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705-805 (Hard)|   Geometry|      
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nick1816
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In the figure, LM and PN are both perpendicular to MN such that LM=3 c 06 Jul 2020, 20:35
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E
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75% (hard)

Question Stats:

57% (03:10) correct 43% (02:47) wrong based on 42 sessions

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In the figure, LM and PN are both perpendicular to MN such that LM=3 cm, MN=6 cm and PN= 5cm . Point O lies somewhere on the line MN. Find the least possible value of the sum of the length of 'LO' and 'OP'

A. 6
B. 7
C. 8
D. 9
E. 10
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E
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In the figure, LM and PN are both perpendicular to MN such that LM=3 c 07 Jul 2020, 19:08
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Invert the NP as shown in figure. Shortest distance between 2 points is a straight line. Hence, LO+OP is least when LOP is a straight line.

Now apply pythagoras theorem

\((LP)^2 = (LM+NP)^2 + (MN)^2\)

\(LP^2 = 8^2 +6^2 = 100\)

LP = 10
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In the figure, LM and PN are both perpendicular to MN such that LM=3 c 06 Jul 2020, 20:54
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Am I missing that LM and PN are perpendicular to MN or sum is like this..

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In the figure, LM and PN are both perpendicular to MN such that LM=3 c 06 Jul 2020, 21:14
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Edited! Thanks
gurmukh
Am I missing that LM and PN are perpendicular to MN or sum is like this..

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Lipun
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In the figure, LM and PN are both perpendicular to MN such that LM=3 c 06 Jul 2020, 22:28
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Min length of LO > 3
Min length of PO > 5

=> Min length of LO + PO > 8
=> options A, B, and C can be discarded.

Check for option E.

\(\sqrt{x^2+9} + \sqrt{(6-x)^2+25} = 10\)
=> \(10 - \sqrt{x^2+9} = \sqrt{(6-x)^2+25}\)
=> \(100 + x^2 + 9 -20\sqrt{x^2+9} = (6-x)^2 + 25 = 61 + x^2 - 12x\)
=> \(5\sqrt{x^2+9} = 12 + 3x\)
=> \(25(x^2+9) = 144 + 72x + 9x^2\)
=> \(16x^2 - 72x + 81 = 0\)
=> \((4x-9)^2 = 0\)
=> \(x = 2.25\)

Option D will result in x < 0 (-ve value not possible) and x > 6 (not possible).

Ans: E

nick1816, the above solution takes a good amount of time. I'm sure you can share a much more logical approach.
Awaiting your solution. :)

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In the figure, LM and PN are both perpendicular to MN such that LM=3 c 25 Mar 2024, 04:11
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