Question

A circle is inscribed in a regular hexagon of Side 2 √3 cm find the circumference of the inscribed circle.

Answer 1

Here

AB  =  23√ cm   and OC  =  radius of inscribed circle

And triangle OAB is a equilateral triangle as we know we can form six equilateral triangle by regular hexagons diagonals .

So, ∠ OAB = ∠ OBA = 60°

And  OC is perpendicular of AB and we know in equilateral triangle altitude is also median , so AC  =  BC  =  2 3√2 = 3√ cm

Now in triangle OAC we get

tan ∠OAC  = OCAC⇒tan 60°  = OC3√⇒3√  = OC3√                                             ( we know tan 60°  =3√  )⇒ OC  = 3 cm

So,
Radius of inscribed circle  =  3 cm

We know area of circle  = π r2 , So

Area of inscribed circle  = 
$22 \div 7 \times 3 {}^{2}$
=28.2857 approx 28.29cm² ( ANS.)
we know the circumference= 2 pi r
$2 \times 22 \div 7 \times 3 = 22 \div 7 \times 6 = 132 \div 7$
=18.857 approx 18.89 (ANS.)

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Answer 2

Answer:

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