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about to only mathematics

A

`3//4`

B

`3sqrt3`

C

3

D

`3sqrt3//2`

Text Solution

Verified by Experts

The correct Answer is:
C
Let O be the center of the circle.
`angleA_0OA_1=(360^(@))/6=60^@`
Thus, `A_0OA_1` is an equilateral triangle. we get
`A_0OA_1 (" radius of circle = 1")`
Also `A_0OA_2=A_0OA_4`
`=2A_0D`
`=2(OA_0sin60^@)`
`=2(1)sqrt3/2=sqrt3`
`:. (A_0A_1)(A_0A_2)(A_0A_4)`
`=(1)(sqrt3)(sqrt3)=3`
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