Find all School-related info fast with the new School-Specific MBA Forum

It is currently 10 Jul 2014, 22:51

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Two squares, each of side lengths 1 unit and having their

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
Retired Moderator
User avatar
Joined: 01 Oct 2009
Posts: 485
Location: Bangalore,India
WE 1: 4yrs in IT Industry
Followers: 21

Kudos [?]: 98 [1] , given: 322

Two squares, each of side lengths 1 unit and having their [#permalink] New post 14 Sep 2010, 06:27
1
This post received
KUDOS
5
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  65% (medium)

Question Stats:

43% (04:56) correct 56% (03:11) wrong based on 82 sessions
Two squares, each of side lengths 1 unit and having their centres at O, are rotated with respect to each other to generate octagon ABCDEFGH, as shown in the figure above. If AB = 43/99, find the area of the octagon.

Attachment:
G86.PNG
G86.PNG [ 5.44 KiB | Viewed 2473 times ]


A. 73/99
B. 86/99
C. 1287/9801
D. 76/99
[Reveal] Spoiler: OA

_________________

One Final Try.......


Last edited by Bunuel on 09 Jul 2013, 09:39, edited 1 time in total.
Added the OA.
Expert Post
11 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18492
Followers: 3188

Kudos [?]: 21216 [11] , given: 2537

Re: geomertry Q [#permalink] New post 14 Sep 2010, 06:52
11
This post received
KUDOS
Expert's post
RaviChandra wrote:
Two squares, each of side lengths 1 unit and having their centres at O, are rotated with respect to each other to generate octagon ABCDEFGH, as shown in the figure above. If AB = 43/99, find the area of the octagon.

Attachment:
G86.PNG


a. 73/99
b. 86/99
c. 1287/9801
d. 76/99


First note that all the triangles will have the same dimensions. Let the legs of these triangles be x and y, also we know that the hypotenuse equals to \frac{43}{99}, so x^2+y^2=(\frac{43}{99})^2.

If we take one side of a square we'll see that x+y+\frac{43}{99}=1 --> x+y=\frac{56}{99} --> square it --> x^2+2xy+y^2=(\frac{56}{99})^2, as from above x^2+y^2=(\frac{43}{99})^2, then: (\frac{43}{99})^2+2xy=(\frac{56}{99})^2 --> 2xy=(\frac{56}{99})^2-(\frac{43}{99})^2=(\frac{56}{99}-\frac{43}{99})(\frac{56}{99}+\frac{43}{99})=\frac{13}{99}*1=\frac{13}{99};

Now, the are of the octagon will be area of a square minus area of 4 triangles --> area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}.

Answer: B.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Retired Moderator
User avatar
Joined: 01 Oct 2009
Posts: 485
Location: Bangalore,India
WE 1: 4yrs in IT Industry
Followers: 21

Kudos [?]: 98 [0], given: 322

Re: geomertry Q [#permalink] New post 14 Sep 2010, 06:56
thanku so much for the reply
_________________

One Final Try.......

9 KUDOS received
Manager
Manager
avatar
Joined: 30 Aug 2010
Posts: 93
Location: Bangalore, India
Followers: 3

Kudos [?]: 92 [9] , given: 27

Re: geomertry Q [#permalink] New post 14 Sep 2010, 07:24
9
This post received
KUDOS
1
This post was
BOOKMARKED
EASY WAY is as shown below

Connect AO and BO to get a AOB. We get totally 8 triangles like this.

Now. Lets calculate the AREA of AOB

AREA = .5*base*hiegt = 0.5*43/99 * 1/2 note: i/2 is the hieght as the length of a side of the given square is 1 and half of that is what is the hieght

hence AREA = 1/4 * 43/99

As told we have 8 similar triangles in this octagon.

