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The figure above shows the lengths of the sides of an equian

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The figure above shows the lengths of the sides of an equian [#permalink] New post 12 Jun 2018, 09:51
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54% (02:15) correct 45% (03:30) wrong based on 11 sessions
Attachment:
polygon.jpg
polygon.jpg [ 8.04 KiB | Viewed 338 times ]


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) \(\frac{14}{2}\)

(E) It cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: The figure above shows the lengths of the sides of an equian [#permalink] New post 13 Jun 2018, 17:57
Image

We have a total of 5 small squares with side of 1 and 4 small right isosceles triangle with diagonal of 2^(1/2) or side of 1
--> Area = 5*1*1 + 4*1/2*1*1 = 7
--> A
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Re: The figure above shows the lengths of the sides of an equian [#permalink] New post 05 Jul 2018, 05:46
Carcass wrote:
Attachment:
The attachment polygon.jpg is no longer available


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) \(\frac{14}{2}\)

(E) It cannot be determined from the information given.


Here,

Let us assume the polygon to be a square, with 4 △'s as in the diagram attached

So the area of the polygon = Area of the square - 4 * Area of the △

Now let us take any one △,

We can see the hypotenuse = \(\sqrt2\)and it is a 45 - 45 - 90 triangle

Therefore each side is of equal length = 1

Now the area of the △ = \(\frac{1}{2} * 1 * 1 = \frac{1}{2}\)

Area of the square = \(side^2 = 3^2 = 9\)

Therefore the area of the polygon = Area of the square - 4 * Area of the △ = \(9 - (4 * \frac{1}{2}) = 7\)
Attachments

FIG 1.jpg
FIG 1.jpg [ 7 KiB | Viewed 226 times ]


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Re: The figure above shows the lengths of the sides of an equian   [#permalink] 05 Jul 2018, 05:46
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The figure above shows the lengths of the sides of an equian

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