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# A regular octagon (a polygon with 8 sides of identical lengt

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A regular octagon (a polygon with 8 sides of identical lengt [#permalink]

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11 Feb 2013, 02:26
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A regular octagon (a polygon with 8 sides of identical length and 8 identical interior angles) is constructed. Next, an equilateral triangle (with sides identical in length to those of the octagon) is attached to each side of the octagon, such that each side of the octagon coincides exactly with the side of the triangle. Finally, each triangle is folded over that coincident side onto the octagon, covering part of the latter’s area. Approximately what proportion of the area of the octagon is left uncovered?

(A) 60%
(B) 50%
(C) 40%
(D) 30%
(E) 20%

[Reveal] Spoiler:
OA will be added later after discussion
[Reveal] Spoiler: OA

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Last edited by emmak on 18 Feb 2013, 00:21, edited 2 times in total.
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Re: A regular octagon (a polygon with 8 sides of identical [#permalink]

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16 Feb 2013, 14:15
1
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This is one of the ways of solving this problem. Since, the octagon is mentioned to be a regular octagon, it can be inscribed inside a square. This can be imagined by knowing that the sides of the octagon are equal and when extended meet each other at same distance at right angles. See figure for example.
Attachment:

octagon_inside_square.png [ 8.19 KiB | Viewed 7420 times ]

Now, consider that the side of the extended triangle is x and since both sides of the triangle are equal, as shown, the triangle is a 45-45-90 triangle. Hence, the hypotenuse is x\sqrt{2}.
Now one side of the octagon will consist of two such sides of triangles and one side of octagon which has length equal to the length of the hypotenuse. Hence,

side of the square= 2x+x\sqrt{2}
= x\sqrt{2}(\sqrt{2} +1)

Now, area of the square=side^2
= (x\sqrt{2}(\sqrt{2} +1))%^2
= 6*x^2+4\sqrt{2}*x^2

Now, the area of the small triangles at the edges of the square when subtracted will give us the area of the octagon.
Area of the triangle= 0.5*x^2
Area of 4 triangles = 2x^2

Hence, the area of the octagon = 4*x^2 + 4\sqrt{2}*x^2 = approx 9.5*x^2 //by estimating \sqrt{2}=1.4

Now, there are 8 equilateral sides with their length equal to the sides of the octagon.
Are of an equilateral triangle=0.5*base(x\sqrt{2})*height (0.5*(\sqrt{3})*x^2)
Hence, the are of the 8 triangles can be solved to be 2\sqrt{3}*x^2
Now, this is approx= 3.5*x^2 //by estimating \sqrt{3}=1.7

Hence, the area left uncovered= 6x^2

finding the percentage reduction

percentage reduction = area left uncovered/total area of the octagon
=6*100/9.5 >60%
Hence, A must be the answer.

Let me know if you need some clarification

emmak wrote:
A regular octagon (a polygon with 8 sides of identical length and 8 identical interior angles) is constructed. Next, an equilateral triangle (with sides identical in length to those of the octagon) is attached to each side of the octagon, such that each side of the octagon coincides exactly with the side of the triangle. Finally, each triangle is folded over that coincident side onto the octagon, covering part of the latter’s area. Approximately what proportion of the area of the octagon is left uncovered?

(A) 60%
(B) 50%
(C) 40%
(D) 30%
(E) 20%

[Reveal] Spoiler:
OA will be added later after discussion
Manager
Joined: 09 Feb 2013
Posts: 121
Re: A regular octagon (a polygon with 8 sides of identical lengt [#permalink]

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18 Feb 2013, 00:24
5
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OE
Let's assume that the octagon's each side length is 1
Figure out this octagon’s area, by splitting up the shape into rectangles and right triangles:

Area = Central square + 4 side rectangles + 4 right triangles (45-45-90)
Area = 1 + 4(√2/2) + 4(½)(√2/2)2 = 2 + 2√2 ≈ 2 + 2(1.4) = 4.8 (we can round because the answer is approximate).
Next, figure out the area of the 8 triangles that will be attached and folded over to cover part of the octagon. (By the way, you can see that the triangles won’t touch each other, because the interior angles of the octagon are 135°, and the two triangles only cover 120° when you fold them in.)

Area of 8 equilateral triangles of side length 1 = 8(s2√3/4) = 2√3 ≈ 2(1.7) = 3.4
The uncovered area equals 4.8 – 3.4 = 1.4, and as a percent of 4.8, that area represents
1.4/4.8 = 14/48 = 7/24 ≈ 7/25 = 28/100 = 28%. The closest answer choice is 30%.

