SOLUTION: I've been asked to explain Leonardo Da Vinci's method of finding the formula for the area of a circle. I've discovered that he claimed that “If you have a wheel whose thickness is

Algebra ->  Surface-area -> SOLUTION: I've been asked to explain Leonardo Da Vinci's method of finding the formula for the area of a circle. I've discovered that he claimed that “If you have a wheel whose thickness is       Log On


   

Question 8895: I've been asked to explain Leonardo Da Vinci's method of finding the formula for the area of a circle. I've discovered that he claimed that “If you have a wheel whose thickness is half its radius and roll it for one full revolution, the area of the track it leaves is equal to the area of the wheel.” But i just cant seem to get my head around it. Please could you try to explain it to me, maybe with a diagram? Thankyou.
Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
Let's say that we have a wheel whose radius is r. We first have to find the circumference of the circle, which is +C+=+2%2Api%2Ar+. This is, as you know, the length of the string you would wrap around the outermost edge of the wheel ONCE AROUND. If you stretched out this string and laid it on the ground, it will be 2%2Api%2Ar long.

Now, if the thickness of the wheel is 1/2 the radius, the thickness +T+=+%281%2F2%29r. If the wheel was rolled out 1 rev, the track's length will be 2%2Api%2Ar and its thickness will be %281%2F2%29r. The track would then be a rectangle whose dimensions are 2%2Api%2Ar and %281%2F2%29r. Since it has those dimensions, it would then make sense to find the area of the track which would be length * width. The length is 2%2Api%2Ar and the width (AKA, the tire's thickness) is %281%2F2%29r. So:

+A+=+%282%2Api%2Ar%29%2A%281%2F2%29r+ <------ Start here

+A+=+pi%2Ar%5E2+ <----- Area of the circle is equal to the area of the track that the wheel would leave behind if it was rolled exactly 1 rev. (That's if the wheel's thickness were 1/2 the radius.