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Question

ABCDEF is a regular hexagon with centre O (Fig). If the area of triangle OAB is 9 cm2. Find the area of (i) the hexagon and (ii) the circle in which the hexagon is inscribed.

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Solution

We know that a regular hexagon is made up of 6 equilateral triangles.

So the Area of a hexagon = 6 × area of one equilateral triangle.

We have given area of the one of the triangles.= 6 x 9cm2

Hence Area of a hexagon= 54cm2

Note that if a regular hexagon is inscribed in the circle, then the radius of the circle is same as the side of the regular hexagon.

We further know that a regular hexagon is made up of 6 equilateral triangles and it is given that the area of an equilateral OAB is 9 cm2

So using the formula of area of an equilateral triangle= 3 4 we have

9 = 3 4× side2

side2= (9)(4)3

Area of a circle= 227 x 363

Now we will find the area of the circle.

Area of a circle = π \({r}^{2}\ )

Area of a circle= 227 x 363

Now we will substitute 3 = 1.732 we get

227 x 361.732

79212.294

65.324 cm2

Therefore, area of the hexagon and area of the circle are 54 cm2 and 65.32 cm2 respectively.


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