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The figure above shows the shape of a flowerbed. If arcs WZ

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The figure above shows the shape of a flowerbed. If arcs WZ [#permalink]

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21 Aug 2013, 17:46
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The figure above shows the shape of a flowerbed. If arcs WZ and XY are semicircles and WXYZ is a square, what is the area of the flowerbed?

(1) The perimeter of square WXYZ is 24.
(2) The diagonal $$WY =6\sqrt{2}$$
[Reveal] Spoiler: OA

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Asif vai.....

Last edited by Bunuel on 22 Aug 2013, 03:09, edited 1 time in total.
Edited the question.

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Re: The figure above shows the shape of a flowerbed. If arcs WZ [#permalink]

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22 Aug 2013, 03:12
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The figure above shows the shape of a flowerbed. If arcs WZ and XY are semicircles and WXYZ is a square, what is the area of the flowerbed?

The area = are of the square + area of the whole circle.

(1) The perimeter of square WXYZ is 24 --> side=24/4=6 --> radius=6/2=3. Sufficient.

(2) The diagonal $$WY =6\sqrt{2}$$[/quote] --> side^2+side^2=WY^2=36*2 --> side=6 --> radius=6/2=3. Sufficient.

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Re: The figure above shows the shape of a flowerbed. If arcs WZ [#permalink]

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22 Aug 2013, 03:58
Asifpirlo wrote:
Attachment:
llllllllllll.png

The figure above shows the shape of a flowerbed. If arcs WZ and XY are semicircles and WXYZ is a square, what is the area of the flowerbed?

(1) The perimeter of square WXYZ is 24.
(2) The diagonal $$WY =6\sqrt{2}$$

In order to find the Area of the flowerbed we need EITHER the radius of the semicircle OR side of the square. Either of these values will give the perimeter.

Statement 1
Provides side of the square, Thus Sufficient

Statement 2
Provides side of the square, Thus Sufficient

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Re: The figure above shows the shape of a flowerbed. If arcs WZ [#permalink]

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16 Oct 2016, 18:14
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Re: The figure above shows the shape of a flowerbed. If arcs WZ   [#permalink] 16 Oct 2016, 18:14
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The figure above shows the shape of a flowerbed. If arcs WZ

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