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In the ||gm, it is given AB=8 and BO=6 find BC

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In the ||gm, it is given AB=8 and BO=6 find BC [#permalink]

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11 Aug 2013, 03:12
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45% (medium)

Question Stats:

82% (02:44) correct 18% (04:00) wrong based on 17 sessions

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In the parallelogram as shown in the figure, it is given AB=8 and BO=6 find BC

See attached all values are in cms:

Please note that 45 degrees as mentioned in attached is not mentioned in the question as I took the image from solutions.

(a). 12cm
(b). 2 sqrt(14) cm
(c). 2 sqrt(7) cm
(d). 10 cm
(e). 2 sqrt(11) cm

Source: Some document I had from internet
[Reveal] Spoiler: OA

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543Img1Temp.png [ 5.21 KiB | Viewed 928 times ]

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Last edited by TGC on 18 Aug 2013, 22:24, edited 2 times in total.
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Kudos [?]: 1194 [0], given: 136

Re: In the ||gm, it is given AB=8 and BO=6 find BC [#permalink]

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11 Aug 2013, 03:26
targetgmatchotu wrote:
In the ||gm, it is given AB=8 and BO=6 find BC

See attached all values are in cms:

(a). 12cm
(b). 2 sqrt(14) cm
(c). 2 sqrt(7) cm
(d). 10 cm
(e). 2 sqrt(11) cm

Source: Some document I had from internet

$$\angle AOD = 90$$ degree. Thus, $$\angle OAD = 45$$degrees$$\to$$ OD = AO(sides opposite equal angles are equal).

Also, $$AO^2 = 8^2-6^2 = 28$$. Now, in triangle AOD, $$AD^2 = AO^2+OD^2 = 2*AO^2 \to AD = 2\sqrt{14}.$$
As it is mentioned that ABCD is a parallelogram, AD = BC =$$2\sqrt{14}$$
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Director
Joined: 03 Aug 2012
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Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Followers: 24

Kudos [?]: 768 [0], given: 322

Re: In the ||gm, it is given AB=8 and BO=6 find BC [#permalink]

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11 Aug 2013, 04:09
mau5 wrote:
targetgmatchotu wrote:
In the ||gm, it is given AB=8 and BO=6 find BC

See attached all values are in cms:

(a). 12cm
(b). 2 sqrt(14) cm
(c). 2 sqrt(7) cm
(d). 10 cm
(e). 2 sqrt(11) cm

Source: Some document I had from internet

$$\angle AOD = 90$$ degree. Thus, $$\angle OAD = 45$$degrees$$\to$$ OD = AO(sides opposite equal angles are equal).

Also, $$AO^2 = 8^2-6^2 = 28$$. Now, in triangle AOD, $$AD^2 = AO^2+OD^2 = 2*AO^2 \to AD = 2\sqrt{14}.$$
As it is mentioned that ABCD is a parallelogram, AD = BC =$$2\sqrt{14}$$

Sorry for that the angle given in the diagram as 45 degrees was not given in the question stem.I think I took the image from the solution and apologies for that.

You may now proceed again

Rgds,
TGC !
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Rgds,
TGC!
_____________________________________________________________________
I Assisted You => KUDOS Please
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Kudos [?]: 147 [0], given: 242

Re: In the ||gm, it is given AB=8 and BO=6 find BC [#permalink]

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13 Sep 2013, 09:35
i still can't understand how they got 45 degrees..but after that its a simple answer...
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Re: In the ||gm, it is given AB=8 and BO=6 find BC [#permalink]

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13 Sep 2013, 15:04
Agreed, I don't see how you are getting 45 degrees from what has been discussed so far.
Re: In the ||gm, it is given AB=8 and BO=6 find BC   [#permalink] 13 Sep 2013, 15:04
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