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Two sprinklers, Q and R, are located feet apart in a rectangular law

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Two sprinklers, Q and R, are located feet apart in a rectangular law  [#permalink]

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New post Updated on: 17 Jul 2018, 10:20
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Originally posted by franteraoka on 17 Jul 2018, 09:07.
Last edited by chetan2u on 17 Jul 2018, 10:20, edited 1 time in total.
Edited and formatted the question
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Two sprinklers, Q and R, are located feet apart in a rectangular law  [#permalink]

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New post 17 Jul 2018, 10:13
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franteraoka wrote:
Question and answer attached! I am not understanding the answer explanation...thank you!




Just for your query, the sketch that can be made is attached..
The sprinkler of radius 5x/3 is inside the other bigger radius sprinkler..
The distance between the O and G is 5x/6 and the edge A is 5x/3 from O, since it is the radius.
Now 5x/6 + 5x/3 = (5x+10x)/6=5x/2, same as radius of larger circle with center G

Now we are looking for chord AD as it is the fence lying within the area of sprinkler

So just concentrate on the sprinkler R...
When you drop a perpendicular from centre G on fence at C..
GC=3x/2, the distance from fence
AG=5x/2, the radius
So AC in ∆AGC = √[(5x/2)^2-(3x/2)^2]=√(25-9)x^2/4=√(16x^2/4)=2x
Fence in the circle is twice of this =2*2x=4x
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Re: Two sprinklers, Q and R, are located feet apart in a rectangular law  [#permalink]

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New post 17 Jul 2018, 11:18
I will! Thank you chetan2u
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Re: Two sprinklers, Q and R, are located feet apart in a rectangular law &nbs [#permalink] 17 Jul 2018, 11:18
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