PDA

View Full Version : Geometry - apothems regular polygon area stuff



kidchaor
02-04-2012, 06:15 PM
Can someone explain to me how to find the length of a side and the apothem of a polygon (hexagon for instance), given the area of the hexagon, (841.8units2)?

bamag
02-04-2012, 06:24 PM
Apothems are lines from the center of the polygon to a side such that it creates a right angle. For example:
[Only registered and activated users can see links]
Regular polygons can be cut up into a bunch of right triangles.
The pentagon pictured above can be cut into 6 triangles. Let a = length of apothem and s = length of side.
Each triangle has area (1/2)(a)(s)=(as/2). Since there's 5 triangles, you multiply that by 5 to get (5as)/2.
In general, the area would be (as/2)(n) with n being the number of sides.
I think that's how it works, lemme go google to verify myself xD

kidchaor
02-04-2012, 06:26 PM
Apothems are lines from the center of the polygon to a side such that it creates a right angle. For example:
[Only registered and activated users can see links]
Regular polygons can be cut up into a bunch of right triangles.
The pentagon pictured above can be cut into 12 right triangles. Let a = length of apothem and s = length of side.
Each right triangle has area (1/2)(a)(s/2)=(as/4). Since there's 10 triangles, you multiply that by 10 to get (5as)/2.
In general, the area would be (as/4)(2n) with n being the number of sides.
I think that's how it works, lemme go google to verify myself xD

im looking to find the measure of the apothem and side....with a given area...not trying to find the area

bamag
02-04-2012, 06:31 PM
My bad.
[Only registered and activated users can see links]
where
A is the length of the apothem (inradius)
N is the number of sides

841.8=A^2(6)tan30=6*sqrt3*A^2
A^2 = 841.8/(6*sqrt3)= around 81
A=9 units