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# In the figure, LM and PN are both perpendicular to MN such that LM=3 c

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DS Forum Moderator
Joined: 19 Oct 2018
Posts: 2053
Location: India
In the figure, LM and PN are both perpendicular to MN such that LM=3 c  [#permalink]

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06 Jul 2020, 19:35
1
5
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35% (medium)

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69% (03:32) correct 31% (03:24) wrong based on 16 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
Senior Manager
Joined: 18 Dec 2017
Posts: 307
Re: In the figure, LM and PN are both perpendicular to MN such that LM=3 c  [#permalink]

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06 Jul 2020, 19:54
3
Am I missing that LM and PN are perpendicular to MN or sum is like this..

Posted from my mobile device
DS Forum Moderator
Joined: 19 Oct 2018
Posts: 2053
Location: India
Re: In the figure, LM and PN are both perpendicular to MN such that LM=3 c  [#permalink]

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06 Jul 2020, 20:14
1
Edited! Thanks
gurmukh wrote:
Am I missing that LM and PN are perpendicular to MN or sum is like this..

Posted from my mobile device
Manager
Joined: 05 Jan 2020
Posts: 142
Re: In the figure, LM and PN are both perpendicular to MN such that LM=3 c  [#permalink]

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06 Jul 2020, 21:28
1
1
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.

Lipun
DS Forum Moderator
Joined: 19 Oct 2018
Posts: 2053
Location: India
Re: In the figure, LM and PN are both perpendicular to MN such that LM=3 c  [#permalink]

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07 Jul 2020, 18:08
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1
Attachment:

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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
Re: In the figure, LM and PN are both perpendicular to MN such that LM=3 c   [#permalink] 07 Jul 2020, 18:08

# In the figure, LM and PN are both perpendicular to MN such that LM=3 c

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