GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Feb 2019, 17:04

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar

February 21, 2019

February 21, 2019

10:00 PM PST

11:00 PM PST

Kick off your 2019 GMAT prep with a free 7-day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.
• Free GMAT RC Webinar

February 23, 2019

February 23, 2019

07:00 AM PST

09:00 AM PST

Learn reading strategies that can help even non-voracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT

The figure shown is a regular hexagon with center H

Author Message
TAGS:

Hide Tags

Intern
Joined: 10 Jun 2014
Posts: 18
The figure shown is a regular hexagon with center H  [#permalink]

Show Tags

11 Sep 2016, 15:09
13
00:00

Difficulty:

45% (medium)

Question Stats:

70% (02:05) correct 30% (02:10) wrong based on 225 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 2947 times ]

Math Expert
Joined: 02 Aug 2009
Posts: 7334
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
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html

GMAT Expert

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13574
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
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Intern
Joined: 18 Jun 2015
Posts: 39
Re: The figure shown is a regular hexagon with center H  [#permalink]

Show Tags

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.
Senior SC Moderator
Joined: 22 May 2016
Posts: 2491
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 1860 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

_________________

To live is the rarest thing in the world.
Most people just exist.

Oscar Wilde

Non-Human User
Joined: 09 Sep 2013
Posts: 9878
Re: The figure shown is a regular hexagon with center H  [#permalink]

Show Tags

16 Dec 2018, 23:16
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: The figure shown is a regular hexagon with center H   [#permalink] 16 Dec 2018, 23:16
Display posts from previous: Sort by