Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: holmes221b on January 11, 2011, 05:28:06 pm
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So I've always had problems remembering the formula for the area of a triangle. Until last night, when my mom pointed out that a triangle is half of a square, and suddenly it all made sense (she also remarked that she had had the same problem with that formula until someone had pointed that fact out to her).
It's not that we're missing the obvious. The reason is that math teachers don't (usually) explain the formula for the area of a triangle in those terms, and most people's brains are not wired to make the connection on their own, especially at the point in their development when they are first introduced to these formulas (in fact, the way math classes are often structured can even aggravate the problem, as the formula for the area of a square is introduced much earlier than the formula for the area of a triangle).
L x W = Area, where L is the length of the square and W is the width of the square
1/2 x B x H = Area, where B is the base of the triangle and H is the height of the triangle
For those of you who are more visually inclined:
(http://www.calculateme.com/cArea/area-triangle-base-height.gif)
(http://www.basic-mathematics.com/images/area2.gif)
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Or, use Hero's formula:
Area = the square root of ( s (s-a) (s-b) (s-c) ) when a, b, and c are the three sides and s is the semi-perimeter ( 1/2(a+b+c) )
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this is exactly how my teachers in elementary school taught the formula. you can also point out that the sum of a triangle's angles are 180 and a square is 360 (360/180 = 2). a pentagon's angles sum to 540. it can also be divided into 3 triangles. 540/180 = 3
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this is exactly how my teachers in elementary school taught the formula.
I didn't learn about finding the area of a triangle until Prealgebra in 8th grade. No wonder America's gotten behind in mathematics.
you can also point out that the sum of a triangle's angles are 180 and a square is 360 (360/180 = 2). a pentagon's angles sum to 540. it can also be divided into 3 triangles. 540/180 = 3
Interesting.
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Or, use Hero's formula:
Area = the square root of ( s (s-a) (s-b) (s-c) ) when a, b, and c are the three sides and s is the semi-perimeter ( 1/2(a+b+c) )
you mean Heron's? :P
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Umm, that's exactly how I learned it too. .5*b*h. I live in Utah, where we have the 3rd highest education rating, but also the lowest education funding from taxes :P learned that in, let me think, somewhere between 4-6 grade.
I dont' remember well, because it's been 5-7 years now.
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Or, use Hero's formula:
Area = the square root of ( s (s-a) (s-b) (s-c) ) when a, b, and c are the three sides and s is the semi-perimeter ( 1/2(a+b+c) )
you mean Heron's? :P
This can be refered to as both Hero's and Heron's formula. When I was in geometry, I remember reading ahead in my book to get this formula, and it said that both names are commonly used.
I think there also is a formula called the "Shoelace Formula" for finding the area of a triangle, but I never really investigated on how to go about using that formula. EDIT: Er, I guess the Shoelace Formula can be used to find the area of any polygon where the verticices are given ordered pairs as points.
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Or, use Hero's formula:
Area = the square root of ( s (s-a) (s-b) (s-c) ) when a, b, and c are the three sides and s is the semi-perimeter ( 1/2(a+b+c) )
you mean Heron's? :P
Somebody forgot their V8 this morning....?
edit: @willrandship maybe it's a coastal thing to not teach it that way?
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or this formula?
V=A*B*sin[theta]
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Heck, I've never lived anywhere but Utah, so I couldn't tell you :P
i'm afraid one thing they tend to lack aroun' here is teachin' how to talk withou' a Utah accen'. We pronounce my town Tremonuhn. :P
Edit: but that takes trig :P not teaching that to twelve year olds.
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for clarification, i didnt post it as being a "heron's formula"
I love this formula because this pretty much get all the triangle's area.
EDIT: I messed around with trigs when I was 11 >:D
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i remember learning the formula for a triangle in 3rd or 4th grade, along with a square and the approximation of a circle being 3 * r * r. in sixth grade i began learning formulas for 3-D shapes like cones, any-base pyramids, spheres. even had to describe relationships. volume of a cone with radius 1 height 1 is 1/3 the volume of a unit cube. volume of a sphere of radius 1 is 2/3 of a unit cube. completely forget why though.
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I was in Honors math, and had an amazing teacher from 5th-8th. before that, i had had one other good math teacher in fourth. So i've been pretty lucky.
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I learned a similar explanation for the volume of a sphere: if you enclose the sphere in the smallest cylinder possible, it takes up exactly 2/3 of the space, so then you just need to find the volume of the cylinder and multiply it by 2/3.
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Or, just do 4/3piR^3.....
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There is a really amazing algorithm for finding the area of ANY closed shape, not just triangles, if you know the coordinates of all the verticies. For instance, take any pentagon with verticies (X1,Y1) through (X5,Y5). You can find the area by listing the pairs clockwise and making the first point appear at the end too:
X1 Y1
X2 Y2
X3 Y3
X4 Y4
X5 Y5
X1 Y1
Then you multiply diagonally down on each side and find the sums of those 2 columns:
X1 Y1
Y1*X2 / X2 Y2 \ X1*Y2
Y2*X3 / X3 Y3 \ X2*Y3
Y3*X4 / X4 Y4 \ X3*Y4
Y4*X5 / X5 Y5 \ X4*Y5
Y5*X1 / X1 Y1 \ X5*Y1
______ ______
SumLeft SumRight
The area of the shape is simply the difference of these 2 sums divided by 2: |SumLeft - SumRight|/2
It even works if the lines pass through each other or the shape has holes in it.
