Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 22 May 2017, 16:03

### 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 June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

BSchool Forum Moderator
Joined: 02 Oct 2009
Posts: 593
GMAT 1: 530 Q47 V17
GMAT 2: 710 Q50 V36
Followers: 38

Kudos [?]: 350 [3] , given: 412

Two squares, each of side lengths 1 unit and having their [#permalink]

### Show Tags

14 Sep 2010, 07:27
3
KUDOS
9
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

46% (04:30) correct 54% (03:01) wrong based on 122 sessions

### HideShow timer Statistics

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 [ 5.44 KiB | Viewed 4027 times ]

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

Last edited by Bunuel on 09 Jul 2013, 10:39, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 38798
Followers: 7714

Kudos [?]: 105765 [12] , given: 11581

### Show Tags

14 Sep 2010, 07:52
12
KUDOS
Expert's post
2
This post was
BOOKMARKED
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 --> $$area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}$$.

_________________
BSchool Forum Moderator
Joined: 02 Oct 2009
Posts: 593
GMAT 1: 530 Q47 V17
GMAT 2: 710 Q50 V36
Followers: 38

Kudos [?]: 350 [0], given: 412

### Show Tags

14 Sep 2010, 07:56
thanku so much for the reply
Manager
Joined: 30 Aug 2010
Posts: 91
Location: Bangalore, India
Followers: 5

Kudos [?]: 169 [10] , given: 27

### Show Tags

14 Sep 2010, 08:24
10
KUDOS
2
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
BSchool Forum Moderator
Joined: 02 Oct 2009
Posts: 593
GMAT 1: 530 Q47 V17
GMAT 2: 710 Q50 V36
Followers: 38

Kudos [?]: 350 [0], given: 412

### Show Tags

14 Sep 2010, 09:34
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 [ 5.65 KiB | Viewed 3888 times ]
Retired Moderator
Joined: 02 Sep 2010
Posts: 803
Location: London
Followers: 110

Kudos [?]: 1020 [0], given: 25

### Show Tags

14 Sep 2010, 09:40
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.
_________________
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2786
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 235

Kudos [?]: 1722 [0], given: 235

### Show Tags

14 Sep 2010, 12:48
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

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Manager
Joined: 15 Apr 2010
Posts: 195
Followers: 2

Kudos [?]: 17 [0], given: 29

### Show Tags

14 Sep 2010, 13:34
great solutions, all.. this helps a lot.

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

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

### Show Tags

17 Jun 2011, 18:48
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?
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15378
Followers: 648

Kudos [?]: 204 [0], given: 0

Re: Two squares, each of side lengths 1 unit and having their [#permalink]

### Show Tags

29 Nov 2013, 06:39
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.
_________________
Manager
Joined: 28 Apr 2013
Posts: 158
Location: India
GPA: 4
WE: Medicine and Health (Health Care)
Followers: 1

Kudos [?]: 71 [0], given: 84

### Show Tags

29 Nov 2013, 06:40
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 --> $$area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}$$.

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

Thnaks for such comprehensible posts.

_________________

Thanks for Posting

LEARN TO ANALYSE

+1 kudos if you like

Senior Manager
Joined: 13 May 2013
Posts: 468
Followers: 3

Kudos [?]: 172 [0], given: 134

Re: Two squares, each of side lengths 1 unit and having their [#permalink]

### Show Tags

12 Dec 2013, 12:28
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 [ 18.91 KiB | Viewed 2224 times ]

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

Kudos [?]: 7 [0], given: 491

### Show Tags

23 Dec 2013, 15:11
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 --> $$area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}$$.

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.
Director
Joined: 25 Apr 2012
Posts: 727
Location: India
GPA: 3.21
Followers: 44

Kudos [?]: 742 [0], given: 723

### Show Tags

24 Dec 2013, 00:26
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 --> $$area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}$$.

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 [ 4.17 KiB | Viewed 2175 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.”

Current Student
Joined: 06 Sep 2013
Posts: 2005
Concentration: Finance
Followers: 68

Kudos [?]: 643 [0], given: 355

### Show Tags

08 May 2014, 09:18
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?

Thanks!

Cheers
J
Math Expert
Joined: 02 Sep 2009
Posts: 38798
Followers: 7714

Kudos [?]: 105765 [1] , given: 11581

### Show Tags

09 May 2014, 02:04
1
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?

Thanks!

Cheers
J

Here they are:
Attachment:

Untitled.png [ 6.85 KiB | Viewed 1819 times ]

_________________
Senior Manager
Joined: 08 Apr 2012
Posts: 455
Followers: 2

Kudos [?]: 61 [0], given: 58

Re: Two squares, each of side lengths 1 unit and having their [#permalink]

### Show Tags

06 Jul 2014, 09:29
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 --> $$area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}$$.

Hi Bunuel,

I don't understand why we subtract only 4 triangle?
Aren't we supposed to subtract 8 triangles?
Math Expert
Joined: 02 Sep 2009
Posts: 38798
Followers: 7714

Kudos [?]: 105765 [0], given: 11581

Re: Two squares, each of side lengths 1 unit and having their [#permalink]

### Show Tags

06 Jul 2014, 12:18
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 --> $$area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}$$.

Hi Bunuel,

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

Attachment:

Untitled.png [ 7.68 KiB | Viewed 1587 times ]

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

Hope it's clear.
_________________
Senior Manager
Joined: 08 Apr 2012
Posts: 455
Followers: 2

Kudos [?]: 61 [0], given: 58

Re: Two squares, each of side lengths 1 unit and having their [#permalink]

### Show Tags

23 Sep 2014, 08:04
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

Can anyone input if this method is correct?
I am also not sure that the line has to be straight.
If the length of the hypotenuse of the triangle was more that half of the side of the square than I could agree. but it's not...
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15378
Followers: 648

Kudos [?]: 204 [0], given: 0

Re: Two squares, each of side lengths 1 unit and having their [#permalink]

### Show Tags

17 Jan 2016, 00:24
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: Two squares, each of side lengths 1 unit and having their   [#permalink] 17 Jan 2016, 00:24
Similar topics Replies Last post
Similar
Topics:
2 If each square in the preceding figure has a side of length 3, what is 2 09 Apr 2016, 14:42
The length of each side of square A is increased by 100 percent to 3 03 Dec 2015, 12:33
12 In the figure, each side of square ABCD has length 1, the length of li 8 30 Sep 2014, 00:40
17 In the figure, each side of square ABCD has length 1, the 10 08 Sep 2014, 04:57
208 In the figure, each side of square ABCD has length 1, the length of li 30 13 May 2017, 14:13
Display posts from previous: Sort by