Maths question- area of a quadrilateral
27-09-2006, 12:26 PM
#2
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Divide it up so you have a large rectangle in the centre, which is easy to work out.
Then you should have a maximum of 2 right angled triangles at either end, which are also easily worked out.
The above assumes 2 walls are parallel. If they are not, you'll have 3 right angled triangles.
Then you should have a maximum of 2 right angled triangles at either end, which are also easily worked out.
The above assumes 2 walls are parallel. If they are not, you'll have 3 right angled triangles.
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27-09-2006, 12:49 PM
#9
if it was a squareish shape it would be 1083.643 m2
but you have to knock off the triangular end bits where the sides aren't parellel
on one side you would have to lose 4.73 m2 and on the other you would have to lose 3.4365 m2
the tricky part is losing both sides but not losing the bit wehre the cover each other
this would be 0.06 m2
so your toatl floor area would be:
1083.643 - 4.73 -3.4365 + 0.06 = 1075.5365 m2
i thank you
but you have to knock off the triangular end bits where the sides aren't parellel
on one side you would have to lose 4.73 m2 and on the other you would have to lose 3.4365 m2
the tricky part is losing both sides but not losing the bit wehre the cover each other
this would be 0.06 m2
so your toatl floor area would be:
1083.643 - 4.73 -3.4365 + 0.06 = 1075.5365 m2
i thank you
27-09-2006, 12:51 PM
#11
Originally Posted by dojj
if it was a squareish shape it would be 1083.643 m2
but you have to knock off the triangular end bits where the sides aren't parellel
on one side you would have to lose 4.73 m2 and on the other you would have to lose 3.4365 m2
the tricky part is losing both sides but not losing the bit wehre the cover each other
this would be 0.06 m2
so your toatl floor area would be:
1083.643 - 4.73 -3.4365 + 0.06 = 1075.5365 m2
i thank you
but you have to knock off the triangular end bits where the sides aren't parellel
on one side you would have to lose 4.73 m2 and on the other you would have to lose 3.4365 m2
the tricky part is losing both sides but not losing the bit wehre the cover each other
this would be 0.06 m2
so your toatl floor area would be:
1083.643 - 4.73 -3.4365 + 0.06 = 1075.5365 m2
i thank you
27-09-2006, 01:03 PM
#14
Originally Posted by Shings
You dont need to know the angles to work out the area of the triangles.
Area of a triangle is 1/2 base X Height.
To work out the missing side you used pythagorus (sp) a2+b2=c2
Area of a triangle is 1/2 base X Height.
To work out the missing side you used pythagorus (sp) a2+b2=c2
27-09-2006, 01:06 PM
#17
Originally Posted by Shings
Originally Posted by Shings
Originally Posted by chip-3door
Originally Posted by dojj
if it was a squareish shape it would be 1083.643 m2
but you have to knock off the triangular end bits where the sides aren't parellel
on one side you would have to lose 4.73 m2 and on the other you would have to lose 3.4365 m2
the tricky part is losing both sides but not losing the bit wehre the cover each other
this would be 0.06 m2
so your toatl floor area would be:
1083.643 - 4.73 -3.4365 + 0.06 = 1075.5365 m2
i thank you
but you have to knock off the triangular end bits where the sides aren't parellel
on one side you would have to lose 4.73 m2 and on the other you would have to lose 3.4365 m2
the tricky part is losing both sides but not losing the bit wehre the cover each other
this would be 0.06 m2
so your toatl floor area would be:
1083.643 - 4.73 -3.4365 + 0.06 = 1075.5365 m2
i thank you
Area of a triangle is 1/2 base X Height.
To work out the missing side you used pythagorus (sp) a2+b2=c2
http://mathforum.org/library/drmath/view/54868.html
I am off to uni now to go and ask them!!!!
Jake
Why didnt i think of that?
27-09-2006, 01:07 PM
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Originally Posted by Shings
I now know I cannot do it correctly.. The area of the an irregular quadrilateral can only be worked out if you have 2 of the angles or 1 of the diagonals.
http://mathforum.org/library/drmath/view/54868.html
I am off to uni now to go and ask them!!!!
Jake
http://mathforum.org/library/drmath/view/54868.html
I am off to uni now to go and ask them!!!!
Jake
27-09-2006, 01:13 PM
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I cant give you a formula for it. I had to draw out your building in autocad using circles, and then take the area from that.
Its not possible to work it out with a formula using only the dimensions.
I think you might need the angles
Its not possible to work it out with a formula using only the dimensions.
I think you might need the angles
27-09-2006, 01:21 PM
#21
Originally Posted by vroomtshh
I cant give you a formula for it. I had to draw out your building in autocad using circles, and then take the area from that.
Its not possible to work it out with a formula using only the dimensions.
I think you might need the angles
Its not possible to work it out with a formula using only the dimensions.
