pair E = rotate(270,A)*B; pair D = rotate(270,E)*A; pair F = rotate(90,A)*C; pair G = rotate(90,F)*A; pair I = rotate(270,B)*C; pair H = rotate(270,I)*B; pair J = rotate(60,I)*D; pair K = rotate(60,E)*F; pair L = rotate(60,G)*H; draw(A--B--C--cycle); draw(A--E--D--B); draw(A--F--G--C); draw(B--I--H--C); draw(E--F); draw(D--I); draw(I--H); draw(H--G); draw(I--J--D); draw(E--K--F); draw(G--L--H); label("A",A,N); label("B",B,SW); label("C",C,SE); label("D",D,W); label("E",E,W); label("F",F,E); label("G",G,E); label("H",H,SE); label("I",I,SW); label("J",J,SW); label("K",K,N); label("L",L,SE); [/asy]$

Triangles $\triangle{DIJ}$, $\triangle{EFK}$, and $\triangle{GHL}$ have sides of $\sqrt{3}$, so their total area is $3(\dfrac{\sqrt{3}}{4}(\sqrt{3})^2)=\dfrac{9\sqrt{3}}{4}$.

Now, you subtract their total area from the area of $\triangle{JKL}$: $\dfrac{12+13\sqrt{3}}{4}-\dfrac{9\sqrt{3}}{4}=\dfrac{12+13\sqrt{3}-9\sqrt{3}}{4}=\dfrac{12+4\sqrt{3}}{4}=3+\sqrt{3}\implies\boxed{\textbf{(C)}\ 3+\sqrt3}$

## Solution 3

We will use, $\frac{1}{2}ab\sin x$ to find the area of the following triangles. Since $\angle A=360$, $\angle EAF=360-90-90-60=120$.

The Area of $\triangle AEF$ is $\frac{1}{2} \cdot 1 \cdot 1 \cdot \sin(120)$. Noting, $\sin (2x) = 2\sin (x)\cos (x)$,

Area of $\triangle AEF = \frac{1}{2} \cdot 1 \cdot 1 \cdot 2 \cdot \sin(60) \cdot \cos(60) = \dfrac{\sqrt{3}}{4}$,

Area of $\triangle ABC = \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin(60) = \dfrac{\sqrt{3}}{4}$,

Area of square ABDE = 1,

Therefore the composite area of the entire figure is, $$3 \cdot [\triangle AEF] + [\triangle ABC] + 3 \cdot [ABDE] = 3 \dfrac{\sqrt{3}}{4} + \dfrac{\sqrt{3}}{4} + 3 \cdot 1 = 4 \dfrac{\sqrt{3}}{4} + 3 = \sqrt{3} + 3 \implies\boxed{\textbf{(C)}\ 3+\sqrt3}$$

## Solution 4

We know that the area is equal to 3*EAF+3*ACGF+ABC. We also know that ACGF and the rest of the squares' area is equal to 1. Therefore the answer is 3*EAF+ABC+3. The only one with "+3" or "3+" is C, our answer. Very unreliable. -Reality Writes

## Solution 5 $\angle{AEF} = 180- \angle{BAC} = 120$

The area of the obtuse triangle is $\frac{1}{2}\sin{120} = \frac{\sqrt{3}}{4}$

The total area is $3\left(1 + \frac{\sqrt{3}}{4}\right) + \frac{\sqrt{3}}{4} = \sqrt{3} + 3$

~mathboy282

## Solution 6

The total area is the sum of the three squares, the three (congruent) obtuse triangles, and the equilateral triangle. The area of the equilateral triangle is $\frac{\sqrt{3}}{4}$ and the area of each square is $1$. The area of a triangle in general is $\frac{1}{2}ab\sin(c)$ where $a$ and $b$ are two sides and $c$ is the included angle. $\angle EAF$ measures $120^{\circ}$ because $\angle EAB$ and $\angle FAC$ are right, and $m\angle CAB=60^{\circ}$. So the area of the obtuse triangle is $\frac{1}{2}\cdot1\cdot1\cdot\sin\left(120^{\circ}\right)=\frac{\sqrt{3}}{4}$. The total area is $3\left(\frac{\sqrt{3}}{4}\right)+3\left(1\right)+\frac{\sqrt{3}}{4}=\sqrt{3}+3 \Longrightarrow \boxed{\textbf{(C )}\sqrt{3}+3}$.

~JH. L

## Solution 7

Since $\angle C=360$, $\angle GCH=360-90-90-60=120.$ Applying the Law of Cosines on $\angle GCH$ gives us $GH = 1.$ Since $\triangle GCH$ is isosceles, the perpendicular bisector of $\angle C$ also intersects segment $\overline{GH}$ in its median, which we can call point $M.$ Hence, we can apply the Pythagorean theorem on $\triangle CMG$ or $\triangle CMH$ to get $CM = \frac{\sqrt{3}}{4}.$ We can use this to get the area of the triangle and multiply that by three since the triangles are congruent. The result follows. ~peelybonehead

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