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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # The figure shown is a regular hexagon with center H  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Intern Joined: 10 Jun 2014 Posts: 19 The figure shown is a regular hexagon with center H [#permalink] ### Show Tags 11 Sep 2016, 15:09 11 00:00 Difficulty: 35% (medium) Question Stats: 72% (02:01) correct 28% (02:17) wrong based on 194 sessions ### HideShow timer Statistics 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) $$256cm^2$$ Attachments gmf25.574B986.jpg [ 8.38 KiB | Viewed 2505 times ] Math Expert Joined: 02 Aug 2009 Posts: 7025 Re: The figure shown is a regular hexagon with center H [#permalink] ### Show Tags 11 Sep 2016, 16:30 1 2 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 _________________ 1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html GMAT online Tutor EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12844 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The figure shown is a regular hexagon with center H [#permalink] ### Show Tags 11 Sep 2016, 17:47 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. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: The figure shown is a regular hexagon with center H  [#permalink]

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12 Sep 2016, 17:18
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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Joined: 22 May 2016
Posts: 2087
The figure shown is a regular hexagon with center H  [#permalink]

### Show Tags

13 Nov 2017, 11:53
2
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 [ 34.8 KiB | Viewed 1423 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

(Hexagon area) - (parallelogram area) =$$(96\sqrt{3}- 32\sqrt{3}) = 64\sqrt{3}$$ sq cm

*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
The figure shown is a regular hexagon with center H &nbs [#permalink] 13 Nov 2017, 11:53
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