I think you might need the angles
The problem is that you can have an infinite number of different area shapes using those same size sides, so you HAVE to have the fucking god damn degrees of at least one bastard angle
27-09-2006, 01:32 PM
#23
seeing as i can't see any pics, you'd have to draw a picture, with the angles involved and the lengths of each side
then you can work out the area no matter what, because both sides aren't parallel you have to assume that the angles won't be the same on any 2 corners, none of this oppositte andgles and that
then you can work out the area no matter what, because both sides aren't parallel you have to assume that the angles won't be the same on any 2 corners, none of this oppositte andgles and that
27-09-2006, 01:48 PM
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Can't you just divide the shape into two triangles, then use Pythagorus? Then work out the area of the two triangles?
Been a while since i've done any maths more complex than basic sums
I'm doing QS at uni aswell, my first bit of coursework i've just started now, involves researching Biomass Fuels
EDIT: sorry, just read the thread properly, totally forgot it had to be right angled triangles to use that method of getting triangle areas.
I'll have another think.
Been a while since i've done any maths more complex than basic sums
I'm doing QS at uni aswell, my first bit of coursework i've just started now, involves researching Biomass Fuels
EDIT: sorry, just read the thread properly, totally forgot it had to be right angled triangles to use that method of getting triangle areas.
I'll have another think.
27-09-2006, 01:57 PM
#26
Originally Posted by Adrenaline Junkie
Can't you just divide the shape into two triangles, then use Pythagorus? Then work out the area of the two triangles?
Been a while since i've done any maths more complex than basic sums
I'm doing QS at uni aswell, my first bit of coursework i've just started now, involves researching Biomass Fuels
Been a while since i've done any maths more complex than basic sums
I'm doing QS at uni aswell, my first bit of coursework i've just started now, involves researching Biomass Fuels
NOT WITHOUT THE FUCKING ANGLES YOU CANT YOU USELESS CUNT
Im sure i mentioned that before on the thread somewhere?
27-09-2006, 02:00 PM
#27
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Originally Posted by Shings
Originally Posted by dojj
seeing as i can't see any pics, you'd have to draw a picture, with the angles involved and the lengths of each side
then you can work out the area no matter what, because both sides aren't parallel you have to assume that the angles won't be the same on any 2 corners, none of this oppositte andgles and that
then you can work out the area no matter what, because both sides aren't parallel you have to assume that the angles won't be the same on any 2 corners, none of this oppositte andgles and that
Yes - I agree I have found out the correct formulae to work out the area of an irregular quadrilateral BUT I do not know how to work out the angles at each corner, I have the measurments of the sides though and I know that the 4 angles must add up to 360'
Originally Posted by Adrenaline Junkie
Can't you just divide the shape into two triangles, then use Pythagorus? Then work out the area of the two triangles?
Been a while since i've done any maths more complex than basic sums
I'm doing QS at uni aswell, my first bit of coursework i've just started now, involves researching Biomass Fuels
Been a while since i've done any maths more complex than basic sums
I'm doing QS at uni aswell, my first bit of coursework i've just started now, involves researching Biomass Fuels
Bio mass fuels...
Basically, as it would create two scalene triangles, you'll have to divide the two scalene triangles into two, forming two triangles with right angles.
Problem solved.
27-09-2006, 02:01 PM
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Originally Posted by chip-3door
Originally Posted by Adrenaline Junkie
Can't you just divide the shape into two triangles, then use Pythagorus? Then work out the area of the two triangles?
Been a while since i've done any maths more complex than basic sums
I'm doing QS at uni aswell, my first bit of coursework i've just started now, involves researching Biomass Fuels
Been a while since i've done any maths more complex than basic sums
I'm doing QS at uni aswell, my first bit of coursework i've just started now, involves researching Biomass Fuels
NOT WITHOUT THE FUCKING ANGLES YOU CANT YOU USELESS CUNT
Im sure i mentioned that before on the thread somewhere?
READ MY FUCKING EDIT... CUNT!
27-09-2006, 02:05 PM
#29
Originally Posted by Adrenaline Junkie
Originally Posted by Shings
Originally Posted by dojj
seeing as i can't see any pics, you'd have to draw a picture, with the angles involved and the lengths of each side
then you can work out the area no matter what, because both sides aren't parallel you have to assume that the angles won't be the same on any 2 corners, none of this oppositte andgles and that
then you can work out the area no matter what, because both sides aren't parallel you have to assume that the angles won't be the same on any 2 corners, none of this oppositte andgles and that
Yes - I agree I have found out the correct formulae to work out the area of an irregular quadrilateral BUT I do not know how to work out the angles at each corner, I have the measurments of the sides though and I know that the 4 angles must add up to 360'
Originally Posted by Adrenaline Junkie
Can't you just divide the shape into two triangles, then use Pythagorus? Then work out the area of the two triangles?
Been a while since i've done any maths more complex than basic sums
I'm doing QS at uni aswell, my first bit of coursework i've just started now, involves researching Biomass Fuels
Been a while since i've done any maths more complex than basic sums
I'm doing QS at uni aswell, my first bit of coursework i've just started now, involves researching Biomass Fuels
Bio mass fuels...
Basically, as it would create two scalene triangles, you'll have to divide the two scalene triangles into two, forming two triangles with right angles.
Problem solved.