Hence ANSWER = 8*1/4 * 43/99 = 86/99

Hope it is clear
Retired Moderator
User avatar
Joined: 01 Oct 2009
Posts: 485
Location: Bangalore,India
WE 1: 4yrs in IT Industry
Followers: 21

Kudos [?]: 98 [0], given: 322

Re: geomertry Q [#permalink] New post 14 Sep 2010, 08:34
muralimba wrote:
EASY WAY is as shown below
AREA = .5*base*hiegt = 0.5*43/99 * 1/2 note: i/2 is the hieght as the length of a side of the given square is 1 and half of that is what is the hieght

This doesnt looks like a straight line i dont think i can be 0.5

Attachment:
G86.PNG
G86.PNG [ 5.65 KiB | Viewed 2393 times ]

_________________

One Final Try.......

Retired Moderator
User avatar
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 77

Kudos [?]: 433 [0], given: 25

GMAT ToolKit User GMAT Tests User Reviews Badge
Re: geomertry Q [#permalink] New post 14 Sep 2010, 08:40
RaviChandra wrote:
muralimba wrote:
EASY WAY is as shown below
AREA = .5*base*hiegt = 0.5*43/99 * 1/2 note: i/2 is the hieght as the length of a side of the given square is 1 and half of that is what is the hieght

This doesnt looks like a straight line i dont think i can be 0.5

Attachment:
G86.PNG


It is the length of the perpendicular from the center of the square to its side AB. It has to be half the length of the side, hence 1/2.
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

Get the best GMAT Prep Resources with GMAT Club Premium Membership

CEO
CEO
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2793
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 173

Kudos [?]: 879 [0], given: 235

GMAT Tests User Reviews Badge
Re: geomertry Q [#permalink] New post 14 Sep 2010, 11:48
Nice solution shrouded....It has to be perpendicular with half the length of square.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned :)

Jo Bole So Nihaal , Sat Shri Akaal

:thanks Support GMAT Club by putting a GMAT Club badge on your blog/Facebook :thanks

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html

Manager
Manager
User avatar
Joined: 15 Apr 2010
Posts: 196
Followers: 3

Kudos [?]: 14 [0], given: 29

GMAT Tests User
Re: geomertry Q [#permalink] New post 14 Sep 2010, 12:34
great solutions, all.. this helps a lot.

thanks
Intern
Intern
avatar
Joined: 24 Jan 2011
Posts: 8
Followers: 0

Kudos [?]: 0 [0], given: 13

GMAT Tests User
Re: geomertry Q [#permalink] New post 17 Jun 2011, 17:48
I dont understand why the solutions remove the area of the four triangles outside the stationary square? Should not the area of octagon add the area of those four triangles rather than subtract? Bunuel ,Shrouded?
SVP
SVP
User avatar
Joined: 09 Sep 2013
Posts: 1535
Followers: 154

Kudos [?]: 32 [0], given: 0

Premium Member
Re: Two squares, each of side lengths 1 unit and having their [#permalink] New post 29 Nov 2013, 05:39
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.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Manager
Manager
User avatar
Joined: 28 Apr 2013
Posts: 174
Location: India
GPA: 4
WE: Medicine and Health (Health Care)
Followers: 0

Kudos [?]: 22 [0], given: 84

Re: geomertry Q [#permalink] New post 29 Nov 2013, 05:40
Bunuel wrote:
RaviChandra wrote:
Two squares, each of side lengths 1 unit and having their centres at O, are rotated with respect to each other to generate octagon ABCDEFGH, as shown in the figure above. If AB = 43/99, find the area of the octagon.

Attachment:
G86.PNG


a. 73/99
b. 86/99
c. 1287/9801
d. 76/99


First note that all the triangles will have the same dimensions. Let the legs of these triangles be x and y, also we know that the hypotenuse equals to \frac{43}{99}, so x^2+y^2=(\frac{43}{99})^2.

If we take one side of a square we'll see that x+y+\frac{43}{99}=1 --> x+y=\frac{56}{99} --> square it --> x^2+2xy+y^2=(\frac{56}{99})^2, as from above x^2+y^2=(\frac{43}{99})^2, then: (\frac{43}{99})^2+2xy=(\frac{56}{99})^2 --> 2xy=(\frac{56}{99})^2-(\frac{43}{99})^2=(\frac{56}{99}-\frac{43}{99})(\frac{56}{99}+\frac{43}{99})=\frac{13}{99}*1=\frac{13}{99};

Now, the are of the octagon will be area of a square minus area of 4 triangles --> area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}.

