I was set a question in my maths GCSE class about finding the volume of an octagonal prism, i know whow i would do this; area of the end face times the length but my problem is how on earth do you find the area of that end face. All i know is that its a regular octagon with all sides being 0.8 metres long.
Any prods in the right direction would be very helpful thanks
well octogon is made up of 8 isos triangles. inter angles of oct add to 1080 ((sides - 2)x180). therefor each int angle is 144. so base angles of each isos tringle are 72. using trig you can find ht of isos triangle (0.4tan72). Area of triangle half base x ht. Multiplied by 8 gives area of octogon. hope helped.
Originally posted by HurricaneConway125the area of any regular polygon can be calculated by the formula
I was set a question in my maths GCSE class about finding the volume of an octagonal prism, i know whow i would do this; area of the end face times the length but my problem is how on earth do you find the area of that end face. All i know is that its a regular octagon with all sides being 0.8 metres long.
Any prods in the right direction would be very helpful thanks
1/2 * (Apothem) * (Perimeter of the polygon)
The apothem is calculated by dropping a perpendicular bisector to one of the sides from the center of the polygon then you can complete the triangle by drawing another line from the center to a vertex of the same side that was bisected. Then you can use the sin laws to find your apothem and you are home free...
Originally posted by jim9That looks interesting thankyou very much i will just have to try it for myself
well octogon is made up of 8 isos triangles. inter angles of oct add to 1080 ((sides - 2)x180). therefor each int angle is 144. so base angles of each isos tringle are 72. using trig you can find ht of isos triangle (0.4tan72). Area of triangle half base x ht. Multiplied by 8 gives area of octogon. hope helped.
Originally posted by jim9No its fine im predicted an A* at GCSE so my maths teacher was giving me something outside of the syllabus to practice for A-level, i can do trig and interior angles etc. just wasnt sure of the way to divide the thing up!
very nice but not sure if too advanced for GCSE course. dont know though.
Originally posted by FabianFnasBut he said the VOLUME not the area. So its a 3d thing, not a 2d.
Draw it on a paper and you can see that ia regular octagon can have a grid of 3 by 3. Five of them is quadratic, four of them is triangles. If you pair the triangles together you get two quadrats of a different size. Then add the areas toghether.
Take a square of side length 1.6m
from each corner mark a point .4m both sides of it, join these by a line and then lop em off your square and you have yourself an octagon with sides .8m each
So essentially you have subtracted 2 squares of side length .4m from a square with 1.6m length, that will be the area of your face which you then multiply by perpendicular length.
Originally posted by HurricaneConway125When you get it, try to do with 12 sides, 30 sides, and "x" sides.
I was set a question in my maths GCSE class about finding the volume of an octagonal prism, i know whow i would do this; area of the end face times the length but my problem is how on earth do you find the area of that end face. All i know is that its a regular octagon with all sides being 0.8 metres long.
Any prods in the right direction would be very helpful thanks
Then try to find the area of the polygon with infinite sides (or a large enough number).
What do you get?
Originally posted by AgergThis is manifestly rubbish.
Take a square of side length 1.6m
from each corner mark a point .4m both sides of it, join these by a line and then lop em off your square and you have yourself an octagon with sides .8m each
So essentially you have subtracted 2 squares of side length .4m from a square with 1.6m length, that will be the area of your face which you then multiply by perpendicular length.
Your octagon will not be regular.
Four sides will have length 0.8
Four sides will have length sqroot(0.32) (by Pythagoras)
Originally posted by serigadon sides, distance from centre to a corner is r.
When you get it, try to do with 12 sides, 30 sides, and "x" sides.
Then try to find the area of the polygon with infinite sides (or a large enough number).
What do you get?
Internal angle = 180 - 360/n
Dividing up into triangles OAB (where O is the centre, A and B are vertices), you have an isosceles triangle with angles x = 90 - 180/n.
The area of the triangle is r^2 sin x cos x, or (r^2 sin 2x)/2,
= 0.5 r^2 sin (180 - 360/n)
= 0.5 r^2 sin (360/n)
Total area of the polygon is 0.5 n r^2 sin (360/n)
Now take the limit as n -> infinity.
sin (360/n) ~ 2pi/n, because sin x ~ x if x is measured in radians.
Total area ~ pi r^2
I.e. the area of a circle with this radius, which is what you'd expect.
