Question

How long is the arc intersected by a central angle of startfraction pi over 3 endfraction radians in a circle with a radius of 6 ft? round your answer to the nearest tenth. use 3.14 for pi. 1.0 ft 5.7 ft 6.3 ft 7.0 ft

Answer 1

The length of the arc which subtends a $\pi/3$ radians angle on a circle with 6 ft radius is given by: Option C: 6.3 feet approximately.

How to find the relation between angle subtended by the arc, the radius and the arc length?

$2\pi^c = 360^\circ = \text{Full circumference}$

The superscript 'c' shows angle measured is in radians.

If radius of the circle is of r units, then:

$1^c \: \rm covers \: \dfrac{circumference}{2\pi} = \dfrac{2\pi r}{2\pi} = r\\\\or\\\\\theta^c \: covers \:\:\: r \times \theta \: \rm \text{units of arc}$

For this case, we have:

• Radius of the circle = r = 6  ft
• Angle subtended by the considered arc of the circle on its center = $\theta^c = \dfrac{\pi}{3}^c$

Thus, if we take:

Length of the arc = L feet, then:

$L =r \times \theta = 6 \times \dfrac{\pi}{3} = 2\pi \: \rm ft \approx 6.28 \approx 6.3 \: ft$

Thus, the length of the arc which subtends a $\pi/3$ radians angle on a circle with 6 ft radius is given by: Option C: 6.3 feet approximately.

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