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

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(( $\triangle AEF,\triangle DBI,\triangle HCG$)=($\frac{\sqrt 3}{4}+3+3\times \frac{\sqrt 3}{4}$)=$3+{\sqrt 3}$

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## Hexagon Problem | Geometry | AMC-10A, 2010 | Problem 19

Try this beautiful problem on area of Hexagon from Geometry.

## Hexagon Problem – AMC-10A, 2010- Problem 19

Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$. The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$?

• $6$
• $\frac{2}{\sqrt5}$
• $8$
• $9$
• $9$

### Key Concepts

Geometry

Hexagon

Triangle

Answer: $6$

AMC-10A (2010) Problem 19

Pre College Mathematics

## Try with Hints

Given that area of $\triangle ACE$ is $70$% of the area of the hexagon.so at first we have to find out the area of the $\triangle ACE$.Clearly $\triangle ACE$ is an equilateral triangle.Now from the cosines law we can say that $AC^2=r^2+1^2-2r cos \frac{2\pi}{3}=r^2+r+1$.Therefore area of $\triangle ACE=\frac{\sqrt 3}{4}(r^2+r+1)$.Can you find out area of the Hexagon $ABCDEF$?

Can you now finish the problem ……….

Area Of The Hexagon $ABCDEF$ :

If we extend $BC$, $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$, $BC$ and $DE$ meet at $Y$, and $DE$ and $FA$ meet at $Z$,Now we can find out area of the hexagon $ABCDEF$ which is formed by taking equilateral triangle $XYZ$ of side length $r+2$. So if we remove three equilateral triangles, $ABX$, $CDY$ and $EFZ$, of side length $1$.

Therefore The area of $ABCDEF$ is $\frac{\sqrt 3}{4} (r+2)^2 – \frac{3\sqrt 3}{4}=\frac{\sqrt 3}{4} (r^2+4r+1)$

can you finish the problem……..

Given that area of $\triangle ACE$ is $70$% of the area of the hexagon.Therefore

$\frac{\sqrt 3}{4}(r^2+r+1)$=$\frac{7}{10}.\frac{\sqrt 3}{4}(r^2+4r+1)$

$\Rightarrow r^2-6r+1=0$.Now from Vieta’s Relation the sum of the possible value of $r$ is $6$

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## Area of a Regular Hexagon | AMC-8, 2012 | Problem 23

Try this beautiful problem from Geometry: Area of the Regular Hexagon – AMC-8, 2012 – Problem 23.

## Area of the Regular Hexagon – AMC-8, 2012- Problem 23

An equilateral triangle and a regular hexagon have equal perimeters. If the triangle’s area is 4, what is the area of the hexagon?

• $8$
• $6$
• $10$

### Key Concepts

Geometry

Triangle

Hexagon

Answer: $6$

AMC-8 (2012) Problem 23

Pre College Mathematics

## Try with Hints

To find out the area of the Regular hexagon,we have to find out the side length of it.Now the perimeter of the triangle and Regular Hexagon are same….from this condition you can easily find out the side length of the regular Hexagon

Can you now finish the problem ……….

Let the side length of an equilateral triangle is$x$.so the perimeter will be $3x$ .Now according to the problem the perimeter of the equiliteral triangle and regular hexagon are same,i.e the perimeter of regular hexagon=$3x$

So the side length of be $\frac{3x}{6}=\frac{x}{2}$

can you finish the problem……..

Now area of the triangle $\frac{\sqrt 3}{4}x^2=4$

Now the area of the Regular Hexagon=$\frac{3\sqrt3}{2} (\frac{x}{2})^2=\frac{3}{2} \times \frac{\sqrt{3}}{4}x^2=\frac{3}{2} \times 4$=6

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## Hexagon and Triangle |AMC 8- 2015 -|Problem 21

Try this beautiful problem from Geometry based on hexagon and Triangle.

## Area of Triangle | AMC-8, 2015 |Problem 21

In The given figure hexagon ABCDEF is equiangular ,ABJI and FEHG are squares with areas 18 and 32 respectively.$\triangle JBK$ is equilateral and FE=BC. What is the area of $\triangle KBC$?

• 9
• 12
• 32

### Key Concepts

Geometry

Triangle

hexagon

Answer:$12$

AMC-8, 2015 problem 21

Pre College Mathematics

## Try with Hints

Clearly FE=BC

Can you now finish the problem ……….

$\triangle KBC$ is a Right Triangle

can you finish the problem……..

Clearly ,since FE is a side of square with area 32

Therefore FE=$\sqrt 32$=$4\sqrt2$

Now since FE=BC,We have BC=$4\sqrt2$

Now JB is a side of a square with area 18

so JB=$\sqrt18$=$3\sqrt2$. since $\triangle JBK$ is equilateral BK=$3\sqrt2$

Lastly $\triangle KBC$ is a right triangle ,we see that

$\angle JBA + \angle ABC +\angle CBK +\angle KBJ$ =$360^\circ$

i.e$90^\circ + 120^\circ +\angle CBK + 60^\circ=360^\circ$

i.e $\angle CBK=90^\circ$

So $\triangle KBC$ is a right triangle with legs $3\sqrt 2$ and $4\sqrt2$

Now its area is $3\sqrt2 \times 4\sqrt 2 \times \frac {1}{2}$=$\frac{24}{2}$=12