What is the length of an arc of a circle that subtends 2\(\frac{1}{2}\) radians at the centre when the radius of the circle is 8cm?
Given: \(\theta\) = 2\(\frac{1}{2}\) radians, r = 8cm
but 360º = 2 \(\pi\) radian.
Using: L = length of an arc = \(\frac{\theta}{2 \pi}\) x 2\(\pi\) r
= \(\frac{2.5}{2\pi}\) x 2 x \(\pi\) x 8
= 2.5 x 8 = 20cm.
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}