# 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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