Sorry Kris, you answer is incorrect, but still nice try.
Attachments

1.gif [ 1.99 KiB | Viewed 7403 times ]

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Last edited by emmak on 18 Feb 2013, 03:18, edited 1 time in total.
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Re: A regular octagon (a polygon with 8 sides of identical lengt [#permalink]

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18 Feb 2013, 02:51
emmak wrote:
OE
Figure out this octagon’s area, by splitting up the shape into rectangles and right triangles:

Area = Central square + 4 side rectangles + 4 right triangles (45-45-90)
Area = 1 + 4(√2/2) + 4(½)(√2/2)2 = 2 + 2√2 ≈ 2 + 2(1.4) = 4.8 (we can round because the answer is approximate).
Next, figure out the area of the 8 triangles that will be attached and folded over to cover part of the octagon. (By the way, you can see that the triangles won’t touch each other, because the interior angles of the octagon are 135°, and the two triangles only cover 120° when you fold them in.)
Area of 8 equilateral triangles of side length 1 = 8(s2√3/4) = 2√3 ≈ 2(1.7) = 3.4
The uncovered area equals 4.8 – 3.4 = 1.4, and as a percent of 4.8, that area represents
1.4/4.8 = 14/48 = 7/24 ≈ 7/25 = 28/100 = 28%. The closest answer choice is 30%.

Sorry Kris, you answer is incorrect, but still nice try.

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Re: A regular octagon (a polygon with 8 sides of identical lengt [#permalink]

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18 Feb 2013, 20:45
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Expert's post
emmak wrote:
A regular octagon (a polygon with 8 sides of identical length and 8 identical interior angles) is constructed. Next, an equilateral triangle (with sides identical in length to those of the octagon) is attached to each side of the octagon, such that each side of the octagon coincides exactly with the side of the triangle. Finally, each triangle is folded over that coincident side onto the octagon, covering part of the latter’s area. Approximately what proportion of the area of the octagon is left uncovered?

(A) 60%
(B) 50%
(C) 40%
(D) 30%
(E) 20%

Alternatively, to find the area of the octagon, think of a square.
Area of square = Area of octagon + Area of 4 isosceles right triangles with hypotenuse 1.

Attachment:

Ques4.jpg [ 7.3 KiB | Viewed 7316 times ]

The two legs of the isosceles triangles will be $$1/\sqrt{2}$$ each (Since sides of an isosceles right triangle are in the ratio $$1:1:\sqrt{2}$$
Area of the isosceles right triangle = $$(1/2)*1/\sqrt{2}*1/\sqrt{2} = 1/4$$
Side of square = $$1/\sqrt{2} + 1 + 1/\sqrt{2} = 1 + \sqrt{2}$$
Area of square = ($$1 + \sqrt{2})^2 = 3 + 2\sqrt{2} = 5.8$$

Area of octagon = $$5.8 - 4*(1/4) = 4.8$$

Area of the equilateral triangles = $$(\sqrt{3}/4)*1^2$$ (because area of an equilateral triangle with side s = $$(\sqrt{3}/4)*s^2$$
Area of 8 equilateral triangles = $$8*\sqrt{3}/4) = 2*\sqrt{3} = 3.464$$
3.6 is 75% of 4.8 so 3.464 will be a little less than 75% of 4.8.
So the leftover area will be a little more than 25%.
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Re: A regular octagon (a polygon with 8 sides of identical lengt [#permalink]

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13 Apr 2014, 23:22
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Re: A regular octagon (a polygon with 8 sides of identical lengt [#permalink]

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21 Apr 2014, 18:44
Hi,

how do we derive.. Side of square = 1/\sqrt{2} + 1 + 1/\sqrt{2} = 1 + \sqrt{2} ?
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Re: A regular octagon (a polygon with 8 sides of identical lengt [#permalink]

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18 Feb 2015, 07:16
The best way is to divide the octagon into a rectangle, 2 squares and 4 triangles .

The triangles are right angled , hence if the hypotenuse is 1 , then the side will be 1/root(2) .

The whole area of octagon can be obtained from the same .
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Re: A regular octagon (a polygon with 8 sides of identical lengt [#permalink]

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18 Feb 2015, 08:17
emmak wrote:
A regular octagon (a polygon with 8 sides of identical length and 8 identical interior angles) is constructed. Next, an equilateral triangle (with sides identical in length to those of the octagon) is attached to each side of the octagon, such that each side of the octagon coincides exactly with the side of the triangle. Finally, each triangle is folded over that coincident side onto the octagon, covering part of the latter’s area. Approximately what proportion of the area of the octagon is left uncovered?

