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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]  12 Jun 2018, 09:51
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Question Stats:

58% (02:18) correct 41% (03:30) wrong based on 12 sessions
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#GREpracticequestion The figure above shows the lengths.jpg [ 8.04 KiB | Viewed 197 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]  13 Jun 2018, 17:57

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]  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 [ 7 KiB | Viewed 464 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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