Coordinate geometry

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dkumar.83: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Thu Jan 14, 2010 5:42 am; Thanked: 2 times

by dkumar.83 » Sun May 09, 2010 10:30 am

Kindly solve the following question:

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Source: Beat The GMAT — Problem Solving | Sun May 09, 2010 10:30 am

nakul_anand: Senior | Next Rank: 100 Posts; Posts: 74; Joined: Fri Oct 09, 2009 4:20 pm; Thanked: 4 times

by nakul_anand » Sun May 09, 2010 10:37 am

Is B the answer?

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iamseer: Master | Next Rank: 500 Posts; Posts: 170; Joined: Wed Jun 10, 2009 5:59 am; Thanked: 13 times

by iamseer » Sun May 09, 2010 10:42 am

IMO B

drop a perpendicular from B on AD.Let that point be X. BX has length of 3. AX =3 b'cos 45-45-90 triangle. Also, AB =3sqrt2
also, BC=8-3=5

therefore perimeter = 2*(5+3sqrt2) = 10 + 6sqrt2

HTH!!

"Choose to chance the rapids and dance the tides"

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dkumar.83: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Thu Jan 14, 2010 5:42 am; Thanked: 2 times

by dkumar.83 » Sun May 09, 2010 10:42 am

I dont know the answer actually. If you have solved it, let me know the calculation.

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harshavardhanc: Legendary Member; Posts: 526; Joined: Sat Feb 21, 2009 11:47 pm; Location: India; Thanked: 68 times; GMAT Score:680

by harshavardhanc » Sun May 09, 2010 10:45 am

nakul_anand wrote:Is B the answer?

IMO B

as <BAD is 45, coordinates of B are (3,3) BA becomes 3 V2 and BC becomes 5. Hence, perimeter is 6 V2 + 10 .

Regards,
Harsha

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dkumar.83: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Thu Jan 14, 2010 5:42 am; Thanked: 2 times

by dkumar.83 » Sun May 09, 2010 10:45 am

Thanks man.

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