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worthawholebean
Posts: 2334
Location: New Haven, CT
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Posted: Apr 02, 2009, 9:22 am •
# 1
In rectangle  ,  . Let  be the midpoint of  . Given that line  and line  are perpendicular, find the greatest integer less than  .
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123456789
Posts: 1343
Location: NJ
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Posted: Apr 02, 2009, 12:15 pm •
# 2
Let  be the origin, and let  have coordinates  . Let  have coordinates  and  have coordinates  . Then the equation of line is  and the equation of line is  .
If the lines are perpendicular, then the slopes of the two lines are negative reciprocals of each other. Then  and  . |
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azjps
Posts: 1209
Location: NJ
Blog: View Blog
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Posted: Apr 02, 2009, 2:27 pm •
# 3
Hidden Text Let  ,  be the midpoint of  ; then  is the midline of  and  . Thus  , and by the Pythagorean Theorem on   Hence  , and  .
Last edited by azjps on Apr 02, 2009, 3:04 pm, edited 1 time in total.
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Sephiroth
Posts: 697
Location: Terra Azucar
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Posted: Apr 02, 2009, 2:32 pm •
# 4
outline:
the quadrilateral ABCE has area 3/4 of ABCD,but we can also find the area with d_1d_2/2, since its diagonals are perpendicular. setting the expressions equal we get x=100\sqrt{2}
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math154
Posts: 3683
Location: St. Louis, MO
Blog: View Blog
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Posted: Apr 02, 2009, 3:00 pm •
# 5
SolutionLook at azjps's diagram.  , so  .
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zserf
Posts: 180
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Posted: Apr 02, 2009, 6:40 pm •
# 6
123456789 wrote: If the lines are perpendicular, then the slopes of the two lines are negative reciprocals of each other. Then  I made it all the way to here, then forgot that my variable only represented half of the side. I put 70. |
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dgs01
Posts: 43
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Posted: Apr 03, 2009, 8:05 am •
# 7
i did square root of 200000 instead of 20000 ...fail...
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matkitarini
Posts: 5
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Posted: Oct 06, 2009, 6:15 pm •
# 8
worthawholebean wrote: In rectangle ,  . Let be the midpoint of . Given that line and line are perpendicular, find the greatest integer less than .   Quote:
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rectABCD_AB100.gif [ 2.79 KiB | Viewed 311 times ]
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