1 Answer
Solution :-
The dimensions of the circle are
Diameter = 40 cm
Radius = \( \dfrac{40}{2}\) = 20 cm
Consider AB be as the chord of the circle i.e. length = 20 cm
In ΔOAB,
Radius of circle = OA = OB = 20 cm
Similarly AB = 20 cm
Hence, ΔOAB is an equilateral triangle.
θ = 60° = \( \dfrac{\pi}{3}\) radian
We know that,
in a circle of radius r unit, if an angle θ radian at the centre is subtended by an arc of length l unit then
We get θ = \( \dfrac{1}{r}\)
\( \dfrac{\pi}{3}=\dfrac{arc \;AB}{20}\)
arc AB= \( \dfrac{20\pi}{3}\)cm
Therefore, the length of the minor arc of the chord is \( \dfrac{20\pi}{3}\) cm.
Answered by Abhisek | 1 year agoRelated Questions
prove that \(sin \dfrac{8π}{3} cos \dfrac{23π}{6} + cos \dfrac{13π}{3} sin \dfrac{35π}{6} = \dfrac{1}{2}\)
prove that \( 3 sin \dfrac{π}{6} sec \dfrac{π}{3} – 4 sin \dfrac{5π}{6} cot \dfrac{π}{4} = 1\)
prove that \( tan \dfrac{11π}{3} – 2 sin \dfrac{4π}{6} – \dfrac{3}{4} cosec^2 \dfrac{π}{4} + 4 cos^2 \dfrac{17π}{6} = \dfrac{(3 – 4\sqrt{3})}{2}\)
prove that cos 570° sin 510° + sin (-330°) cos (-390°) = 0
prove that tan (-125°) cot (-405°) – tan (-765°) cot (675°) = 0