Question

a wire is in the shape of a regular hexagon encloses an area of 726root3 cm2. if the same wire is bent into form a circle. find the area of the circle

Answer 1

area of hexagon=

$\binom{3 \sqrt{3} }{2} \times {a}^{2}$

$726 \sqrt{3} {cm}^{2} = \binom{3 \sqrt{3} }{2} {a}^{2}$

726×2/3=a²

242×2=a²

484=a²

√484=22

side=22

πr²

Answer 2

The area of the circle is 1386 cm².

Step-by-step explanation:

Consider the provided information.

It is given that the area of regular hexagon is $726\sqrt{3}$ cm².

The area of regular hexagon is: $A=\dfrac{3\sqrt{3} }{2}a^2$

$\dfrac{3\sqrt{3} }{2}a^2=726\sqrt{3}$

$a^2=\dfrac{726\times2}{3}\\\\a^2=484\\\\a=22$

The perimeter of the hexagon is: $P=6a$

Substitute the respective value in above formula.

$P=6(22)=132$

Hence, the length of the wire is 132 cm which must be equal to the circumference of the circle.

$2\pi r=132\\\\r=\dfrac{132\times7}{2\times22}=21$

Now find the area of circle using the formula: $A=\pi r^2$

$A=\dfrac{22}{7} \times(21)^2\\\\A=22\times63\\\\A=1386$

Hence, the area of the circle is 1386 cm².

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