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In the figure above, two security lights, L1 and L2, are loc [#permalink]
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25 Aug 2013, 16:57
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In the figure above, two security lights, L1 and L2, are located 100 feet apart. Each illuminates an area of radius 100 feet, and both are located 60 feet from a chainlink fence. What is the total length s of fence, in feet, illuminated by the two lights? (A) 260 (B) 240 (C) 220 (D) 200 (E) 180
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Last edited by Bunuel on 25 Aug 2013, 22:24, edited 1 time in total.
Edited the question.



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Re: In the figure above, two security lights, L1 and L2, are loc [#permalink]
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25 Aug 2013, 21:22
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Consider S = AB+BC+CD+DE And, we also know AB=BC=DE = SQRT(100^260^2) = 80 Hence S = 3*(80)+CD Since CD is definitely >0, from the answer choices we could safely choose option .
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Re: In the figure above, two security lights, L1 and L2, are loc [#permalink]
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26 Aug 2013, 00:17
Smallwonder wrote: Consider S = AB+BC+CD+DE And, we also know AB=BC=DE = SQRT(100^260^2) = 80 Hence S = 3*(80)+CD Since CD is definitely >0, from the answer choices we could safely choose option . Very nice solution my friend..... A good post always worth a kudos. Thus you got one for this
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Re: In the figure above, two security lights, L1 and L2, are loc [#permalink]
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11 Sep 2013, 23:04
i don't think this is 700+ sum. It was very easy. Would have been tought if options were like 260,265,270. In that case you have to calculate CD. Can we do it?



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Re: In the figure above, two security lights, L1 and L2, are loc [#permalink]
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14 Sep 2013, 06:54
b2bt wrote: i don't think this is 700+ sum. It was very easy. Would have been tought if options were like 260,265,270. In that case you have to calculate CD. Can we do it? hi b2bt, yes you can ,, it is alread given that the distance between lights is 100m so CD=10080=20
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Re: In the figure above, two security lights, L1 and L2, are loc [#permalink]
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14 Sep 2013, 08:16
total distance= AB+DE+L1L2 AB=DE=80 L1L2 = 100 Hence total = 80*2+100 = 260.
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Re: In the figure above, two security lights, L1 and L2, are loc [#permalink]
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30 Mar 2017, 06:58
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