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Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

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02 Jun 2013, 01:49

In an regular hexagon inscribed in a circle, its side is equal the radius. We can divide the hexagon in 6 triangles each with the base of 4. The heigth will equal \(\sqrt{4^2-2^2}=\sqrt{12}=2\sqrt{3}\). To obtain this just use Pythagoras, the hypotenuse of each triangle it's the radius, and the bases it's \(\frac{4}{2}=2\).

Now you have the height of each triangle, so \(A_t=(4*2\sqrt{3})/2=4\sqrt{3}\).

\(A_h=6*A_t=6*4\sqrt{3}=24\sqrt{3}\)

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Regular hexagon will have six traingle regions. Central angle will be 60. (360/6) Equilateral triangle will be formed as we have three 60 angles. Length of each side is equal to radius 4. Area of hexagon = area of 6 eq. triangles = 6 * root 3/4 * 4^2 = 24 root3
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A regular hexagon is essentially composed of 6 equilateral trianlges...and the line joining the opposite vertices is the diameter of the circle in which the hexagon is inscribed...So the radius of the circle forms the side of the equilateral triangle... Area is 6* (3^1/2)/4(a^2) where a = radius of the circle. 6*sq. rt3/4 * 4^2=24 root 3

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

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03 Jun 2013, 13:24

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fozzzy wrote:

Attachment:

screen_shot_2010_12_23_at_3.41.13_pm.png

If a regular hexagon is inscribed in a circle with a radius of 4, the area of the hexagon is

A. 12 root 3 B. 8 pi C. 18 root 2 D. 24 root 3 E. 48

The Hexagon can be divided into 6 equilateral triangles , each with side = radius of the circle. Since area of equilateral triangle is (root(3)*a^2)4, for 6 such triangles we will get 24root3

WE: General Management (Non-Profit and Government)

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

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05 Jun 2013, 07:15

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1) Each of the hexagon's angles 120 degrees : formula if you don't know it is [(# of sides - 2) x 180] = (6 - 2) x 180 = 720 ÷ 6 = 120. 2) Next, split up the hexagon into 6 equilateral triangles with 4 for each of its sides. 3) Find the area of one of the triangles: - Base = 4 - find the height by splitting the triangle in half so that it becomes a 30/60/90 triangle and find the height using the pythagorean theorem or knowing the 1,√3,2 triangle. Height = 2√3 - one triangle's area = (1/2)bh = (1/2)(4)(2√3) = 4√3 4) find the area of the hexagon by multiplying the one triangle's area by 6: - 6 x 4√3 = 24√3 Answer is D

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

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12 Jun 2013, 05:22

lchen wrote:

1) Each of the hexagon's angles 120 degrees : formula if you don't know it is [(# of sides - 2) x 180] = (6 - 2) x 180 = 720 ÷ 6 = 120. 2) Next, split up the hexagon into 6 equilateral triangles with 4 for each of its sides. 3) Find the area of one of the triangles: - Base = 4 - find the height by splitting the triangle in half so that it becomes a 30/60/90 triangle and find the height using the pythagorean theorem or knowing the 1,√3,2 triangle. Height = 2√3 - one triangle's area = (1/2)bh = (1/2)(4)(2√3) = 4√3 4) find the area of the hexagon by multiplying the one triangle's area by 6: - 6 x 4√3 = 24√3 Answer is D

Great method we could use the formula for the area of equilateral triangle and save a few steps Root 3/4 a^2 where a = 4
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Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

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04 Feb 2016, 12:21

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Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

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07 Jun 2017, 03:55

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If a regular hexagon is inscribed in a circle with a radius of 4, the area of the hexagon is

A. 12 root 3 B. 8 pi C. 18 root 2 D. 24 root 3 E. 48

We may recall that a regular hexagon can be divided into 6 equilateral triangles, and thus the area of a regular hexagon is:

6[(s^2√3)/4], where s = side of the equilateral triangle and (s^2√3)/4 = the area of the equilateral triangle. Since the radius of 4 also represents one side of the equilateral triangle, we can now determine the area of the hexagon.

Area = 6[(4^2√3)/4]

Area = 6[(16√3)/4]

Area = 6(4√3)

Area = 24√3

Answer: D
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