Categories
AMC 10 Geometry Math Olympiad USA Math Olympiad

Area of the Octagon | AMC-10A, 2005 | Problem 20

Try this beautiful problem from Geometry:Area of Octagon.AMC-10A, 2005. You may use sequential hints to solve the problem

Try this beautiful problem from Geometry based on the Area of the Octagon from AMC 10A, 2005, Problem 2005

Area of the octagon – AMC-10A, 2005- Problem 20


An equiangular octagon has four sides of length 1 and four sides of length \(\frac{\sqrt{2}}{2}\), arranged so that no two consecutive sides have the same length. What is the area of the octagon?

  • \(\frac{4+5\sqrt 2}{2}\)
  • \(\frac{7}{2}\)
  • \(7\)

Key Concepts


Geometry

Triangle

Octagon

Check the Answer


Answer: \(\frac{7}{2}\)

AMC-10A (2005) Problem 20

Pre College Mathematics

Try with Hints


Octagon figure

We have to find out the equiangular octagon whose four sides of length 1 and four sides of length \(\frac{\sqrt{2}}{2}\),

we join \(AD\),\(HE\),\(BG\) and \(CF\).We assume that side lengths of \(AB=CD=EF=GH=1\) and side lengths of \(AH=BC=DE=GF=\frac{\sqrt{2}}{2}\)( As no two consecutive sides have the same length). Now

Can you now finish the problem ……….

finding the Area of Octagon


There are 5 squares with side lengths \(\frac{\sqrt{2}}{2}\) and 4 Triangles of side lengths \(1\)

Now area of \(5\) squares=\( 5 \times (\frac{\sqrt{2}}{2})^2\)=\(\frac{5}{2}\) and area of each Triangle is half of the area of a square.so the area of \(4\) Triangles=\(4 \times \frac{1}{2} \times \frac{1}{2}\)=\(1\)

can you finish the problem……..

Therefore the Total area of the required octagon=Total area of Five squares + Total areas of Four Triangles=\(\frac{5}{2} +1\)=\(\frac{7}{2}\)

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *