298 Answersfact, it would be 1 to l/y'IO, or 1 to 3.1622+. The method we recommend is
the following:
In the diagram, AB is the diameter. Bisect the semicircle in D. Now, with
the radius AC mark off the points E and F from A and B, and draw the lines
DE and DF. The distance DG, added to the distance GH, gives a quarter of
the length of the circumference (IK), correct within a five-thousandth part.
IKLM is the length of complete straight line.
There is another way, correct to a seventeen-thousandth part, but it is a
little more difficult. [See W. W. Rouse Ball, Mathematical Recreations and
Essays, revised 11th edition, Macmillan, 1960, p. 348.-M. G.)
- THE CIRCLING CAR
Since the outside wheels go twice as fast as the inside ones, the circle they
describe is twice the length of the inner circle. Therefore, one circle is twice
the diameter of the other, and, since the wheels are 5 ft. apart, the diameter
of the larger circle is 20 ft. Multiply 20 ft. by 3.1416 (the familiar approximate
value for "pi") to get 62.832 ft. as the length of the circumference of the
larger circle.
- SHARING A GRINDSTONE
The first man should use the stone until he has reduced the radius by 1.754
in. The second man will then reduce it by an additional 2.246 in., leaving the
last man 4 in. and the aperture. This is a very close approximation.
- THE WHEELS OF THE CAR
The circumference of the front wheel and the rear wheel respectively must
have been 15 ft. and 18 ft. Thus IS ft. goes 24 times in 360 ft., and 18 ft. 20
times-a difference of four revolutions. But if we reduced the circumference
by 3 ft., then 12 goes 30 times, and IS goes 24 times-a difference of 6
revolutions.