How do you find the dimensions of a rectangle if the area is #63cm^2# and the perimeter is #32cm# where #x# represents the width of the rectangle?

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Answer 1 · 2017-04-03T08:06:31

Width = #7cm#; Length = #9cm#

let #color(red)x# be the width and #color(red)y# be the length of the rectangle.

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Given
#area# #= x*y = 63# -----------#(1)#

#perimeter# #=x+y+x+y = 2x+2y = 32# ----------#(2)#

from #(1), x=63/y#

substituting this in #(2),#

#63/y+y=16#

#=>y^2-16y+63=0#

#=> y^2-7*y-9*y+63=0#

#=>y(y-7)-9(y-7)=0#

#=>(y-9)(y-7)=0#

#=># either #y-9=0# i.e. #y=9#

or #y-7=0# i.e. #y=7#

therefore #x=63/9=7# or #x=63/7=9#

Hence, width = #7cm#; length = #9cm#