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DS - Geometry

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Joined: 12 Apr 2006
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DS - Geometry [#permalink]

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New post 24 Jun 2009, 03:49
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A
B
C
D
E

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Re: DS - Geometry [#permalink]

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New post 24 Jun 2009, 10:31
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The answer is A.

From the figure:

\(\angle BDC = 2x\)

\(\Rightarrow \angle BDA = 180 - 2x\)

Also

\(\angle BAD = x\)

\(then \angle ABD = 180 - \angle BDA - \angle BAD = 180 - (180-2x) - x = 180-180+2x-x = x\)

\(as \angle BAD = \angle ABD\)

This make \(\triangle ABD\) an isosceles triangle

hence

\(side (AD) = side (BD)\) --- 1

Also

\(\angle BDC = \angle BCD = 2x\)

This make\(\triangle DBC\) an isosceles triangle

Hence

\(side (BD) = side (BC)\) --- 2

from 1 and 2

\(side (AD) = side (BD) = side (BC)\)

Hence we need length of \(side (AD)\) or \(side (BD)\) to get length of \(side (BC)\)

Looking at statement 1.

\(side(AD) = 6\) ---- Sufficient

Looking at Statement 2

\(x = 36\)

this does not tell us anything about the lengths of \(side(AD)\) or \(side(BD)\) ---- Insufficient.

Kudos [?]: 305 [1], given: 3

Re: DS - Geometry   [#permalink] 24 Jun 2009, 10:31
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