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Regular hexagon ABCDEF has a perimeter of 36. O is the center of the h
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31 Mar 2021, 21:15
31 Mar 2021, 21:15
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Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?
A. \(108-18\pi\)
B. \(54\sqrt{3}-9\pi\)
C. \(54\sqrt{3}-18\pi\)
D. \(108-27\pi\)
E. \(54\sqrt{3}-27\pi\)
Regular hexagon ABCDEF has a perimeter of 36. O is the center of the h
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31 Mar 2021, 22:08
31 Mar 2021, 22:08
1
A regular hexagon is comprised of 6 equilateral triangles. For our convenience, let's take any one of those triangles and multiple the measure for the whole figure.
In triangle FAO, there are 3 equal sectors of a circle and the shaded region in the centre.
Since the perimeter of hexagon is 36, each side of triangle must be 6.
The radius of the circle of which the sectors are a part must be 3.
To find the area of the shaded part in triangle FAO, let's calculate the area of sectors and subtract it from the area of triangle.
Each sector has 60° for it is at the corner of equilateral triangle. The area of each sector must be 60/360 = 1/6 th the area of circle of which it is a part.
Area of each sector \(= (1/6) * pi * (3^2)\)
Area of the three sectors \(= 3 * (1/6) * pi * (3^2) = (9/2) * pi\)
Area of triangle FAO \(= (\frac{\sqrt{3}}{4} * 6^2)\)
Area of shaded region in triangle FAO = Area of triangle FAO - Area of the three sectors \(= 9 * \sqrt{3} - (9/2) * pi\)
This is area of the shaded regions in one of the 6 equilateral triangles in hexagon.
Area of 6 such shaded regions in the hexagon \(= 6 * [9 * \sqrt{3} - (9/2) * pi]\)
E is the correct answer.
In triangle FAO, there are 3 equal sectors of a circle and the shaded region in the centre.
Since the perimeter of hexagon is 36, each side of triangle must be 6.
The radius of the circle of which the sectors are a part must be 3.
To find the area of the shaded part in triangle FAO, let's calculate the area of sectors and subtract it from the area of triangle.
Each sector has 60° for it is at the corner of equilateral triangle. The area of each sector must be 60/360 = 1/6 th the area of circle of which it is a part.
Area of each sector \(= (1/6) * pi * (3^2)\)
Area of the three sectors \(= 3 * (1/6) * pi * (3^2) = (9/2) * pi\)
Area of triangle FAO \(= (\frac{\sqrt{3}}{4} * 6^2)\)
Area of shaded region in triangle FAO = Area of triangle FAO - Area of the three sectors \(= 9 * \sqrt{3} - (9/2) * pi\)
This is area of the shaded regions in one of the 6 equilateral triangles in hexagon.
Area of 6 such shaded regions in the hexagon \(= 6 * [9 * \sqrt{3} - (9/2) * pi]\)
E is the correct answer.
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the center of the h
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31 Mar 2021, 22:14
31 Mar 2021, 22:14
GeminiHeat wrote:
Attachment:
Untitled.png
Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?
A. \(108-18\pi\)
B. \(54\sqrt{3}-9\pi\)
C. \(54\sqrt{3}-18\pi\)
D. \(108-27\pi\)
E. \(54\sqrt{3}-27\pi\)
Let each side of the Hexagon be \(a\)
\(6a = 36\)
\(a = 6\)
Area of Regular Hexagon \(= 6(\frac{\sqrt{3}}{4} a^2) = 54\sqrt{3}\)
Now, we can notice that each side \(a\) is equal to twice the radius \(r\) of the circles
i.e. \(6 = 2r\)
\(r = 3\)
Now, Since each interior angle of Regular Hexagon is \(120\) degrees, \((120)(6) = 720\) degrees
Therefore we will have \(2\) circles from \(720\) degrees and \(1\) circle in the centre.
Required Area = Area of Hexagon - Area of 3 Circles
\(= 54\sqrt{3} - 3π(3)^2\)
\(= 54\sqrt{3} - 27π\)
Hence, option E
