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Geometry - center of ceiling to farthest pt in a room

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by goalevan » Sat Jan 10, 2009 5:00 pm
Can anybody give me the answer for this fairly simple problem?

I'm coming up with 16, but I must be missing something because this isn't an available choice. Maybe there was a mistype in the book... it's from Manhattan's GMAT Guide.


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Re: Geometry - center of ceiling to farthest pt in a room

by logitech » Sat Jan 10, 2009 5:08 pm
so the height = x = 12

then you have 12 - 8 rectangle in the bottom the mid point has a triangle with sides 12/2 and 8/2 = 6 and 4

so the diagonal is is SQRT(6^+4^2) = SQRT(52)

the closest distance is than the right angle triangle with sides SQRT(52) and 12

Answer is SQRT [ SQRT(52) ^2 + 12^2] = SQRT196 = 14

I dont think this question is "fairly" simple by the way :) You have to know your space geometry and play with a rectangular box during your spare time :)
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Diagram for understanding

by kanha81 » Sat Jan 10, 2009 7:13 pm
Logitech,

Grateful, if you can please include a rough sketch or diagram for understanding purposes. I would really appreciate that, especially with the geometry problems. I have a hard time grasping the concepts.

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Re: Diagram for understanding

by logitech » Sat Jan 10, 2009 7:52 pm
kanha81 wrote:Logitech,

Grateful, if you can please include a rough sketch or diagram for understanding purposes. I would really appreciate that, especially with the geometry problems. I have a hard time grasping the concepts.

Best regards.
Sure thing. Let me know if you still have doubts.
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by goalevan » Sat Jan 10, 2009 8:44 pm
Having recently started my studying, for some reason I had thought sqrt(196) was equal to 16. I have forgotten more than I expected! Thank you Logitech, really appreciate the help.
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