Ok. Some of these answers are pathetic. pi is not 22/7 whatever the Indians say. Its close, but still far away,
You want the proof for A = pi * r^2
Pi is defined as the ratio of the circle's circumf to diameter .
To get at the proof imagine the circle cut into four piepiecelike parts and placed like a rectangle. If you cut it it even further the figure becomes even more like a rectangle. If you were to cut it up really small, it would be a rectangle.
What is the area of this rectangle:
If you look at it, you will see that the height is r (radius) and the length 1/2 the circumference.
A = 1/2 r c
but pi = c /2 r
Therefore, A = 1/2 * r * 2 pi r
A = pi r^2
Without diagrams it is a bit difficult to explain but I hope you understood how I cut up the circle and arranged it like a rectangle
2006-08-02 22:58:02
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answer #1
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answered by blind_chameleon 5
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I'm not completely sure what you're asking, but here goes anyway:
If you know that the radius is one/half the diameter, or
r = D/2
you can just plug this into the equation for area given the radius:
A = PI * r**2
or
A = PI * (D/2)**2
squaring D/2 gives
A = PI * D**2/4
If you need to actually prove that the area of a circle is
A = PI * r**2,
You can do it using what is called an integral. You have to start by measuring angles in radians, not degrees, and assume that
360 degrees = 2* PI radians
That is, there are 2 *PI radians in a circle of 360 degrees.
Now take a small triangle which has two sides of length r, the radius of your circle. Define "alpha" as the angle between the two sides, measured in radians. By the definition of what a radian is, the length of the third side is just one of the two sides (r) times an angle of one radian in size (alpha)
or
Base = r * alpha
The area of such a traingle is one half the base times the height, or
Area = 1/2 * Base * r
Substituting for "Base" gives
Area = 1/2 * (r * alpha) * r
Combining r terms gives
Area = 1/2 * alpha * r**2
Subsituting the size of alpha as 1 radian gives
Area = 1/2 * (1) * r**2
or
Area of triangle = 1/2 * r**2
Now add up the areas of all of these triangles it takes to fill up your circle. Start by placing the first triangle in the circle with the "top" of the triangle (where the two sides of length r come to a point) on the center. Now add another triangle right next to it, with its point on the center of the circle, and keep going until you have added triangles all the way "around" the circle and are back at the first triangle. You've effectively "swept out the area", or "taken the integral" of a circle from 0 to 2*PI radians.
Now add up all the areas of the traingles. Since there are 2*PI radians in a circle, the total area of all the triangles is
Area of circle = 2*PI * 1/2 * r**2
the "2" terms cancel to give
Area of circle = PI * r**2
Obviously this is a self-referencing proof, because it only works if you define an angle of 1 radian to be 1/(2*PI) of a circle; but it's valid as long as you're consistent in your use of the definition of the radian in all applicable branches of Cartesian geometry and trigonometry.
2006-08-02 22:16:08
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answer #2
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answered by almintaka 4
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One visual proof is to imagine you cut a round slice of watermelon. Make two cuts near the middle of a spherical watermelon and put away the two semi-spheres, only use the circular middle.
This circular slice will have a green skin all round (circumference) and the red flesh (area).
Now cut the circular slice into 4 pieces just like cutting a cake and arrange the pieces like in the crude diagram below with all the pieces touching each other.
(>
<)
(>
<)
You will notice that this is "very roughly" a Rectangle. The Width is "roughly" the Radius and the Length is "roughly" 1/2 the Circumference.
Since the circumference is C = 2*PI*R, the Width of this rectangle would be 1/2 (2*PI*R) = PI*R.
Therefore the Area of the Rectangle is L x W = PI*R x R
Area = PI*R^2
To refine the "crude rectangle" we can cut our original circle to many more parts and you'll see that it will become the rectangle of Length = PI*R and Width = R, giving the area = PI*R^2.
2006-08-02 23:07:32
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answer #3
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answered by ideaquest 7
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1. Divide the circle into 4 equal triangular parts like a "pizza"(you can divide it as many parts as you want, but this is the simpliest).
2.Arrange this parts to make a parallelogram.
3.The radius of the circle will become the length of the formed paralellogram.1/2 of the circumference will be the width(circumference of a circle is=2r x pi. 1/2 circumference will be r x pi).
4.Substitute the data to the formula for the area of a rectangle, A=l x w, therefore A=(r)(r x pi)=r^2 x pi, the area of a circle.
2006-08-02 23:04:22
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answer #4
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answered by blitz2.0 1
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If I understand your question correctly you are asking how to find the Area of a circle by only knowing the Circumference.
Divide Circumference by Pi, that will give you Diameter. Divide Diameter by 2, that will give you radius, by which now you can plug into the equation A = Pi times r squared.
2006-08-02 22:06:39
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answer #5
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answered by Anonymous
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for example, you have a circle.
divide it into 16 parts
then try to form a figure
you will form a parallelogram
the formula of the are of a parallelogram is A=bh
if you will imagine the base of the parallelogram is actually the one-half of the circumference of the circle and the height is the radius of the circle
A=bh or in a circle A=(pi)r^2
2006-08-03 02:05:05
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answer #6
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answered by xavierbondoc_15 1
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as we know area of a circle is
a=pi(r*r)
circumference is 2(pi)(r)
pi = 22/7 this value is been introduced by "BHASKARACHARYA" of indian origin a way long back.
circumference is the whole length of the circle that is you put a thread on a circle and from a point and cover the circle
till it comes to the point you started now measure it it gives you cicircumference.you can get circumference from area by
cir=2(area)/r
i.e area=pi(r*r)
2006-08-02 22:06:24
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answer #7
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answered by brightstar 2
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A=LW
Area=Length x Width
Example:
6lx6w=36a
Pi
3.141592654
Example:
36axTT=113.0973355
113.0973355x18r=2035.752039c
2006-08-03 05:54:04
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answer #8
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answered by Sk8erboi83 3
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this can be done by integration and if you require to sum infinite series as mentioned in follwowing
2006-08-05 15:35:14
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answer #9
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answered by Mein Hoon Na 7
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http://mathforum.org/library/drmath/view/57660.html
2006-08-02 22:14:40
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answer #10
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answered by DigitalManic 2
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