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What is the area of the hexagonal region shown above ?

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What is the area of the hexagonal region shown above ? [#permalink] New post 27 Jun 2018, 10:14
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70% (01:14) correct 29% (00:51) wrong based on 77 sessions
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What is the area of the hexagonal region shown above ?

A. \(54 \sqrt{3}\)

B. \(108\)

C. \(108 \sqrt{3}\)

D. \(216\)

E. It cannot be determined from the information given
[Reveal] Spoiler: OA

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Re: What is the area of the hexagonal region shown above ? [#permalink] New post 28 Jun 2018, 00:54
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Carcass wrote:
Attachment:
The attachment ex.jpg is no longer available


What is the area of the hexagonal region shown above ?

A. \(54 \sqrt{3}\)

B. \(108\)

C. \(108 \sqrt{3}\)

D. \(216\)

E. It cannot be determined from the information given


There are no. of ways to solve it:

From the diagram we know it is a regular hexagon

and formula of the area of a regular hexagon = \((3\sqrt{3} * side^2)\) * \(\frac{1}{2}\)

= \((3\sqrt{3} * 6^2)\) * \(\frac{1}{2}\) = \(54\sqrt{3}\).


Now if we donot know the formula then, please refer to the diagram below:

Now we can write : y° + y° + y° + y° + y° + y° = 360°
or y° = 60°.

Now we can see the regular hexagon is divided into 6 triangle.
So each triangle will be equilateral triangle since it is inscribed in a regular hexagon.

Now area of Equilateral triangle = \(\sqrt{3} * (side^2)\) * \(\frac{1}{4}\) = \(\sqrt{3} * (6^2)\) * \(\frac{1}{4}\)

=\(9\sqrt{3}\)

Therefore for 6 triangles we have = 6* \(9\sqrt{3}\) = \(54\sqrt{3}\).

Now if we donot know the area of equilateral triangle then we can always find out:

Take any one triangle inscribed in the regular hexagon:-
SO the area of th triangle = \(\frac{1}{2} * base * altitude\)

Now we need to find out the altitude, which can be found out by Pythagoras theorem, as we draw a line from the midpoint of the hexagon to the base of the triangle as shown in diagram,It divides the base at the midpoint as it is a equilateral triangle. (Also a 30 - 60 - 90 triangle will also have the same result)

So \(altitude^2 = 6^2 - 3^2\) = 27

or altitude = \(3\sqrt{3}\)

Now area = \(\frac{1}{2} * 6 * 3\sqrt{3}\) = \(9\sqrt{3}\)

Now area of all the six triangle = 6 * \(9\sqrt{3}\) = \(54\sqrt{3}\) = Area of the hexagon.

** Please note this applies only for a regular hexagon
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Re: What is the area of the hexagonal region shown above ? [#permalink] New post 01 Jun 2020, 21:42
pranab223 wrote:
Carcass wrote:
Attachment:
ex.jpg


What is the area of the hexagonal region shown above ?

A. \(54 \sqrt{3}\)

B. \(108\)

C. \(108 \sqrt{3}\)

D. \(216\)

E. It cannot be determined from the information given


There are no. of ways to solve it:

From the diagram we know it is a regular hexagon

and formula of the area of a regular hexagon = \((3\sqrt{3} * side^2)\) * \(\frac{1}{2}\)

= \((3\sqrt{3} * 6^2)\) * \(\frac{1}{2}\) = \(54\sqrt{3}\).


Now if we donot know the formula then, please refer to the diagram below:

Now we can write : y° + y° + y° + y° + y° + y° = 360°
or y° = 60°.

Now we can see the regular hexagon is divided into 6 triangle.
So each triangle will be equilateral triangle since it is inscribed in a regular hexagon.

Now area of Equilateral triangle = \(\sqrt{3} * (side^2)\) * \(\frac{1}{4}\) = \(\sqrt{3} * (6^2)\) * \(\frac{1}{4}\)

=\(9\sqrt{3}\)

Therefore for 6 triangles we have = 6* \(9\sqrt{3}\) = \(54\sqrt{3}\).

Now if we donot know the area of equilateral triangle then we can always find out:

Take any one triangle inscribed in the regular hexagon:-
SO the area of th triangle = \(\frac{1}{2} * base * altitude\)

Now we need to find out the altitude, which can be found out by Pythagoras theorem, as we draw a line from the midpoint of the hexagon to the base of the triangle as shown in diagram,It divides the base at the midpoint as it is a equilateral triangle. (Also a 30 - 60 - 90 triangle will also have the same result)

So \(altitude^2 = 6^2 - 3^2\) = 27

or altitude = \(3\sqrt{3}\)

Now area = \(\frac{1}{2} * 6 * 3\sqrt{3}\) = \(9\sqrt{3}\)

Now area of all the six triangle = 6 * \(9\sqrt{3}\) = \(54\sqrt{3}\) = Area of the hexagon.

** Please note this applies only for a regular hexagon




Hexagon has 720 degrees, not 360... 720 / 6 = 120
Re: What is the area of the hexagonal region shown above ?   [#permalink] 01 Jun 2020, 21:42
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