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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




Check the Answer

Answer: \(6\)

AMC-10A (2010) Problem 19

Pre College Mathematics

Try with Hints

Hexagon Problem - figure

Given that area of \(\triangle ACE\) is \(70\)% of the area of the 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 \(ABCDEF\) :

Hexagon in a triangle

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 $1$.

Shaded hexagon figure

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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