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# If a regular octagon is inscribed in a square

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Senior Manager
Joined: 17 Mar 2014
Posts: 432
If a regular octagon is inscribed in a square  [#permalink]

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28 Jul 2017, 12:31
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Difficulty:

95% (hard)

Question Stats:

16% (03:29) correct 84% (02:41) wrong based on 47 sessions

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If a regular octagon is inscribed in a square , what is the ratio of the square’s area to the regular octagon’s area?

(A) $$(\sqrt{2}+1)$$ to $$2$$
(B) $$(\sqrt{2}+2)$$ to $$2$$
(C) $$(\sqrt{2}-1)$$ to $$2$$
(D) $$(\sqrt{2}-1)$$ to $$1$$
(E) 9 to 7
Math Expert
Joined: 02 Aug 2009
Posts: 7575
If a regular octagon is inscribed in a square  [#permalink]

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29 Jul 2017, 03:03
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ammuseeru wrote:
If a regular octagon is inscribed in a square , what is the ratio of the square’s area to the regular octagon’s area?

(A) $$(\sqrt{2}+1)$$ to $$2$$
(B) $$(\sqrt{2}+2)$$ to $$2$$
(C) $$(\sqrt{2}-1)$$ to $$2$$
(D) $$(\sqrt{2}-1)$$ to $$1$$
(E) 9 to 7

Hi..

look at the att figure..

say each side of Octagon is $$a\sqrt{2}$$, then the sides of the square will be $$2a+a\sqrt{2}$$..
this is because each vertices has a isosceles right angled triangle with hypotenuse $$a\sqrt{2}$$..

area of square = $$(2a+a\sqrt{2})^2=2a^2(1+\sqrt{2})^2$$
area of octagon = $$(2a+a\sqrt{2})^2-4*\frac{1}{2}*a^2=a^2(2+\sqrt{2})^2-2a^2=a^2(4+2+4\sqrt{2}-2)=a^2*4*(1+\sqrt{2})$$..
ratio of areas = $$2a^2(1+\sqrt{2})^2/a^2*4*(1+\sqrt{2})=1+\sqrt{2}/2$$
A
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octagon.png [ 8.31 KiB | Viewed 5470 times ]

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Joined: 07 Dec 2014
Posts: 1179
If a regular octagon is inscribed in a square  [#permalink]

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29 Jul 2017, 21:09
ammuseeru wrote:
If a regular octagon is inscribed in a square , what is the ratio of the square’s area to the regular octagon’s area?

(A) $$(\sqrt{2}+1)$$ to $$2$$
(B) $$(\sqrt{2}+2)$$ to $$2$$
(C) $$(\sqrt{2}-1)$$ to $$2$$
(D) $$(\sqrt{2}-1)$$ to $$1$$
(E) 9 to 7

let area of each small square=1
area of large square=3*3=9
area of octagon=9-(4*1/2)=7
ratio of large square area to octagon area=9 to 7
E
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inscribed octagon.png [ 8.13 KiB | Viewed 5376 times ]

Math Expert
Joined: 02 Aug 2009
Posts: 7575
Re: If a regular octagon is inscribed in a square  [#permalink]

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29 Jul 2017, 21:31
gracie wrote:
ammuseeru wrote:
If a regular octagon is inscribed in a square , what is the ratio of the square’s area to the regular octagon’s area?

(A) $$(\sqrt{2}+1)$$ to $$2$$
(B) $$(\sqrt{2}+2)$$ to $$2$$
(C) $$(\sqrt{2}-1)$$ to $$2$$
(D) $$(\sqrt{2}-1)$$ to $$1$$
(E) 9 to 7

let area of each small square=1
area of large square=3*3=9
area of octagon=9-(4*1/2)=7
ratio of large square area to octagon area=9 to 7
E

Hi...

A REGULAR octagon is an octagon with EQUAL sides..

but in your sketch, that is not the case.
4 sides are equal to SIDE of square and other 4 sides are equal to DIAGONAL of square..

that is why the answer is wrong.
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Joined: 07 Dec 2014
Posts: 1179
Re: If a regular octagon is inscribed in a square  [#permalink]

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29 Jul 2017, 22:50
chetan2u wrote:
gracie wrote:
ammuseeru wrote:
If a regular octagon is inscribed in a square , what is the ratio of the square’s area to the regular octagon’s area?

(A) $$(\sqrt{2}+1)$$ to $$2$$
(B) $$(\sqrt{2}+2)$$ to $$2$$
(C) $$(\sqrt{2}-1)$$ to $$2$$
(D) $$(\sqrt{2}-1)$$ to $$1$$
(E) 9 to 7

let area of each small square=1
area of large square=3*3=9
area of octagon=9-(4*1/2)=7
ratio of large square area to octagon area=9 to 7
E

Hi...

A REGULAR octagon is an octagon with EQUAL sides..

but in your sketch, that is not the case.
4 sides are equal to SIDE of square and other 4 sides are equal to DIAGONAL of square..

that is why the answer is wrong.

hi chetan2u,
thank you for pointing out my error.
gracie
Re: If a regular octagon is inscribed in a square   [#permalink] 29 Jul 2017, 22:50
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# If a regular octagon is inscribed in a square

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