The original "Ferris" wheel, designed by George Washington Gale Ferris Jr., was built for the World's Columbian Exposition of 1893 in Chicago, Illinois. There were 36 fully enclosed gondolas (passenger cars) that carried up to 40 passengers.

This "Ferris" wheel eventually left Chicago and was relocated in St. Louis, Missouri in time for the 1904 Louisiana Purchase Exposition. On May 11, 1906, it was demolished using a dynamite charge and sold for scraps.

The circumference of the "Ferris" wheel had to be large enough to allow for an arc length of 21.94 ft between each gondola. What was the circumference of the original "Ferris" wheel?*

Round to the nearest whole number.

Question

Mathematics

Taffy

19 February, 02:09

The original "Ferris" wheel, designed by George Washington Gale Ferris Jr., was built for the World's Columbian Exposition of 1893 in Chicago, Illinois. There were 36 fully enclosed gondolas (passenger cars) that carried up to 40 passengers.

This "Ferris" wheel eventually left Chicago and was relocated in St. Louis, Missouri in time for the 1904 Louisiana Purchase Exposition. On May 11, 1906, it was demolished using a dynamite charge and sold for scraps.

The circumference of the "Ferris" wheel had to be large enough to allow for an arc length of 21.94 ft between each gondola. What was the circumference of the original "Ferris" wheel?*

Round to the nearest whole number.

+3

Answers (1)

Know the Answer?

Arnold · Answer 1

790 ft

Step-by-step explanation:

If there were 36 gondolas separated by an arc length of 21.94 ft, the circumference of the Ferris wheel was

36 * 21.94 ft = 790 ft