An easy area

Let's divide the polygon into 12 isosceles triangles, each with side $10$ and the opposite angle with a measure of $\dfrac{360°}{12}=30°$ . Now, bisect that angle in order to form a right triangle and name that segment $a$ (apothem), now with side $5$ and the opposite angle with a measure of $15°$ .

Now, let's apply the tangent function:

$\tan 15°=\dfrac{5}{a}$

$a=\dfrac{5}{\tan 15°}$

Calculate the area of the isosceles triangle, with base $10$ and height $a$ :

$A_{t}=\dfrac{25}{\tan 15°}$

And the total area:

$A=12A_{t}=\dfrac{300}{\tan 15°}$

Also, we know that $\tan 15°=\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$ , so:

$A=\dfrac{300}{\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}}=600+300\sqrt{3}$ , so $a=600$ , $b=300$ and $c=3$ , and $a-b-c=\boxed{297}$

I did it by using the formulas.

I didn't knew the formulas but surfed the internet and luckily found it on this page

which states that the formula to calculate the area of regular do-decagon.( a figure having 12 sides) is

The area of a regular dodecagon ( a figure with 12 sides) with side a is given by:

3 ( 2 + square root 3) a * a

= 3 ( 2 + square root 3) 10 * 10

= 300 ( 2 + square root 3 )

= 600 + 300 square root 3

= a + b square root c

So,

a = 600

b = 300

c = 3

Hence,

a - b -c

= 600 - 300 - 3

= 297

Great problem Alan!

Area of do-decagon = 3(2 + sqrt(3))d^2 where, d is the side of regular do-decagon. Thus, if d=10, Then A = 600 +300 sqrt(3). Thus, a-b-c = 600-300-3 = 297