Answer: B.


Is the above question follow the pattern of word problems in GMAT?

Thnaks for such comprehensible posts.

:lol:
_________________

Thanks for Posting

LEARN TO ANALYSE

+1 kudos if you like

Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 476
Followers: 1

Kudos [?]: 41 [0], given: 134

Re: Two squares, each of side lengths 1 unit and having their [#permalink] New post 12 Dec 2013, 11:28
Basically, we have to find the area of the square and the eight triangles that border it.

We are told that measure AB = 43/99. Therefore, we know that measure XA +BZ (see attached diagram) = 56/99. We also know that each triangle formed is a right triangle so we can use the Pythagorean theorem: a^2 + b^2 = c^2 or in this case, a^2 + b^2 = (43/99)^2. Also, keep in mind that the lengths XA + BA are also the leg lengths of the two right triangles.

What we know:

a+b = 56/99
a^2 + b^2 = (43/99)^2

From the first equation, we could isolate a variable to get a=(56/99) - b then plug it into the Pythagorean theorem but that might get messy. Instead, we notice that in both equations we have an a + b: we can square a and b in the first equation to make it easier to substitute into the second equation.

a+b = 56/99
(a+b)^2 = (56/99)^2
a^2 + 2ab + b^2 = (56/99)^2
a^2 + b^2 + 2ab = (56/99)^2
Now we have an equation that can easily plug into the Pythagorean theorem

a^2 + b^2 = (43/99)^2 (substitute in a^2 + b^2 + 2ab = (56/99)^2)

(56/99)^2 - 2ab = (43/99)^2
(56/99)^2 = (43/99)^2 + 2ab
√(56/99)^2 = √(43/99)^2 + √2ab
(56/99) = (43/99) + √2ab
(56/99) - (43/99) = √2ab
13/99 = √2ab

Clearly I made a mistake here. If I were to square both sides I would get 169/9801 which is much smaller than 13/99. If I were to square (56/99)^2 and (43/99)^2 then reduce, I would get 13/99 but this is far too time consuming for the test. Where did I go wrong with my equation?

Furthermore, why would the area of this square be the square MINUS the area of the triangles? Wouldn't it be the area of the square PLUS the area of the triangles?

Thanks!
Attachments

EXAMPLE SEVEN.png
EXAMPLE SEVEN.png [ 18.91 KiB | Viewed 738 times ]

Intern
Intern
User avatar
Joined: 07 Jan 2013
Posts: 26
Location: Poland
GPA: 3.8
Followers: 0

Kudos [?]: 5 [0], given: 491

Re: geomertry Q [#permalink] New post 23 Dec 2013, 14:11
Bunuel wrote:
RaviChandra wrote:
Two squares, each of side lengths 1 unit and having their centres at O, are rotated with respect to each other to generate octagon ABCDEFGH, as shown in the figure above. If AB = 43/99, find the area of the octagon.

Attachment:
G86.PNG


a. 73/99
b. 86/99
c. 1287/9801
d. 76/99


First note that all the triangles will have the same dimensions. Let the legs of these triangles be x and y, also we know that the hypotenuse equals to \frac{43}{99}, so x^2+y^2=(\frac{43}{99})^2.

If we take one side of a square we'll see that x+y+\frac{43}{99}=1 --> x+y=\frac{56}{99} --> square it --> x^2+2xy+y^2=(\frac{56}{99})^2, as from above x^2+y^2=(\frac{43}{99})^2, then: (\frac{43}{99})^2+2xy=(\frac{56}{99})^2 --> 2xy=(\frac{56}{99})^2-(\frac{43}{99})^2=(\frac{56}{99}-\frac{43}{99})(\frac{56}{99}+\frac{43}{99})=\frac{13}{99}*1=\frac{13}{99};

Now, the are of the octagon will be area of a square minus area of 4 triangles --> area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}.

