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In the triangle above, lengths AC and BD are both equal to the same in

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In the triangle above, lengths AC and BD are both equal to the same in  [#permalink]

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New post 24 Aug 2017, 23:35
2
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

73% (01:51) correct 27% (01:37) wrong based on 57 sessions

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Re: In the triangle above, lengths AC and BD are both equal to the same in  [#permalink]

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New post 24 Aug 2017, 23:43
1
Image

If the integer the base and height are equal to is x,
the area for the triangle is \(\frac{1}{2}*x^2 = \frac{x^2}{2}\)

Of the answer options available (after converting the mixed fractions to proper fraction)
(A) \(\frac{1}{2}\)
(B) \(4\frac{1}{2}\) = \(\frac{9}{2}\)
(C) \(6 = \frac{12}{2}\)
(D) \(8 = \frac{16}{2}\)
(E) \(12\frac{1}{2} = \frac{25}{2}\)

The numerator is not a square only in Option C and is our correct answer!
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Re: In the triangle above, lengths AC and BD are both equal to the same in  [#permalink]

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New post 25 Aug 2017, 00:03
Bunuel wrote:
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In the triangle above, lengths AC and BD are both equal to the same integer. Which of the following could NOT be the area of ∆ ABC?

(A) 1/2
(B) 4 1/2
(C) 6
(D) 8
(E) 12 1/2

Attachment:
2017-08-24_1251_002.png


1/2 is 1/2 * 1^2
9/2 is 1/2 * 3^2
12/2 is 1/2 * not an integer
16/2 is 1/2 * 4^2
25/2 is 1/2 * 5^2
C
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Re: In the triangle above, lengths AC and BD are both equal to the same in  [#permalink]

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New post 27 Aug 2018, 02:34
Let AC and BD be x, then the area \(A = x^2/2\)

A*2 = x^2 \(\)
This means that \(A*2\) has to be a perfect square. Of the answer choices only \(6*2 = 12\) is not a perfect square.

Thus, the answer is C
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Re: In the triangle above, lengths AC and BD are both equal to the same in   [#permalink] 27 Aug 2018, 02:34
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