Problem: Find the area of a regular hexagon whose sides measures 5 cm

Sol'n 1: I can cut the hexagon into 6 small triangles, so the area of triangle times 6 will be equal to the area of the polygon. Since the triangles are equilateral I can use the formula for it.

Area of triangle = $\frac{\sqrt{3}}{4}(5)^2= 10.83$

Area of hexagon = (6)(10.83) = 64.95

Sol'n 2: Using the formula $\frac{1}{2}(base)(height)$ for the area of triangle. The angle of triangle (angle at the radius) is equal to 60 degrees (From 360 / 6). Cutting the triangle into half (to get the base and height) will result to an angle of 30 degrees and base of 2.5.

To get the height: $tan(30) = \frac{2.5}{height} = 4.33$, so the area of the triangle: $\frac{1}{2}(2.5)(4.33) = 5.41$

Area of hexagon = (6)(5.41) = 32.475 which does not equal to the area in solution 1.

Question: I noticed that the area calculated on solution 2 is half the area calculated in solution 1. I don't know why. I don't think plugging in the original length of the base is right or logical? I have answered a related problem like this and I have gotten the polygon's area with 1/2(base)(height) and not plugging the original base length back to the formula of area after getting the height. Topic is easy but I don't get why I get so confused. :( Any help will be appreciated.


2 Answers 2


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You've created your diagram like this, calculated the area of the triangle I've labelled, and then multiplied that area by $6$. Notice however, by your divisions, we've split the hexagon into $12$ triangles, so we would need to multiply the area by $12$, not $6$, which explains why your area for Sol $2$ is half the area for Sol $1$ (which is correct)

The area of one triangle is given by $\frac 12\cdot2.5\cdot2.5\tan(60)=\frac{25\sqrt3}{8}$. Multiplying this area by $12$ (the amount of triangles) gives us $\frac{75\sqrt3}{2}\approx64.9519$

  • $\begingroup$ Please excuse the awful quality of the diagram but I hope that it is clear enough $\endgroup$ May 19, 2018 at 14:29
  • $\begingroup$ I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I also dissected the octagon into 8 triangles and like in the problem above, I formed a right triangle with hypotenuse = 12, angle = 22.5 and base = x/2. The calculated area of the octagon using this method equals the calculated area using the other triangle formula: $absin\theta$ and I did not need to multiply the area by 16 but just by 8 which confuses me. $\endgroup$
    – Jayce
    May 19, 2018 at 14:36
  • $\begingroup$ Typo on my comment above. Should be $\frac{1}{2}absin\theta$. $\endgroup$
    – Jayce
    May 19, 2018 at 14:52
  • $\begingroup$ Octagons have a regular angle of $135^0$. When you divide the octagon into eight triangles, you get an angle of $45^0$ at the centre. Thus for the whole octagon you should use $8\cdot\frac{1}{2}\cdot12^2\cdot\sin45=288\sqrt2 \approx407.2935$ $\endgroup$ May 19, 2018 at 14:58
  • $\begingroup$ You are right. Try solving it via $\frac{1}{2}(base)(height)$. It will result to 407.29 even though the area is only multiplied by 8 which should be by 16 as per your explanation. I used 22.5 as the angle for the right triangle, "x/2" for the base and the hypotenuse, 12. Then Via "soh-cah-toa", base is 9.184 and height is 11.087. Plugging it in: $\frac{1}{2}(9.184)(11.087) = 50.911$ then multiplied by 8 which equals to 407.29, the same with $absin\theta$ but I only multiplied it by 8 and not by 16. I need an answer for this, so I can move on. :( $\endgroup$
    – Jayce
    May 19, 2018 at 15:14

Solution 1 gives the correct answer. In solution 2, the triangle that you're calculating is actually a right-angled triangle with interior angles $30°$, $60°$ and $90°$, and it is in fact half of the triangle calculated in solution 1. This explains why the numerical result obtained in solution 2 is half of that in solution 1.

  • $\begingroup$ I answered a similar problem. Find the area of a regular octagon inscribed in a circle whose radius is 12. I used 2 formulas of triangle: $\frac{1}{2}absin\theta$ and $\frac{1}{2}(base)(height)$. For $\frac{1}{2}(base)(height)$, I did the same thing, I formed a right triangle to get the height then calculated the area then multiplied it by 8 to get the octagon area. This equals the area calculated using $\frac{1}{2}absin\theta$. It equals even though I multiplied it just by 8 which is supposed to be by 16 just like you explained. :( $\endgroup$
    – Jayce
    May 19, 2018 at 14:50
  • $\begingroup$ The solution for the above, @The right triangle: angle = 22.5 (From 360/8 = 45/2 = 22.5), side = 12 (From the given radius) and base = x/2. Via "soh-cah-toa", base = 9.184 and height = 11.087. Plugging it into $8*[\frac{1}{2}(base)(height)] = 407.89$ which is the area of octagon. Via $\frac{1}{2}absin\theta$ where a = b = 12 and angle = 45 (from 360/8). Plugging it in the formula and multiplying it by 8 to get the octagon area, we get 407.29. The areas are equal even though in solution 1, I just multiplied it by 8. $\endgroup$
    – Jayce
    May 19, 2018 at 15:05
  • $\begingroup$ @Jayce Sorry for late reply as I was outside. It seems that another answerer is following your problem. Please leave a comment if you need further help :-) $\endgroup$ May 19, 2018 at 19:52

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