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PS-geometry

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PS-geometry [#permalink]

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New post 31 Dec 2008, 12:00
00:00
A
B
C
D
E

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ABCD has an area equal to 28. BC is parallel to AD. BA is perpendicular to AD. IF BC is 6 and AD is 8, then what is CD?

1] 2√ 2
2] 2√ 3
3] 4
4] 2√ 5
5] 6

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Re: PS-geometry [#permalink]

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New post 31 Dec 2008, 12:13
vscid wrote:
ABCD has an area equal to 28. BC is parallel to AD. BA is perpendicular to AD. IF BC is 6 and AD is 8, then what is CD?

1] 2√ 2
2] 2√ 3
3] 4
4] 2√ 5
5] 6


28 = (ab x bc) + (1/2) (ab) (ad-bc)
28 = 6ab + (1/2) (ab) (8-6)
28 = 6ab + ab
28 = 7ab
ab = 4

(cd)^2 = (ab)^2 + (ad-bc)^2
(cd)^2 = (4)^2 + (8-6)^2
(cd)^2 = 16 + 4
(cd)^2 = 20
cd = 2√5

so D. (updated for typo).
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Re: PS-geometry [#permalink]

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New post 01 Jan 2009, 22:36
GMAT TIGER wrote:
vscid wrote:
ABCD has an area equal to 28. BC is parallel to AD. BA is perpendicular to AD. IF BC is 6 and AD is 8, then what is CD?

1] 2√ 2
2] 2√ 3
3] 4
4] 2√ 5
5] 6


28 = (ab x bc) + (1/2) (ab) (ad-bc)
28 = 6ab + (1/2) (ab) (8-6)
28 = 6ab + ab
28 = 7ab
ab = 4

(cd)^2 = (ab)^2 - (ad-bc)^2
(cd)^2 = (4)^2 - (8-6)^2
(cd)^2 = 16 - 4
(cd)^2 = 12
cd = 2√3

------------------------

just one correction here - (cd)^2 = (ab)^2 + (ad-bc)^2
ans - 2√5
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Re: PS-geometry [#permalink]

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New post Updated on: 03 Jan 2009, 11:29
area of the rectangle + area of the triangle = 28
= 7 * area of triangle = 28 => area of triangle = 4

1/2*2* ( height of triangle ) = 4=> height = 4

CD = v ( base^2 + height ^2) = v ( 2^2 + 4 ^2) = 2v5

Originally posted by LAKSH0328 on 01 Jan 2009, 22:51.
Last edited by LAKSH0328 on 03 Jan 2009, 11:29, edited 1 time in total.
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Re: PS-geometry [#permalink]

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New post 02 Jan 2009, 17:27
mar2hathoda wrote:
GMAT TIGER wrote:
vscid wrote:
ABCD has an area equal to 28. BC is parallel to AD. BA is perpendicular to AD. IF BC is 6 and AD is 8, then what is CD?

1] 2√ 2
2] 2√ 3
3] 4
4] 2√ 5
5] 6


28 = (ab x bc) + (1/2) (ab) (ad-bc)
28 = 6ab + (1/2) (ab) (8-6)
28 = 6ab + ab
28 = 7ab
ab = 4

(cd)^2 = (ab)^2 - (ad-bc)^2
(cd)^2 = (4)^2 - (8-6)^2
(cd)^2 = 16 - 4
(cd)^2 = 12
cd = 2√3

------------------------

just one correction here - (cd)^2 = (ab)^2 + (ad-bc)^2
ans - 2√5


Thanks for correction. updated the typo.
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Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


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Re: PS-geometry [#permalink]

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New post 02 Jan 2009, 20:01
given fig is a trapezium.

1/2 (b1+b2) * h = Area

we get h = 4

base of right angled triangle = 2

another side = Sqrt (20)

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: PS-geometry   [#permalink] 02 Jan 2009, 20:01
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