About this sample
About this sample
Words: 589 |
3 min read
Published: Jan 4, 2019
Words: 589|Page: 1|3 min read
Archimedes compared the area enclosed by a circle to a right triangle whose base has the length of the circle’s circumference and whose height equals the circle’s radius. If the area of the circle is not equal to that of the triangle, then it must be either greater or less. He then eliminates each of these by contradiction, leaving equality as the only possibility.
Archimedes’ proof consists of constructing a circle ABCD and a triangle K.
Archimedes starts by inscribing a square in the circle and bisects the segments of arc AB, BC, CD, DE subtended by the sides of the square. Afterwards he proceeds to inscribe another polygon on the bisected points. He repeats this process until the difference in area between the circle and the inscribed polygon is smaller than the difference between the area of the circle and the area of the triangle.
The polygon is then greater than the triangle K.
Archimedes then proceeds to explain that a line from the center of the polygon to the bisection of one of its sides is shorter than the radius of the circle, and its circumference is smaller than the circumference of the circle. This disproves the statement that the polygon is greater than the triangle, since the legs of the triangle are made up of the radius and circumference of the circle.
The triangle K cannot be both smaller and larger than the polygon, and thus cannot be smaller than the circle.
After Archimedes proved that that the triangle cannot be smaller than the circle, he continues to prove that the triangle cannot be larger than the circle, either. This is accomplished by first assuming the triangle K to be larger than the circle ABCD. Then, a square is circumscribed around the circle so that lines drawn from the center of the circle will go through the points A, B, C, and D and bisect the corners of the square, one of which Archimedes labels T.
Archimedes then connects the sides of the square with a tangent line and labels the points at which the line meets the square G and F. He goes on to say that because TG > GA > GH, the triangle formed by FTG is larger than half the area of the difference in area between the square and the circle. Archimedes uses the fact that continual bisecting of the arc of a circle will produce a polygon with this characteristic to assert that continuing this method will ultimately produce a polygon around the circle such that the difference in area between the polygon and the circle is less than the difference in area between the triangle K and the circle.
The polygon is thus less in area than the triangle K
The length of a line from the center of the circle to a side of the polygon is equal to the radius of the circle. However, the perimeter of the polygon is larger in length than the circumference of the circle, and since the circumference of the circle is equal to the length of the longer leg of the triangle, the polygon must be larger in area than the triangle K. Again, the triangle cannot be both larger and smaller than the polygon, so the triangle cannot be larger than the circle.
Archimedes accomplished to prove his theory by using contradiction. After he proved that the triangle with legs equal to the radius and circumference of a given circle is not greater or less in area than that circle he concludes that the two must be equal in area.
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