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

## Check the Answer

But try the problem first…

Answer: \(3+{\sqrt 3}\)

AMC-10A (2014) Problem 13

Pre College Mathematics

## Try with Hints

First hint

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……..

Second Hint

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……..

Final Step

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

## Other useful links

- https://www.cheenta.com/roots-of-equations-prmo-2016-problem-8/
- https://www.youtube.com/watch?v=d9WXHQFqbWs

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