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#1 2018-06-07 05:58:07

Joe walsh
Registered: 2018-06-07
Posts: 1

finding areas of polygons given only sides

I have had relatively no trouble this year until I encountered trig earlier in the year which if I'm honest I cheated my way through which is out of character for me but I was in a rush and now I'm paying for it because I now need to use trig to find the area of polygons given only a side measure, please help. I don't need or want straight up answers i just need to know how to do the problems I was given.

Q1. find the area of a equilateral triangle using trigonometry given a side length of 1.


#2 2018-06-07 13:55:36

Registered: 2016-02-01
Posts: 81

Re: finding areas of polygons given only sides

If we drop a vertical line on the side that is opposite to any vertex of the equilateral triangle it will bisect it, so since the are of a triangle is half the base (b) times the height (h), we would get:

and since a=60 degrees and l=1 we have:

Note: What do you what to do exactly? It is easier to calculate the Area using the original definition than using trigonometry!! It is like trying to kill a sparrow by a cannon!!

Last edited by Grantingriver (2018-06-07 14:08:18)


#3 2018-06-09 19:52:36

bob bundy
Registered: 2010-06-20
Posts: 8,354

Re: finding areas of polygons given only sides

hi Joe walsh,

Welcome to the forum.

Sounds to me like you also need a general way to do this for any regular polygon.

Here's a diagram for a heptagon (7 sides).  You can easily adapt it for n sides.


(1) Provided the polygon is regular you can calculate the central angle (AOB).  Divide 360 by 7 ( or n).

(2) Half this to get AOH and then subtract from 90 to get OAH.

(3) In the green triangle AOH, AH is adjacent to angle OAH and OH is opposite so OH = AH x tan(OAH)

(4) The yellow area is then 1/2 AB x OH

(5) Multiply by 7 (n) to get the area of the whole polygon.

Hope that is useful for you,


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei


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