Question

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

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

$\dfrac92$

$3+\sqrt3$

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

$6$

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 $\angle EAF = 360^{\circ} - 60^{\circ} - 2 \cdot 90^{\circ}$ $= 120^{\circ}.$

We also have that \triangleriangle EAF is isosceles, which means that we can rearrange the triangle by splitting it down the middle and recombining it into an equilateral triangle.

This means that the area of the three triangles is then $3 \cdot \dfrac{1^2 \sqrt{3}}{4} = \dfrac{3\sqrt{3}}{4}.$

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

Solution:

Problem 13 in Other Years