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# 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?

• $$\frac{4+5\sqrt 2}{2}$$
• $$\frac{7}{2}$$
• $$7$$

Geometry

Triangle

Octagon

## Check the Answer

But try the problem first…

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

Source

AMC-10A (2005) Problem 20

Pre College Mathematics

## Try with Hints

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

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