document.write( "Question 187800: please help me with world problems i have no clue where to even begin, please help!! \r
\n" ); document.write( "\n" ); document.write( "An adjustable water sprinkler that sprays water in a circular pattern is placed at the center of a square field whose area is 1250 square feet. What is the shortest radius setting that can be used if all of the grass is to be watered?\r
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Algebra.Com's Answer #140801 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Let s = side length of the square, d = diagonal of the square\r
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\n" ); document.write( "\n" ); document.write( "It turns out that the radius must be equal to half the diagonal \"d\" in order to water all of the grass since the distance to the corner (from the center) is the largest.\r
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\n" ); document.write( "\n" ); document.write( "So let's draw the picture:\r
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\n" ); document.write( "\n" ); document.write( "From the drawing, we can see that the radius of the circle \"r\" is half the diagonal which means \"r=%281%2F2%29%2Ad\"\r
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\n" ); document.write( "\n" ); document.write( "Since the area of the square is 1250 square feet. This means that \"A=1250\"\r
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\n" ); document.write( "\n" ); document.write( "\"A=s%5E2\" Start with the area of a square formula\r
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\n" ); document.write( "\n" ); document.write( "\"1250=s%5E2\" Plug in \"A=1250\"\r
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\n" ); document.write( "\n" ); document.write( "\"s%5E2=1250\" Rearrange the equation.\r
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\n" ); document.write( "\n" ); document.write( "Now, by the Pythagorean Theorem \"a%5E2%2Bb%5E2=c%5E2\", we can see that the sides \"s\" form the legs of a triangle with a hypotenuse \"d\"\r
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\n" ); document.write( "\n" ); document.write( "So \"a=s\", \"b=s\", and \"c=d\"\r
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\n" ); document.write( "\n" ); document.write( "Plug this in to get: \"s%5E2%2Bs%5E2=d%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "\"2s%5E2=d%5E2\" Combine like terms.\r
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\n" ); document.write( "\n" ); document.write( "\"2%281250%29=d%5E2\" Plug in \"s%5E2=1250\"\r
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\n" ); document.write( "\n" ); document.write( "\"2500=d%5E2\" Multiply\r
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\n" ); document.write( "\n" ); document.write( "\"d%5E2=2500\" Rearrange the equation.\r
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\n" ); document.write( "\n" ); document.write( "\"d=sqrt%282500%29\" Take the square root of both sides (note: we're only considering the positive square root).\r
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\n" ); document.write( "\n" ); document.write( "\"d=50\" Take the square root of 2500 to get 50\r
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\n" ); document.write( "\n" ); document.write( "\"r=%281%2F2%29%2Ad\" Go to the formula dealing with the radius\r
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\n" ); document.write( "\n" ); document.write( "\"r=%281%2F2%29%2A%2850%29\" Plug in \"d=50\"\r
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\n" ); document.write( "\n" ); document.write( "\"r=50%2F2\" Multiply\r
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\n" ); document.write( "\n" ); document.write( "\"r=25\" Reduce\r
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\n" ); document.write( "\n" ); document.write( "So the radius must be 25 feet in order for the entire lawn to be watered.
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