Answer: B.

hi Banuel. I can't understand why I should subtract the four triangles when it seems that I should add them to the area of the square. Please explain. Thanks.
Moderator
Moderator
User avatar
Joined: 25 Apr 2012
Posts: 508
Location: India
Concentration: Marketing, International Business
GPA: 3.21
WE: Business Development (Other)
Followers: 11

Kudos [?]: 196 [0], given: 552

Re: geomertry Q [#permalink] New post 23 Dec 2013, 23:26
Magdak wrote:
Bunuel wrote:
RaviChandra wrote:
Two squares, each of side lengths 1 unit and having their centres at O, are rotated with respect to each other to generate octagon ABCDEFGH, as shown in the figure above. If AB = 43/99, find the area of the octagon.

Attachment:
The attachment G86.PNG is no longer available


a. 73/99
b. 86/99
c. 1287/9801
d. 76/99


First note that all the triangles will have the same dimensions. Let the legs of these triangles be x and y, also we know that the hypotenuse equals to \frac{43}{99}, so x^2+y^2=(\frac{43}{99})^2.

If we take one side of a square we'll see that x+y+\frac{43}{99}=1 --> x+y=\frac{56}{99} --> square it --> x^2+2xy+y^2=(\frac{56}{99})^2, as from above x^2+y^2=(\frac{43}{99})^2, then: (\frac{43}{99})^2+2xy=(\frac{56}{99})^2 --> 2xy=(\frac{56}{99})^2-(\frac{43}{99})^2=(\frac{56}{99}-\frac{43}{99})(\frac{56}{99}+\frac{43}{99})=\frac{13}{99}*1=\frac{13}{99};

Now, the are of the octagon will be area of a square minus area of 4 triangles --> area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}.

Answer: B.

hi Banuel. I can't understand why I should subtract the four triangles when it seems that I should add them to the area of the square. Please explain. Thanks.



Look at the figure below...the area of octagon is the coloured portion which is 1 - area of 4 similar triangles


The drawing is not to scale but I hope you get the point
Attachments

untitled.PNG
untitled.PNG [ 4.17 KiB | Viewed 678 times ]


_________________


“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

SVP
SVP
User avatar
Joined: 06 Sep 2013
Posts: 1629
Location: United States
Concentration: Finance
GMAT 1: 710 Q48 V39
WE: Corporate Finance (Investment Banking)
Followers: 10

Kudos [?]: 128 [0], given: 254

GMAT ToolKit User
Re: geomertry Q [#permalink] New post 08 May 2014, 08:18
muralimba wrote:
EASY WAY is as shown below

Connect AO and BO to get a AOB. We get totally 8 triangles like this.

Now. Lets calculate the AREA of AOB

AREA = .5*base*hiegt = 0.5*43/99 * 1/2 note: i/2 is the hieght as the length of a side of the given square is 1 and half of that is what is the hieght

hence AREA = 1/4 * 43/99

As told we have 8 similar triangles in this octagon.

Hence ANSWER = 8*1/4 * 43/99 = 86/99

Hope it is clear


Cool method, but where do you get the 8 triangles from? I don't see them?

Could you please clarify?
Thanks!

Cheers
J :)
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18492
Followers: 3188

Kudos [?]: 21216 [1] , given: 2537

Re: geomertry Q [#permalink] New post 09 May 2014, 01:04
1
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
jlgdr wrote:
muralimba wrote:
EASY WAY is as shown below

Connect AO and BO to get a AOB. We get totally 8 triangles like this.

Now. Lets calculate the AREA of AOB

AREA = .5*base*hiegt = 0.5*43/99 * 1/2 note: i/2 is the hieght as the length of a side of the given square is 1 and half of that is what is the hieght

hence AREA = 1/4 * 43/99

As told we have 8 similar triangles in this octagon.

Hence ANSWER = 8*1/4 * 43/99 = 86/99

Hope it is clear


Cool method, but where do you get the 8 triangles from? I don't see them?

Could you please clarify?
Thanks!

