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

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

Answer: $\frac{7}{2}$

AMC-10A (2005) Problem 20

Pre College Mathematics

## Try with Hints

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 ..........

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=$\frac{5}{2} +1$=$\frac{7}{2}$

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