document.write( "Question 159747: 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 the field is to be completely enclosed within the circle? \n" ); document.write( "

Algebra.Com's Answer #117785 by jojo14344(1513) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "Remember: $\"A%5Bsq%5D=s%5E2\"$
\n" ); document.write( " $\"1250=s%5E2\"$ ------> $\"s=sqrt%281250%29\"$
\n" ); document.write( " $\"s=35.355339ft\"$
\n" ); document.write( "Now, the shortest radius will point to one corner of the square, so the field will be enclosed to the circle, oks.(draw it to better visualize it I guess)
\n" ); document.write( "That radius forms a right triangle where that computed side $\"s\"$ is cut into half to complete the triangle. Thus the radius now will be the hypotenuse.
\n" ); document.write( ".
\n" ); document.write( "By Trigo function, in reference to the 45 degree formed now in the corner of the square field: $\"cos%2845%29=adj%2Fhyp\"$
\n" ); document.write( "where $\"hyp=radius\"$ , and
\n" ); document.write( " $\"adj=35.355339%2F2=17.67767ft\"$
\n" ); document.write( "therefore, $\"cos%2845%29=17.67767%2Fr\"$
\n" ); document.write( " $\"r=17.67767%2Fcos%2845%29=25ft\"$ , ANSWER
\n" ); document.write( "thank you,
\n" ); document.write( "Jojo \n" ); document.write( "

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