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Posers and Puzzles

Posers and Puzzles

  1. Standard member HurricaneConway125
    SUPREMO OF SOMERSET
    17 Mar '08 20:12
    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
  2. 17 Mar '08 20:23
    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.
  3. 17 Mar '08 20:26
    Originally posted by HurricaneConway125
    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
    the area of any regular polygon can be calculated by the formula

    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...
  4. Standard member HurricaneConway125
    SUPREMO OF SOMERSET
    17 Mar '08 20:29
    Originally posted by jim9
    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.
    That looks interesting thankyou very much i will just have to try it for myself
  5. 17 Mar '08 20:30
    very nice but not sure if too advanced for GCSE course. dont know though.
  6. Standard member HurricaneConway125
    SUPREMO OF SOMERSET
    17 Mar '08 20:39
    Originally posted by jim9
    very nice but not sure if too advanced for GCSE course. dont know though.
    No 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!
  7. 17 Mar '08 20:47
    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.
  8. Subscriber sonhouse
    Fast and Curious
    17 Mar '08 22:45
    Originally posted by FabianFnas
    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.
    But he said the VOLUME not the area. So its a 3d thing, not a 2d.
  9. Standard member Agerg
    The 'edit'or
    17 Mar '08 23:08 / 3 edits
    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.
  10. 18 Mar '08 02:23
    Originally posted by HurricaneConway125
    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
    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?
  11. 18 Mar '08 10:20
    Originally posted by sonhouse
    But he said the VOLUME not the area. So its a 3d thing, not a 2d.
    It's an octagonal prism - find the area of the octagon then multiply by the length of the prism-bit. He was wondering how you found the area of the octagon.
  12. Standard member wolfgang59
    Infidel
    18 Mar '08 11:00
    Originally posted by Agerg
    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.
    This is manifestly rubbish.

    Your octagon will not be regular.

    Four sides will have length 0.8

    Four sides will have length sqroot(0.32) (by Pythagoras)
  13. Standard member Agerg
    The 'edit'or
    18 Mar '08 12:58
    good point...I made a mistake
  14. 18 Mar '08 13:09
    Originally posted by serigado
    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?
    n sides, distance from centre to a corner is r.

    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.
  15. Subscriber AThousandYoung
    It's about respect
    18 Mar '08 14:25
    Originally posted by sonhouse
    But he said the VOLUME not the area. So its a 3d thing, not a 2d.
    That's why you multiply that area by the height...