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In the figure above, ∆ ABC has sides x, y and (x + y)/2. On each side

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In the figure above, ∆ ABC has sides x, y and (x + y)/2. On each side [#permalink]

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New post 10 Nov 2017, 09:09
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In the figure above, ∆ ABC has sides x, y and (x + y)/2. On each side, a square is constructed as shown. What is the sum of the lengths of the sides of the resulting 9-sided figure in terms of x and y?

(A) 6(x + y)
(B) 9(x + y)/2
(C) 4(x + y)
(D) 3(x + y)
(E) 3(x + y)/2


[Reveal] Spoiler:
Attachment:
2017-11-10_1034_002.png
2017-11-10_1034_002.png [ 13.38 KiB | Viewed 273 times ]
[Reveal] Spoiler: OA

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Re: In the figure above, ∆ ABC has sides x, y and (x + y)/2. On each side [#permalink]

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New post 10 Nov 2017, 09:19
Option B

3 sides of the square between BC=3y
3 sides of the square between AB=3x
3 sides of the square between AC=(x+y)/2
Hence 9(x+y)/2


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In the figure above, ∆ ABC has sides x, y and (x + y)/2. On each side [#permalink]

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New post 10 Nov 2017, 10:46
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Since the triangle has the lengths of sides as x,y, and (x+y)/2
and the lengths of the three sides in the triangle are the square's sides,

the perimeter of the figure formed is 4x+4y+2(x+y) - 3/2(x+y)

This can be further simplified as 6(x+y) - \(\frac{3}{2}\)(x+y) = (\(6-\frac{3}{2}\))(x+y) = \(\frac{9(x+y)}{2}\)(Option B)

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Re: In the figure above, ∆ ABC has sides x, y and (x + y)/2. On each side [#permalink]

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New post 10 Nov 2017, 11:17
Agree with the above comments.

I thought it is going to be more complicated, and spent some extra seconds to realize that it is not actually a tough question.

basically, we have 3 squares, with 3 different sizes of sides. new figure's perimeter will be:
3x + 3y + 3(x+y)/2
we can rewrite it as:
(6x+6y+3x+3y)/2
that's (9x+9y)/2 or 9(x+y)/2
the answer is B.

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Re: In the figure above, ∆ ABC has sides x, y and (x + y)/2. On each side   [#permalink] 10 Nov 2017, 11:17
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