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

Dec 17, 2017

$\left(\sqrt{3} + 3\right) \text{ sq.unit}$.

#### Explanation:

We will use, to find the Area of $\Delta A B C$, the Formula,

$\text{Area of "DeltaABC=} \frac{1}{2} \cdot A B \cdot A C \cdot \sin \angle B A C .$

Observe that ${\Delta}^{s} A F E , B D I , \mathmr{and} C H G$ are all congruent, so,

they all have the same Area,

$= \frac{1}{2} \cdot A F \cdot A E \cdot \sin \angle E A F ,$

$= \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin \left\{{360}^{\circ} - \left({90}^{\circ} + {90}^{\circ} + {60}^{\circ}\right)\right\} ,$

$= \frac{1}{2} \cdot \sin {120}^{\circ}$,

$= \frac{1}{2} \cdot \sin \left({180}^{\circ} - {60}^{\circ}\right)$,

$= \frac{1}{2} \cdot \sin {60}^{\circ}$,

$= \frac{1}{2} \cdot \frac{\sqrt{3}}{2} ,$

$= \frac{\sqrt{3}}{4.}$

Also, Area of the equilateral $\Delta A B C = \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin {60}^{\circ} = \frac{\sqrt{3}}{4}$.

$\text{Area of each square=1}$.

Hence, The Area of the Hexagon $D E F G H I$,

$= 3 \times \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} + 3 \times 1$,

$= \left(\sqrt{3} + 3\right) \text{ sq.unit}$.

Dec 17, 2017

Area of hexagon$=$Area of $\Delta A B C +$Area of three squares$+$Area of three triangles $E A F , D B I , H C G$

1. Area of $\Delta A B C$
Draw a line perpendicular from vertex $A$ on side $B C$. This is altitude of the triangle $A B C$. This perpendicular also bisects $\angle B A C$. As each side of equilateral triangle is $= 1$ and each angle $= {60}^{\circ}$
Altitude $= 1 \times \cos {30}^{\circ} = \frac{\sqrt{3}}{2}$
Area of $\Delta A B C = \frac{1}{2} \times \text{base"xx"altitude}$
$= \frac{1}{2} \times 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$

2. Area of three squares.
Each square has side $= 1$ and therefore has area $= {1}^{2} = 1$
Total area of three squares$= 3 \times 1 = 3$

3. Area of three $\Delta$s $E A F , D B I , H C G$
For $\Delta E A F$
Note that in angle at $A = {360}^{\circ}$
This angle is equal to four angles $= {60}^{\circ} + {90}^{\circ} + {90}^{\circ} + \angle E A F$
Equating both we get $\angle E A F = {360}^{\circ} - {240}^{\circ} = {120}^{\circ}$.
Altitude of $\Delta E A F$ can be found as explained in case of $\Delta A B C$ above
Altitude of $\Delta E A F = 1 \times \cos {60}^{\circ} = \frac{1}{2}$
Half of side $E F = 1 \times \sin {60}^{\circ} = \frac{\sqrt{3}}{2}$
Base $E F = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$
Area of $\Delta E A F = \frac{1}{2} \times \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{4}$
Similarly area of other two triangles is also same.

Area of hexagon$= \frac{\sqrt{3}}{4} + 3 + \left(3 \times \frac{\sqrt{3}}{4}\right)$
$\implies$Area of hexagon$= 3 + \sqrt{3}$