# Area of a regular octagon with a side-length of 10 km

I am asked to calculate the area of a regular octagon given the side-length of 10 km.

I saw some examples saying that I should start by splitting the octagon into eight isosceles triangles, and that the length of the base would be 10 km, since we're given that the sides of the octagon are all 10km. What I don't know is what to do next.

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Can you find the angles of the isosceles triangles? If you know one side and all three angles of a triangle, do you know how to compute its area? –  Micah Jul 9 '12 at 5:04
For your first question, if I can, I do not know how. Second, no I do not. –  Dustin L. Jul 9 '12 at 5:09
Draw a picture if it helps. Put a dot in the center of the regular octagon and draw edges connecting the vertices on the octagon to the dot in the center of the octagon. Notice that you have 8 isosceles triangles inside the regular octagon. I forget the correct terminology but the inner angle should be 360 degrees divided by 8. Use this to find the other angles. –  math-visitor Jul 9 '12 at 5:13
Are you supposed to come up with an exact algebraic expression, or can you use trigonometry? –  André Nicolas Jul 9 '12 at 5:19
@Dustin: If you're willing to listen to your "extremely fast" teacher but not to anyone who answers you here (Brian's answer is wonderful), then why did you ask here in the first place? –  Henning Makholm Jul 9 '12 at 10:48

I think that a different approach is easier: try cutting it up as in the rough sketch below. Once you work out the lengths of the legs of the right triangles, you should be able to calculate the area pretty easily.

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Well, I would love to do that, but I do not know how. –  Dustin L. Jul 9 '12 at 5:16
@Dustin: Can you see that the triangles are isosceles right triangles, with two angles of $45$°? If $x$ is the length of one of the short sides, the Pythagorean theorem tells you that $2x^2=10$. From that you can calculate the areas of the triangles and rectangles, and the square in the centre clearly has area $10^2=100$. –  Brian M. Scott Jul 9 '12 at 5:20
@Brian, this is very clever! Dustin, keep in mind that you are now working with 4 isosceles triangles ("move" two of them together to get a square), 4 rectangles, and 1 square. Repeatedly use the length of the edge of the regular octagon to find the area of each of the smaller pieces. –  math-visitor Jul 9 '12 at 5:23
Thank you, but I am done dicking around with this. I don't understand this, so I'll try and see if my teacher, who won't slow down for anyone, will tell me what to do. –  Dustin L. Jul 9 '12 at 5:32

I'd like to suggest an alternate strategy. A regular octagon has angles of 135$^o$. I don't know if it's true where you're from, but in the US, stop signs are octagon shaped. So picture a stop sign, or look at the drawing in Brian's answer. Now extend the top, bottom, left, and right sides until they meet to form a square with a 45-45-90 right triangle at each corner. The hypoteneuse of each triangle is a side of the octagon.

Can you finish it from here, calculating the area of the square and subtracting the area of the 4 triangles?

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The following trigonometric approach is less elegant (and somewhat harder) than the ones that have been suggested. Its main advantage is that something like it works for any regular polygon.

Let $P$ be the point at the centre of the octagon. Join $P$ to all the vertices. We have divided the octagon into $8$ triangles. Now we find the area of one of these triangles, and multiply by $8$.

Let $PAB$ be one of our $8$ triangles. Draw a perpendicular from $P$ to $AB$. Suppose that this perpendicular meets $AB$ at $Q$. Then the area of $\triangle PAB$ is $\frac{1}{2}(AB)(PQ)$. We know that $AB=10$, and we need to find $PQ$.

Angle $APQ$ is $360^\circ/8$, that is, $45^\circ$. So the two equal angles $PAQ$ and $PBQ$ add up to $180^\circ-45^\circ$, that is, $135^\circ$. So each of them is $67.5^\circ$.

Note that $$\frac{PQ}{AQ}=\tan(\angle PAQ)=\tan(67.5^\circ).$$ But $AQ=5$, and therefore $$PQ=5\tan(67.5^\circ).$$ So the area of $\triangle PAB$ is $\frac{1}{2}(10)(5\tan(67.5^\circ))$. Multiply by $8$, and simplify. We get that the octagon has area $200\tan(67.5^\circ))$. If you want a decimal approximation, the calculator will give you one.

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Area of a $n$-sided regular polygon, $$A=\frac{na^2\cot(\pi/n)}{4}$$ where $a$ is the side length. For your case, $n=8$ and $a=10$ which gives $A=200\cot(\pi/8) km^2$.

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