Question

The area of a circle passing through the vertices of a regular hexagon is ‘x’ units2. Find the area of the region not occupied by hexagon. Express in terms of ‘x’ and ‘π’.​

Answer 1

Answer:

Step-by-step explanation:

the area of a circle passing through the vertices of a regular hexagon is ‘x’ units2. Find the area of the region not occupied by hexagon. Express in terms of ‘x’ and ‘π’.​

Answer 2

Answer:

The area of the region not occupied by hexagon is $x-\sqrt{\frac{3x}{\pi } }$   $units^{2}$.

Step-by-step explanation:

The hexagon gets inscribed in the circle where all the vertices of a hexagon lies on the circle.

A hexagon is inscribed in the circle, then the diameter of the circle will be the longest diagonal in the hexagon.

The area of circle is x $units^{2}$ where $x=\pi r^{2}$ where r is the radius of circle.

Now, solving for r, we get:

$r=\sqrt{ \frac{x}{\pi } }$

Also, we can say that diameter of the circle = 2 * side of the hexagon.

Let the side of hexagon be y.

The length of diameter is $2r$.

Then, the equation becomes:

$2r=2*y\\ \\ 2\sqrt{\frac{x}{\pi } } =2y\\ \\ \sqrt{\frac{x}{\pi } } =y$

Thus, the side of hexagon is $\sqrt{ \frac{x}{\pi } }$.

It should be noted that a regular hexagon is composed of six equilateral triangles of equal sides.

The area of the hexagon will be 6 times the area of an equilateral triangle. Let the area be z. Then,

$z=6*\sqrt{ \frac{3}{4}} *y\\ \\ z=6*\sqrt{ \frac{3}{4}} *\sqrt{\frac{x}{\pi } } \\ \\ z=3\sqrt{\frac{3x}{\pi } }$

The z represents the occupied by hexagon. The area unoccupied by hexagon is : Area of circle - Area of hexagon .

The area unoccupied is:

$x-\sqrt{\frac{3x}{\pi } }$$units^{2}$.

To know more about hexagon and its properties:

https://calcresource.com/geom-hexagon.html

https://collegedunia.com/exams/regular-hexagon-properties-area-and-perimeter-formulas-articleid-4940

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