Ah right, so i see then.
you take a shape who's angles you DONT KNOW, then you get two triangles with angles you DONT KNOW
Then you divide those into 2 so you have two right angled triangles but you DONT KNOW the dimensions of them cause you DONT HAVE THE FUCKING BASTARD CUNTING ANGLES THAT YOU FUCKING WELL HAVE TO HAVE IN THE FIRST CUNTING PLACE YOU USELESS DRIP OF SCROTUM SWEAT
27-09-2006, 02:11 PM
#32
Originally Posted by Shings
(PS Chip, you know you keep on about angles... do you think I might need them? )
)
FUCKING BUNCH OF USELSS FUCKING CUNTS HOW THE FUCK DID ANY OF YOU RETARDS GET SO MUCH AS A GCSE, YOU ARENT EVEN MENTALLY FIT TO CLEAN MY SHOES
27-09-2006, 02:13 PM
#33
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Originally Posted by chip-3door
Originally Posted by Adrenaline Junkie
Originally Posted by Shings
Originally Posted by dojj
seeing as i can't see any pics, you'd have to draw a picture, with the angles involved and the lengths of each side
then you can work out the area no matter what, because both sides aren't parallel you have to assume that the angles won't be the same on any 2 corners, none of this oppositte andgles and that
then you can work out the area no matter what, because both sides aren't parallel you have to assume that the angles won't be the same on any 2 corners, none of this oppositte andgles and that
Yes - I agree I have found out the correct formulae to work out the area of an irregular quadrilateral BUT I do not know how to work out the angles at each corner, I have the measurments of the sides though and I know that the 4 angles must add up to 360'
Originally Posted by Adrenaline Junkie
Can't you just divide the shape into two triangles, then use Pythagorus? Then work out the area of the two triangles?
Been a while since i've done any maths more complex than basic sums
I'm doing QS at uni aswell, my first bit of coursework i've just started now, involves researching Biomass Fuels
Been a while since i've done any maths more complex than basic sums
I'm doing QS at uni aswell, my first bit of coursework i've just started now, involves researching Biomass Fuels
Bio mass fuels...
Basically, as it would create two scalene triangles, you'll have to divide the two scalene triangles into two, forming two triangles with right angles.
Problem solved.
Ah right, so i see then.
you take a shape who's angles you DONT KNOW, then you get two triangles with angles you DONT KNOW
Then you divide those into 2 so you have two right angled triangles but you DONT KNOW the dimensions of them cause you DONT HAVE THE FUCKING BASTARD CUNTING ANGLES THAT YOU FUCKING WELL HAVE TO HAVE IN THE FIRST CUNTING PLACE YOU USELESS DRIP OF SCROTUM SWEAT
And you know that one of those angles is 90 degrees... for each triangle...
Yes?
Right...
If you know the length of two sides of a right angled triangle, using Triganometry, you can calculate the angles.
I haven't got the time to be trying to work this out... but there must be a way of calculating the size of the sides... infact, if you can work out the diaganal length, you can do it.
Please don't call me a drip of scrotum sweat.
27-09-2006, 02:15 PM
#34
Originally Posted by Adrenaline Junkie
I do believe a triangle's angles will add up to 180 degrees...
Originally Posted by Adrenaline Junkie
And you know that one of those angles is 90 degrees... for each triangle...
Originally Posted by Adrenaline Junkie
Yes?
Originally Posted by Adrenaline Junkie
Right...
Originally Posted by Adrenaline Junkie
If you know the length of two sides of a right angled triangle, using Triganometry, you can calculate the angles.
BUT YOU DONT KNOW THE FUCKING LENGTH OF TWO OF THE BASTARDS SIDES DO YOU IN THE FIRST PLACE YOU CUNT!
Cause to do that you would need the fucking angles off the shape in the first cunting place you useless bastard.
Originally Posted by Adrenaline Junkie
I haven't got the time to be trying to work this out... but there must be a way of calculating the size of the sides... infact, if you can work out the diaganal length, you can do it.
Originally Posted by Adrenaline Junkie
Please don't call me a drip of scrotum sweat.
27-09-2006, 02:17 PM
#35
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Mate, you evidentily don't know either.. so shut the fuck up.
And i'd like to add.. this is the first time i've attempted to remember any complex maths, since leaving school a number of years ago.
And i'd like to add.. this is the first time i've attempted to remember any complex maths, since leaving school a number of years ago.
27-09-2006, 02:21 PM
#38
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Originally Posted by Shings
Originally Posted by "Adrenaline Junkie
And i'd like to add.. this is the first time i've attempted to remember any complex maths, since leaving school a number of years ago.
me too...
it hurts my head..
I'd do it for you, but i've got to get my own shizzle done, and i think the other QS's in the office will look at me funny if i start drawing an 'almost' rectangle and proceding to calculate angles and area!
I know for a fact if it was my work that i had to do, i could do it.
27-09-2006, 02:23 PM
#39
Originally Posted by Shings
OI
Calm the funk down...
But seriously thanks for both of your help.
I'll let you know how I get on when its marked.
Jake
Calm the funk down...
But seriously thanks for both of your help.
I'll let you know how I get on when its marked.
Jake
But you do need to know the angles as a nearly square one will have a larger area than an almost totally flattened shape with the same size sides.