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# Area of Hexagon Problem | AMC-10A, 2014 | Problem 13

Try this beautiful problem from Geometry based on Hexagon from AMC-10A, 2014. You may use sequential hints to solve the problem

Try this beautiful problem from Geometry based on Area of Hexagon Problem

## Area of Hexagon Problem – AMC-10A, 2014- Problem 13

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

• $3+{\sqrt 5}$
• $4+{\sqrt 3}$
• $3+{\sqrt 3}$
• $\frac{1}{2}$
• $\frac{1}{9}$

### Key Concepts

Geometry

Triangle

square

Answer: $3+{\sqrt 3}$

AMC-10A (2014) Problem 13

Pre College Mathematics

## Try with Hints

Given that $\triangle ABC$ is an Equilateral Triangle with side length $1$ and squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle.Now we have to find out the area of hexagon $DEFGHI$.Now area of the the Hexagon $DEFGHI$=Area of $\triangle ABC$+Area of three squares( i.e $ABDE$, $BCHI$, $CAFG$)+Area of Triangles(( $\triangle AEF,\triangle DBI,\triangle HCG$)

Since the side length of Equilateral Triangle $\triangle ABC$ is given then we can find out the area of the $\triangle ABC$ and area of the squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle(as side length of one square =side length of the equilateral $\triangle ABC$.Now we have to find out the area of other three Triangles( $\triangle AEF,\triangle DBI,\triangle HCG$)

can you finish the problem……..

Area of the $\triangle ABC$(Red shaded Region)=$\frac{\sqrt 3}{4}$ (as side lengtjh is 1)

Area of 3 squares =$3\times {1}^2=3$

Now we have to find out the area of the $\triangle GCH$.At first draw a perpendicular $CL$ on $HG$. As $\triangle GCH$ is an isosceles triangle (as $HC=CG=1$),Therefore $HL=GL$

Now in the $\triangle CGL$,

$\angle GCL=60^{\circ}$ (as $\angle GCH=360^{\circ}-\angle ACB -\angle ACG-\angle BCH$ $\Rightarrow \angle GCH=360^{\circ}-60^{\circ}-90^{\circ}-90^{\circ}=120^{\circ}$)

So $\angle GCL=60^{\circ}$

So $\angle CGL=30^{\circ}$

$\frac{CL}{CG}$=Sin $30^{\circ}$

$\Rightarrow CL=\frac{1}{2}$ (as CG=1)

And ,

$\frac{GL}{CG}$=Sin $60^{\circ}$

$\Rightarrow GL=\frac{\sqrt 3}{2}$ (as CG=1)

So $GH=\sqrt 3$

Therefore area of the $\triangle CGH=\frac{1}{2}\times \sqrt 3 \times{1}{2}=\frac{\sqrt 3}{4}$

Therefore area of three Triangles ( $\triangle AEF,\triangle DBI,\triangle HCG$)=$3\times \frac{\sqrt 3}{4}$

can you finish the problem……..

Therefore area of the the Hexagon $DEFGHI$=Area of $\triangle ABC$+Area of three squares( i.e $ABDE$, $BCHI$, $CAFG$)+Area of Triangles(( $\triangle AEF,\triangle DBI,\triangle HCG$)=($\frac{\sqrt 3}{4}+3+3\times \frac{\sqrt 3}{4}$)=$3+{\sqrt 3}$