Cheers
J :)


Here they are:
Attachment:
Untitled.png
Untitled.png [ 6.85 KiB | Viewed 327 times ]

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Senior Manager
Senior Manager
avatar
Joined: 07 Apr 2012
Posts: 289
Followers: 0

Kudos [?]: 13 [0], given: 44

Re: Two squares, each of side lengths 1 unit and having their [#permalink] New post 06 Jul 2014, 08:29
Bunuel wrote:
RaviChandra wrote:
Two squares, each of side lengths 1 unit and having their centres at O, are rotated with respect to each other to generate octagon ABCDEFGH, as shown in the figure above. If AB = 43/99, find the area of the octagon.

Attachment:
G86.PNG


a. 73/99
b. 86/99
c. 1287/9801
d. 76/99


First note that all the triangles will have the same dimensions. Let the legs of these triangles be x and y, also we know that the hypotenuse equals to \frac{43}{99}, so x^2+y^2=(\frac{43}{99})^2.

If we take one side of a square we'll see that x+y+\frac{43}{99}=1 --> x+y=\frac{56}{99} --> square it --> x^2+2xy+y^2=(\frac{56}{99})^2, as from above x^2+y^2=(\frac{43}{99})^2, then: (\frac{43}{99})^2+2xy=(\frac{56}{99})^2 --> 2xy=(\frac{56}{99})^2-(\frac{43}{99})^2=(\frac{56}{99}-\frac{43}{99})(\frac{56}{99}+\frac{43}{99})=\frac{13}{99}*1=\frac{13}{99};

Now, the are of the octagon will be area of a square minus area of 4 triangles --> area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}.

Answer: B.

Hi Bunuel,

I don't understand why we subtract only 4 triangle?
Aren't we supposed to subtract 8 triangles?
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18492
Followers: 3188

Kudos [?]: 21216 [0], given: 2537

Re: Two squares, each of side lengths 1 unit and having their [#permalink] New post 06 Jul 2014, 11:18
Expert's post
ronr34 wrote:
Bunuel wrote:
RaviChandra wrote:
Two squares, each of side lengths 1 unit and having their centres at O, are rotated with respect to each other to generate octagon ABCDEFGH, as shown in the figure above. If AB = 43/99, find the area of the octagon.

Attachment:
The attachment G86.PNG is no longer available


a. 73/99
b. 86/99
c. 1287/9801
d. 76/99


First note that all the triangles will have the same dimensions. Let the legs of these triangles be x and y, also we know that the hypotenuse equals to \frac{43}{99}, so x^2+y^2=(\frac{43}{99})^2.

If we take one side of a square we'll see that x+y+\frac{43}{99}=1 --> x+y=\frac{56}{99} --> square it --> x^2+2xy+y^2=(\frac{56}{99})^2, as from above x^2+y^2=(\frac{43}{99})^2, then: (\frac{43}{99})^2+2xy=(\frac{56}{99})^2 --> 2xy=(\frac{56}{99})^2-(\frac{43}{99})^2=(\frac{56}{99}-\frac{43}{99})(\frac{56}{99}+\frac{43}{99})=\frac{13}{99}*1=\frac{13}{99};

Now, the are of the octagon will be area of a square minus area of 4 triangles --> area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}.

Answer: B.

Hi Bunuel,

I don't understand why we subtract only 4 triangle?
Aren't we supposed to subtract 8 triangles?


Attachment:
Untitled.png
Untitled.png [ 7.68 KiB | Viewed 95 times ]

The area of yellow octagon = the area of blue square - the area of 4 red triangles.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Re: Two squares, each of side lengths 1 unit and having their   [#permalink] 06 Jul 2014, 11:18
    Similar topics Author Replies Last post
Similar
Topics:
1 each side of square ABCD has length 1, the length of line haichao 3 20 Nov 2008, 08:12
each side of square ABCD has length 1, the length of line Jcpenny 1 19 Nov 2008, 20:13
4 Experts publish their posts in the topic In the figure, each side of square ABCD has length 1, the jugolo1 9 25 Sep 2008, 02:05
Experts publish their posts in the topic In the figure, each side of square ABCD has length 1, the gregspirited 2 27 Nov 2007, 13:42
18 Experts publish their posts in the topic In the figure, each side of square ABCD has length 1, the singh_amit19 20 19 Oct 2007, 03:37
Display posts from previous: Sort by

Two squares, each of side lengths 1 unit and having their

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.