(A) 60%
(B) 50%
(C) 40%
(D) 30%
(E) 20%

The explanation is available in the attachment. Please check the same
Attachments

121.jpg [ 237.86 KiB | Viewed 4093 times ]

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A regular octagon (a polygon with 8 sides of identical lengt [#permalink]

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02 May 2016, 23:48
Assume length of side = 1
Area of 8 triangles = 8*sqrt(3)/4 = 3.5
Area of a Octagon = 4.8 * side square = 4.8 * 1 = 4.8

Remaining area = 4.8 - 3.5 = Approx(28 %) = Rounding it to 30 %

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Re: A regular octagon (a polygon with 8 sides of identical lengt [#permalink]

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09 May 2016, 01:29
Kris01 wrote:
This is one of the ways of solving this problem. Since, the octagon is mentioned to be a regular octagon, it can be inscribed inside a square. This can be imagined by knowing that the sides of the octagon are equal and when extended meet each other at same distance at right angles. See figure for example.
Attachment:
octagon_inside_square.png

Now, consider that the side of the extended triangle is x and since both sides of the triangle are equal, as shown, the triangle is a 45-45-90 triangle. Hence, the hypotenuse is x\sqrt{2}.
Now one side of the octagon will consist of two such sides of triangles and one side of octagon which has length equal to the length of the hypotenuse. Hence,

side of the square= 2x+x\sqrt{2}
= x\sqrt{2}(\sqrt{2} +1)

Now, area of the square=side^2
= (x\sqrt{2}(\sqrt{2} +1))%^2
= 6*x^2+4\sqrt{2}*x^2

Now, the area of the small triangles at the edges of the square when subtracted will give us the area of the octagon.
Area of the triangle= 0.5*x^2
Area of 4 triangles = 2x^2

Hence, the area of the octagon = 4*x^2 + 4\sqrt{2}*x^2 = approx 9.5*x^2 //by estimating \sqrt{2}=1.4

Now, there are 8 equilateral sides with their length equal to the sides of the octagon.
Are of an equilateral triangle=0.5*base(x\sqrt{2})*height (0.5*(\sqrt{3})*x^2)
Hence, the are of the 8 triangles can be solved to be 2\sqrt{3}*x^2
Now, this is approx= 3.5*x^2 //by estimating \sqrt{3}=1.7

Hence, the area left uncovered= 6x^2

finding the percentage reduction

percentage reduction = area left uncovered/total area of the octagon
=6*100/9.5 >60%
Hence, A must be the answer.

Let me know if you need some clarification

emmak wrote:
A regular octagon (a polygon with 8 sides of identical length and 8 identical interior angles) is constructed. Next, an equilateral triangle (with sides identical in length to those of the octagon) is attached to each side of the octagon, such that each side of the octagon coincides exactly with the side of the triangle. Finally, each triangle is folded over that coincident side onto the octagon, covering part of the latter’s area. Approximately what proportion of the area of the octagon is left uncovered?

(A) 60%
(B) 50%
(C) 40%
(D) 30%
(E) 20%

[Reveal] Spoiler:
OA will be added later after discussion

I think this Q can be solved by simple guess:

Area uncovered has be less than 50%. More over 20% is too less.
Now the options left: 30% & 40%.
40% in nearer to 50%, possibly will not be the answer.

hence 30 percent.
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Re: A regular octagon (a polygon with 8 sides of identical lengt [#permalink]

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01 Aug 2016, 10:25
emmak wrote:
A regular octagon (a polygon with 8 sides of identical length and 8 identical interior angles) is constructed. Next, an equilateral triangle (with sides identical in length to those of the octagon) is attached to each side of the octagon, such that each side of the octagon coincides exactly with the side of the triangle. Finally, each triangle is folded over that coincident side onto the octagon, covering part of the latter’s area. Approximately what proportion of the area of the octagon is left uncovered?

(A) 60%
(B) 50%
(C) 40%
(D) 30%
(E) 20%

[Reveal] Spoiler:
OA will be added later after discussion

Please find the solution in attachment
Attachments

File comment: www.GMATinsight.com

Answer 1.jpg [ 152.81 KiB | Viewed 1573 times ]

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Re: A regular octagon (a polygon with 8 sides of identical lengt   [#permalink] 01 Aug 2016, 10:25
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