EDIT: Oops, forgot the /2 hehe
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There is a really amazing algorithm for finding the area of ANY closed shape, not just triangles, if you know the coordinates of all the verticies. For instance, take any pentagon with verticies (X1,Y1) through (X5,Y5). You can find the area by listing the pairs clockwise and making the first point appear at the end too:
X1 Y1
X2 Y2
X3 Y3
X4 Y4
X5 Y5
X1 Y1
Then you multiply diagonally down on each side and find the sums of those 2 columns:
X1 Y1
Y1*X2 / X2 Y2 \ X1*Y2
Y2*X3 / X3 Y3 \ X2*Y3
Y3*X4 / X4 Y4 \ X3*Y4
Y4*X5 / X5 Y5 \ X4*Y5
Y5*X1 / X1 Y1 \ X5*Y1
______ ______
SumLeft SumRight
The area of the shape is simply the difference of these 2 sums: |SumLeft - SumRight|
It even works if the lines pass through each other or the shape has holes in it.
Ah yes, I could be wrong, but I think I recall something like that done to find an approximate value for Pi
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There is a really amazing algorithm for finding the area of ANY closed shape, not just triangles, if you know the coordinates of all the verticies. For instance, take any pentagon with verticies (X1,Y1) through (X5,Y5). You can find the area by listing the pairs clockwise and making the first point appear at the end too:
X1 Y1
X2 Y2
X3 Y3
X4 Y4
X5 Y5
X1 Y1
Then you multiply diagonally down on each side and find the sums of those 2 columns:
X1 Y1
Y1*X2 / X2 Y2 \ X1*Y2
Y2*X3 / X3 Y3 \ X2*Y3
Y3*X4 / X4 Y4 \ X3*Y4
Y4*X5 / X5 Y5 \ X4*Y5
Y5*X1 / X1 Y1 \ X5*Y1
______ ______
SumLeft SumRight
The area of the shape is simply the difference of these 2 sums: |SumLeft - SumRight|
It even works if the lines pass through each other or the shape has holes in it.
That is the Shoelace formula I was talking about earlier! :D
After writing that post, I made a nice Java program to do that formula for you!
Ah yes, I could be wrong, but I think I recall something like that done to find an approximate value for Pi
Do you mean when they used shapes like a regular 100-gon, found the area of that, and used that to estimate Pi?
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stuff
woaaahhhhh that is awesome
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There is a really amazing algorithm for finding the area of ANY closed shape, not just triangles, if you know the coordinates of all the verticies. For instance, take any pentagon with verticies (X1,Y1) through (X5,Y5). You can find the area by listing the pairs clockwise and making the first point appear at the end too:
X1 Y1
X2 Y2
X3 Y3
X4 Y4
X5 Y5
X1 Y1
Then you multiply diagonally down on each side and find the sums of those 2 columns:
X1 Y1
Y1*X2 / X2 Y2 \ X1*Y2
Y2*X3 / X3 Y3 \ X2*Y3
Y3*X4 / X4 Y4 \ X3*Y4
Y4*X5 / X5 Y5 \ X4*Y5
Y5*X1 / X1 Y1 \ X5*Y1
______ ______
SumLeft SumRight
The area of the shape is simply the difference of these 2 sums: |SumLeft - SumRight|
It even works if the lines pass through each other or the shape has holes in it.
o.o That is awesome!
/me memorizes
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I smell first place in academic team this year with this.
thank you so much.
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I see why its called the shoelace algorithm. How would you list the points for a shape with a hole?
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I assumed you just did the two shapes and subtracted them.
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Oh, I knew that. :P
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Or, use Hero's formula:
Area = the square root of ( s (s-a) (s-b) (s-c) ) when a, b, and c are the three sides and s is the semi-perimeter ( 1/2(a+b+c) )
There's also a formula for cyclic quadrilaterals: area=sqrt(s-a)(s-b)(s-c)(s-d) where s=(a+b+c+d)/2 and a,b,c, and d are the side lengths of the cyclic quadrilateral.
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Jeez, that's just out of my league...
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Or, use Hero's formula:
Area = the square root of ( s (s-a) (s-b) (s-c) ) when a, b, and c are the three sides and s is the semi-perimeter ( 1/2(a+b+c) )
There's also a formula for cyclic quadrilaterals: area=sqrt(s-a)(s-b)(s-c)(s-d) where s=(a+b+c+d)/2 and a,b,c, and d are the side lengths of the cyclic quadrilateral.
That's Brahmagupta's Theorem
BTW, for shoelace you need to divide by 1/2; it's a common mistake to forget that.
Shoelace can also be considered the determinant of the following matrix:
[[1 1 1 ... 1 x1 y1]
[1 1 1 ... 1 x2 y2]
[1 1 1 ... 1 x3 y3]
...
[1 1 1 ... 1 xn yn]]
One half the absolute value of the determinant = area.
This is probably really useless, but if you know the circumradius (radius of circle passing through three points of the triangle) and the sidelengths, then the area is (abc)/4R.
If you have a lot of time to waste, and you know that all vertices are on lattice points (integer coordinates), then use Pick's theorem! Count up the number of interior points and then the boundary points; the area is I+B/2-1.