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# The figure above shows the lengths of the sides of an equiangular poly

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Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132588 [0], given: 12326

The figure above shows the lengths of the sides of an equiangular poly [#permalink]

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20 Sep 2017, 22:57
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(N/A)

Question Stats:

86% (01:36) correct 14% (00:47) wrong based on 21 sessions

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The figure above shows the lengths of the sides of an equiangular polygon. What is the area of the polygon?

(A) 7
(B) 8
(C) 9
(D) 14√2
(E) not determinable

[Reveal] Spoiler:
Attachment:

2017-09-20_1022_001.png [ 6.6 KiB | Viewed 451 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 132588 [0], given: 12326

Senior Manager
Joined: 02 Jul 2017
Posts: 269

Kudos [?]: 79 [2], given: 65

Location: India
Re: The figure above shows the lengths of the sides of an equiangular poly [#permalink]

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20 Sep 2017, 23:55
2
KUDOS
Given figure is Octagon => sum of angle = (8-2)*180
As all angles are equal : (8-2)*180/8 = 135

Find attached figure:
Octagon can be divided into 4 triangle and 5 rectangles.

While drawing rectangles we see angle of triangles formed is 45-45-90
Sides of triangles can be calculated = 1 => 1-1-$$\sqrt{2}$$ triangle

So total area of given figure = 4 triangles + 5 rectangles formed = 4*(1/2 *1*1) + 5* (1*1) = 7

Attachments

Octagon.png [ 11.14 KiB | Viewed 351 times ]

Kudos [?]: 79 [2], given: 65

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7736

Kudos [?]: 17779 [3], given: 235

Location: Pune, India
Re: The figure above shows the lengths of the sides of an equiangular poly [#permalink]

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21 Sep 2017, 23:11
3
KUDOS
Expert's post
Think of the entire figure as a square with corners cut out.
If you draw the corners back, you will see that they are 45-45-90 triangles with sqrt(2) as the hypotenuse. So each side of the triangle will be 1.

Area of figure = Area of square - Area of 4 triangles

Area of figure = 3^2 - 4*(1/2)*1*1 = 7

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Kudos [?]: 17779 [3], given: 235

Re: The figure above shows the lengths of the sides of an equiangular poly   [#permalink] 21 Sep 2017, 23:11
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