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

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

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Updated on: 09 Jul 2013, 09:39
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Difficulty:

95% (hard)

Question Stats:

49% (03:24) correct 51% (03:19) wrong based on 136 sessions

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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 5156 times ]

A. 73/99
B. 86/99
C. 1287/9801
D. 76/99

Originally posted by RaviChandra on 14 Sep 2010, 06:27.
Last edited by Bunuel on 09 Jul 2013, 09:39, edited 1 time in total.
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14 Sep 2010, 06:52
14
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 --> $$area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}$$.

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14 Sep 2010, 07:24
12
2
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
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14 Sep 2010, 06:56
thanku so much for the reply
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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 [ 5.65 KiB | Viewed 4986 times ]
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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.
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14 Sep 2010, 11:48
Nice solution shrouded....It has to be perpendicular with half the length of square.
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14 Sep 2010, 12:34
great solutions, all.. this helps a lot.

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

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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 [ 18.91 KiB | Viewed 3314 times ]

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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 --> $$area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}$$.

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.
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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 --> $$area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}$$.

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 3267 times ]

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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?

Thanks!

Cheers
J
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09 May 2014, 01:04
1
1
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 2909 times ]

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

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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 --> $$area=1-4*(\frac{1}{2}*xy)=1-2xy=1-\frac{13}{99}=\frac{86}{99}$$.

Hi Bunuel,

I don't understand why we subtract only 4 triangle?
Aren't we supposed to subtract 8 triangles?
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Re: Two squares, each of side lengths 1 unit and having their  [#permalink]

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06 Jul 2014, 11:18
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 2677 times ]

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

Hope it's clear.
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Re: Two squares, each of side lengths 1 unit and having their  [#permalink]

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23 Sep 2014, 07:04
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...
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Re: Two squares, each of side lengths 1 unit and having their  [#permalink]

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