Question

Equilateral $\triangle ABC$ has side length $1,$ and squares $ABDE,$ $BCHI,$ $CAFG$ lie outside the triangle. What is the area of hexagon $DEFGHI?$

$\dfrac{12+3\sqrt3}4$

$\dfrac92$

$3+\sqrt3$

$\dfrac{6+3\sqrt3}2$

$6$

Answer: C

Concepts:equilateral triangle square (geometry)area decomposition

Difficulty rating: 1540

Solution:

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.

2014 AMC 10A Problem 13

Answer: C

Difficulty rating: 1540

Solution:

Problem 13 in Other Years