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In the figure above, the area of ∆ ABC is 32, and the total area of

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In the figure above, the area of ∆ ABC is 32, and the total area of [#permalink]

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New post 06 Nov 2017, 22:56
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In the figure above, the area of ∆ ABC is 32, and the total area of the three smaller triangles is 16. If DEFG is a square, what is the length of segment GF?

(A) 4
(B) 5
(C) √34
(D) 6
(E) 5√2


[Reveal] Spoiler:
Attachment:
2017-11-07_0941_002.png
2017-11-07_0941_002.png [ 6.03 KiB | Viewed 333 times ]
[Reveal] Spoiler: OA

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In the figure above, the area of ∆ ABC is 32, and the total area of [#permalink]

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New post 07 Nov 2017, 00:52
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From the figure, we know that the area of the triangle(ABC) is 32.
It has also been given that the sum of the other smaller triangles is 16.

Therefore the figure DEFG has to be having area of 16(which is a square)
Since the segment(GF) is the side of the square DEFG,
the length must be \(\sqrt{16}\) or \(4\)(Option A)
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Re: In the figure above, the area of ∆ ABC is 32, and the total area of [#permalink]

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New post 10 Nov 2017, 09:40
Area of square = Area of ABC - area of three mall triangle.
GF^2 = 32-16 = 16
GF =4.
Hence Answer = A
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Re: In the figure above, the area of ∆ ABC is 32, and the total area of   [#permalink] 10 Nov 2017, 09:40
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In the figure above, the area of ∆ ABC is 32, and the total area of

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