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805+ (Hard)|   Geometry|      
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If a regular octagon is inscribed in a square 28 Jul 2017, 12:31
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36% (02:51) correct 64% (02:27) wrong based on 126 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
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If a regular octagon is inscribed in a square 29 Jul 2017, 03:03
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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
Show more


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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If a regular octagon is inscribed in a square 29 Jul 2017, 21:09
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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
Show more

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
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If a regular octagon is inscribed in a square 29 Jul 2017, 21:31
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gracie
ammuseeru
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
Show more


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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If a regular octagon is inscribed in a square 29 Jul 2017, 22:50
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gracie
ammuseeru
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.
Show more

hi chetan2u,
thank you for pointing out my error.
gracie
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If a regular octagon is inscribed in a square 04 Nov 2019, 22:50
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ammuseeru
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
Show more

Can you explain - how you got the Side of the square after assuming that the side of the Octagon is a√2?

Also, are all sides of Octagons always equal?
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If a regular octagon is inscribed in a square 22 May 2021, 23:25
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chetan2u
ammuseeru
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
Show more

Can you please explain the highlighted part? I think it is wrong. The area of square should be \(a^2(2+\sqrt{2})^2\)
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If a regular octagon is inscribed in a square 23 May 2021, 00:08
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ammuseeru
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

Can you please explain the highlighted part? I think it is wrong. The area of square should be \(a^2(2+\sqrt{2})^2\)
Show more

Both are the same

\((2a+a\sqrt{2})^2=(a(2+\sqrt{2}))^2\)

Take out a

\(a^2(2+\sqrt{2)}^2\)
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If a regular octagon is inscribed in a square 28 May 2021, 09:10
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chetan2u


Both are the same

\((2a+a\sqrt{2})^2=(a(2+\sqrt{2}))^2\)

Take out a

\(a^2(2+\sqrt{2)}^2\)
Show more

But in the solution, you have written
\((2a+a\sqrt{2})^2=2a^2(1+\sqrt{2})^2\)
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If a regular octagon is inscribed in a square 28 May 2021, 09:41
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chetan2u


Both are the same

\((2a+a\sqrt{2})^2=(a(2+\sqrt{2}))^2\)

Take out a

\(a^2(2+\sqrt{2)}^2\)

But in the solution, you have written
\((2a+a\sqrt{2})^2=2a^2(1+\sqrt{2})^2\)
Show more


\((2a+a\sqrt{2})^2=(a(2+\sqrt{2}))^2\)

Take out a

\(a^2(2+\sqrt{2)}^2\)

Take out \(\sqrt{2}\) to get what you are asking about.

\(a^2(\sqrt{2}*\sqrt{2}+\sqrt{2})^2\)
\(a^2(\sqrt{2})^2(\sqrt{2}+1)^2\)
\(2a^2(\sqrt{2}+1)^2\)
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If a regular octagon is inscribed in a square 18 Jan 2024, 07:29
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ammuseeru
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
Show more

how you considered those triangles formed to be right isosceles triangle? please explain
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If a regular octagon is inscribed in a square 18 Jan 2024, 08:04
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ammuseeru
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

how you considered those triangles formed to be right isosceles triangle? please explain
Show more


Hi

Side of Regular octagon will be in the middle of the side of square as also shown in the sketch in initial post. let the portion be 'a' on either end.
Each side of the square will be a+(side of octagon)+a.

Thus, each vertices of square will have small portions 'a' as the two side and third side will be the hypotenuse, which is nothing but the side of Octagon.

As each vertices of a square makes a right angle, the corner triangle formed will be a:a:a\(\sqrt{2}\), an isosceles right angled triangle.
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If a regular octagon is inscribed in a square 18 Jan 2024, 11:03
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But every octagon side trisects every square sides right? so if one of the side of octagon is a square root 2 so the other two portions as shown in the sketch cant they be a square root 2?
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If a regular octagon is inscribed in a square 18 Jan 2024, 11:28
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But every octagon side trisects every square sides right? so if one of the side of octagon is a square root 2 so the other two portions as shown in the sketch cant they be a square root 2?
Show more


No, it does not. Although a sketch would be easier to explain, but I will try through words.

Let us assume that it does trisect.
so you have each side of square is 3x, and is trisected as x+x+x. Also, every side of octagon too is x.
If this is true, take the triangles on any corner of square. The triangle will have two sides x each on its neighboring sides and the third side will also be x, given by the octagonal side. So, the triangle is x-x-x or an equilateral triangle. This means all angle of the triangle including the corner angle is 60. But the square has to have all angles as 90.
Thus, our assumption that each side of square is trisected is wrong.

Ill add a video or a sketch if this doesnt help.
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If a regular octagon is inscribed in a square 18 Jan 2024, 13:38
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But every octagon side trisects every square sides right? so if one of the side of octagon is a square root 2 so the other two portions as shown in the sketch cant they be a square root 2?
Show more

I hope this helps. Please remember, we have to keep the sides of the octagon the same as I have considered x here, but the sides of squares will be obtained based on the sides of the octagon.
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