How do you simplify #sqrt(11/132)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Elvan Burcu K. May 29, 2015 #sqrt(11/132) = sqrt(11/(11*12)) =sqrt (cancel 11/ (cancel 11* 12)) = sqrt (1/12) = 1/ sqrt12 = 1/ ( 2sqrt3) = sqrt3/( 2*sqrt3*sqrt3) = sqrt3/ (2*3) = sqrt3/6 # Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 2150 views around the world You can reuse this answer Creative Commons License