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The figure shown is a regular hexagon with center H

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The figure shown is a regular hexagon with center H [#permalink]

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New post 11 Sep 2016, 16:09
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The figure shown is a regular hexagon with center h. the shaded area is a parallelogram that shares three vertices with the hexagon and its fourth vertex is the center of the hexagon. If the length of one side of the hexagon is 8 centimeters, what is the area of the unshaded region?

A) \(16\sqrt{3}cm^2\)
B) \(96 cm^2\)
C) \(64\sqrt{3}cm^2\)
D) \(96\sqrt{3}cm^2\)
E) \(256cm^2\)
[Reveal] Spoiler: OA

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Re: The figure shown is a regular hexagon with center H [#permalink]

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azamaka wrote:
The figure shown is a regular hexagon with center h. the shaded area is a parallelogram that shares three vertices with the hexagon and its fourth vertex is the center of the hexagon. If the length of one side of the hexagon is 8 centimeters, what is the area of the unshaded region?

A) \(16\sqrt{3}cm^2\)
B) \(96 cm^2\)
C) \(64\sqrt{3}cm^2\)
D) \(96\sqrt{3}cm^2\)
E) \(256cm^2\)


Hi,
The area is equal to 6 equilateral triangle with side 8..
However two of these 6 triangles are shaded..
So area = (square root 3)/4 *8^2 *(6-2)= 64* sqroot 3 cm^2
C
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Re: The figure shown is a regular hexagon with center H [#permalink]

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New post 11 Sep 2016, 18:47
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Hi azamaka,

For future reference, any time a GMAT question includes a 'weird' shape, chances are really good that you can 'break down' that shape into smaller shapes that you DO know (re: right triangles, rectangles, etc.). Here, we have a regular hexagon, which is certain weird. Regular hexagons can be broken down into 6 equilateral triangles and (if you find it necessary), each equilateral triangle can be broken down into two 30/60/90 right triangles. Thankfully, the GMAT won't make you deal with this type of weird Geometry very often. If you do face this type of situation on Test Day, and you decide to spend time on the prompt (as opposed to dumping it), you'll likely need to spend a bit more time (than average) answering the question and you should look for ways to break the shape(s) down into pieces.

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Re: The figure shown is a regular hexagon with center H [#permalink]

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New post 12 Sep 2016, 18:18
chetan2u wrote:
Hi,
The area is equal to 6 equilateral triangle with side 8..
However two of these 6 triangles are shaded..
So area = (square root 3)/4 *8^2 *(6-2)= 64* sqroot 3 cm^2
C

Thanks for the crisp explanation.
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The figure shown is a regular hexagon with center H [#permalink]

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azamaka wrote:
The figure shown is a regular hexagon with center h. the shaded area is a parallelogram that shares three vertices with the hexagon and its fourth vertex is the center of the hexagon. If the length of one side of the hexagon is 8 centimeters, what is the area of the unshaded region?

A) \(16\sqrt{3}cm^2\)
B) \(96 cm^2\)
C) \(64\sqrt{3}cm^2\)
D) \(96\sqrt{3}cm^2\)
E) \(256cm^2\)

Attachment:
hexshade.jpg
hexshade.jpg [ 34.8 KiB | Viewed 742 times ]

This prompt's wording and wordiness could confuse. The figure is better.
Unshaded region area = (hexagon area) - (parallelogram area)

Dividing shapes into triangles is an easy way to find area, especially when the polygons are regular (equal angles, equal sides).

A regular polygon can be divided into congruent isosceles triangles.
A regular hexagon is the only regular polygon that can be divided into congruent equilateral triangles.

Connect the vertices. There are 6 equilateral triangles.

An equilateral triangle has three 60° angles.
Six sides of a hexagon divide 360° at center into six 60° angles at center
The other two angles of each triangle also = 60°; both bisect a vertex of 120° = 60° each

Given: each side of hexagon = 8 cm
From above: the six triangles are equilateral. Each side of each triangle = 8 cm

Find the area of one of these triangles.
Multiply by 6 for the area of the hexagon.
Multiply by 2 for the area of the parallelogram, which = two of those triangles

Area of equilateral triangle* is a handy thing to know, s = side:

\(A =\frac{s^2\sqrt{3}}{4}\)
\(A =\frac{8^2\sqrt{3}}{4}\)
\(A =16\sqrt{3}\)


Area of hexagon (six equilateral triangles):
\(6 * 16\sqrt{3}= 96\sqrt{3}\) sq cm

Area of parallelogram (two equilateral triangles)
\(2 *16\sqrt{3} = 32\sqrt{3}\) sq cm

Area of unshaded region:
(Hexagon area) - (parallelogram area) =\((96\sqrt{3}- 32\sqrt{3}) = 64\sqrt{3}\) sq cm

Answer C

*If you don't remember the formula for the area of an equilateral triangle, draw one. Drop an altitude, which is a perpendicular bisector of the opposite side and of the vertex.

That altitude creates two congruent right 30-60-90 triangles with side lengths that correspond to 30-60-90, in ratio \(x : x\sqrt{3} : 2x\)

Side lengths? Side opposite the 90° angle = \(2x = 8\). Side opposite 30° angle is half of that, i.e., \(x, x = 4\). Side opposite 60° angle = height of triangle = \(x\sqrt{3}\) or \(4\sqrt{3}\)

Area\(=\frac{b*h}{2} = (8*4\sqrt{3})*\frac{1}{2} = 16\sqrt{3}\) square cm

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The figure shown is a regular hexagon with center H   [#permalink] 13 Nov 2017, 12:53
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