A merry-go-round has a radius of 10 ft. To the nearest tenth of a foot, what distance does the merry-go-round cover when it rotates through an angle of 72°?

For my first try, I got 62.9 as my answer, and it was marked wrong. I did this a second time, and got 12.6. I would like to check with anyone else to see if I got the answer right.

The answers for this question were:

62.9, 12.6, 25.1, or 125.7

Yes, the correct answer is 12.6 ft.

Using the formula of arc length, that is
s = rθ
where s = arc length, r = radius, θ = angle (unit must be in radian)
With r = 10 ft and θ = 72 deg ( π/180 deg) = 1.26 rad,
s = (10)(1.26) = 12.6 ft

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jbiain jbiain · Answer 2 · 2020-04-23T04:53:38+02:00

jbiain jbiain

23-04-2020

Answer:

12.6

Step-by-step explanation:

The distance covered by the merry-go-round if makes a complete circle (360°) is its circumference:

circumference = 2*π*radius = 2*π*10 = 62.8 ft

This distance corresponds to 360°, for 72°:

62.8 ft / x ft = 360° / 72°

x = (72/360)*62.8

x = 12.6 ft

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