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Attachments

543Img1Temp.png [ 5.21 KiB | Viewed 883 times ]

_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Last edited by TGC on 18 Aug 2013, 21:24, edited 2 times in total.

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

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11 Aug 2013, 02:26

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 =\(2\sqrt{14}\)
_________________

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

Show Tags

11 Aug 2013, 03:09

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 !
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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