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AMC-8 Math Olympiad USA Math Olympiad

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. You may use sequential hints to solve the problem.

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

Check the Answer


But try the problem first…

Answer: \(6\)

Source
Suggested Reading

AMC-8 (2012) Problem 23

Pre College Mathematics

Try with Hints


First hint

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 ……….

Second Hint

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……..

Final Step

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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