Maths Question
Discussion
Mammasaid said:
Presuming that the curve is a 1/4 of an ellipse with height 29 and width 20.
Area of ellipse is Pi x 14.5 x 10
Area of shaded area is therefore
14.5 x 10 - (Pi x 14.5 x 10 / 4)
= 145 x (1 -Pi/4)
= 145 x 0.215
= 31.175 square units.
It's not a 1/4 circle I have just drawn it that way from a site measurement, it's just an example to show the lengths of the lines where they intersect from the tangent points.Area of ellipse is Pi x 14.5 x 10
Area of shaded area is therefore
14.5 x 10 - (Pi x 14.5 x 10 / 4)
= 145 x (1 -Pi/4)
= 145 x 0.215
= 31.175 square units.
The formula I used to use was something like this
10 squared x 14.5 squared, square root the answer then multiply by 3/14 ths.
I am missing something as the areas I get do not add up visually to what is on site.
flatsix3.6 said:
Mammasaid said:
Presuming that the curve is a 1/4 of an ellipse with height 29 and width 20.
Area of ellipse is Pi x 14.5 x 10
Area of shaded area is therefore
14.5 x 10 - (Pi x 14.5 x 10 / 4)
= 145 x (1 -Pi/4)
= 145 x 0.215
= 31.175 square units.
It's not a 1/4 circle I have just drawn it that way from a site measurement, it's just an example to show the lengths of the lines where they intersect from the tangent points.Area of ellipse is Pi x 14.5 x 10
Area of shaded area is therefore
14.5 x 10 - (Pi x 14.5 x 10 / 4)
= 145 x (1 -Pi/4)
= 145 x 0.215
= 31.175 square units.
The formula I used to use was something like this
10 squared x 14.5 squared, square root the answer then multiply by 3/14 ths.
I am missing something as the areas I get do not add up visually to what is on site.
However without knowing whether it was an ellipse or a circle, the nearest I could get it was to assume it was an ellipse.
BTW 10 squared x 14.5 squared, square root the answer is 10 x 14.5 as I said above.
For reference a right angle triangle with sides of 10 and 14.5 has an area of 10 x 14.5 / 2 = 29 Sq units.
If I recall from my hnd then it's integration. A quick google should help but I found this pretty quick. Might help. Busy at work at the moment.
http://m.intmath.com/integration/3-area-under-curv...
http://m.intmath.com/integration/3-area-under-curv...
Mammasaid said:
I think you're thinking of Pi which is 3.1415926.....
However without knowing whether it was an ellipse or a circle, the nearest I could get it was to assume it was an ellipse.
BTW 10 squared x 14.5 squared, square root the answer is 10 x 14.5 as I said above.
For reference a right angle triangle with sides of 10 and 14.5 has an area of 10 x 14.5 / 2 = 29 Sq units.
I think I have it nowHowever without knowing whether it was an ellipse or a circle, the nearest I could get it was to assume it was an ellipse.
BTW 10 squared x 14.5 squared, square root the answer is 10 x 14.5 as I said above.
For reference a right angle triangle with sides of 10 and 14.5 has an area of 10 x 14.5 / 2 = 29 Sq units.
Square root of 14.5 & 10 added together = 6.97 X 3.142 = 21.90 m2
Does that work ?
flatsix3.6 said:
I think I have it now
Square root of 14.5 & 10 added together = 6.97 X 3.142 = 21.90 m2
Does that work ?
[pic]http://i2.kym-cdn.com/entries/icons/facebook/000/000/554/facepalm.jpg/[pic]Square root of 14.5 & 10 added together = 6.97 X 3.142 = 21.90 m2
Does that work ?
Short answer is no.
BTW, I have a maths degree and worked as a maths teacher for 13 years.
I'll draw you a picture shortly....
Mammasaid said:
[pic]http://i2.kym-cdn.com/entries/icons/facebook/000/000/554/facepalm.jpg/[pic]
Short answer is no.
BTW, I have a maths degree and worked as a maths teacher for 13 years.
I'll draw you a picture shortly....
Short answer is no.
BTW, I have a maths degree and worked as a maths teacher for 13 years.
I'll draw you a picture shortly....
You can see that the curve forms part of an ellipse.
The area of the ellipse is (height / 2) x (width /2) x Pi (3.142) = 455.59 m2
The area you want is the difference from the area of a rectangle with sides of 14.5 and 10, i.e. 145 m2 and a quarter of the area of the ellipse i.e. 455.59 / 4 = 113.8975 m2
so 145 - 113.8975 = 31.1025 m2 (approx)
HTH
Aha, you made me get my calculator out there as the area of the right handed triangle (14.5 * 10) has got to be bigger than the required shaded area (as it has a chunk bitten out) and wasn't in your original.
The numbers all look right now though. (Although the shaded area is a fair bit less than half the triangle size though which it doesn't look eyeballing the diagram.)
The numbers all look right now though. (Although the shaded area is a fair bit less than half the triangle size though which it doesn't look eyeballing the diagram.)
No, good answer and great diagram.
I used to love doing all these "puzzles" at school (a long time ago) even though don't often use them in real life if ever. When I had builders around to do an extension they were great at calculating all sorts of practical stuff like this even though they had no paper qualifications.
(Also got a maths degree but can't remember any of it).
I used to love doing all these "puzzles" at school (a long time ago) even though don't often use them in real life if ever. When I had builders around to do an extension they were great at calculating all sorts of practical stuff like this even though they had no paper qualifications.
(Also got a maths degree but can't remember any of it).
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