2014 AMC 10A Problem 13

Question

Equilateral

\triangle ABC

has side length

1,

and squares

ABDE,

BCHI,

CAFG

lie outside the triangle. What is the area of hexagon

DEFGHI?

\dfrac{12+3\sqrt3}4

\dfrac92

3+\sqrt3

\dfrac{6+3\sqrt3}2

6

Decompose the hexagon into the original triangle, three squares, and three outer triangles

Each outer triangle has sides

1,1

with included angle

120^\circ

We can find the areas of all the individual pieces and then add them up together.

The area of the center equilateral triangle is

\dfrac{1^2 \sqrt{3}}{4} = \dfrac{\sqrt{3}}{4}.

We have that the areas of all the squares is

3 \cdot 1^2 = 3.

We also have that

\angle EAF = 360^{\circ} - 60^{\circ} - 2 \cdot 90^{\circ}

= 120^{\circ}.

Also,

AE=AF=1

and

\angle EAF=120^\circ

. Dropping the altitude from

A

shows that

EF=\sqrt3

and the altitude is

\frac12

, so

[EAF]=\frac{\sqrt3}{4}

. The other two outer triangles have the same area. Thus their combined area is

\frac{3\sqrt3}{4}

.

The total area is then

\dfrac{\sqrt{3}}{4} + \dfrac{3\sqrt{3}}{4} + 3 = 3 + \sqrt{3}.

Thus, C is the correct answer.

Problem 13 in other years