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 $DEFGHI$?

area of hexagon

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

Key Concepts


Geometry

Triangle

square

Check the Answer


But try the problem first…

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

Source
Suggested Reading

AMC-10A (2014) Problem 13

Pre College Mathematics

Try with Hints


First hint

shaded hexagon

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 =\(3\times {1}^2=3\)

solution figure of hexagon problem

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

Shaded area of hexagon

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

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