Cheenta is joining hands with Aditya Birla Education Academy for AMC Training.

Learn MoreContents

[hide]

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

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?

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

Geometry

Triangle

Octagon

But try the problem first...

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

Source

Suggested Reading

AMC-10A (2005) Problem 20

Pre College Mathematics

First hint

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 ..........

Second Hint

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........

Final Step

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

- https://www.cheenta.com/problem-based-on-lcm-amc-8-2016-problem-20/
- https://www.youtube.com/watch?v=7AlfBAPWEMg

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIAL
Google