The area of a rectangle is 27 square meters. If the length is 6 meters less than 3 times the width, then find the dimensions of the rectangle. Round off your answers to the nearest hundredth.?
↳Redirected from
"The area of a rectangle is 27 square meters. If the length is 6 meters less than 3 times the width, then find the dimensions of the rectangle. Round off your answers to the nearest hundredth.?"
#\color{blue}{6.487 m, 4.162m}#
Let #L# & #B# be the length & width of rectangle then as per given conditions,
#L=3B-6 \ .........(1)#
#LB=27 \ .........(2)#
substituting the value of L from (1) into (2) as follows
#(3B-6)B=27#
#B^2-2B-9=0#
#B=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-9)}}{2(1)}#
#=1\pm\sqrt{10}#
since, #B>0#, hence we get
#B=1+\sqrt{10}# &
#L=3(1+\sqrt{10})-6#
#L=3(\sqrt{10}-1)#
Hence, length & width of given rectangle are
#L=3(\sqrt{10}-1)\approx 6.486832980505138\ m#
#B=\sqrt{10}+1\approx 4.16227766016838\ m#
length = m = 6.49
width = n = 4.16
Assume that length = #m# and width = #n#.
The rectangle's area will hence be #mn#.
The first statement states "The area of a rectangle is 27 square meters.
Hence #mn=27#.
The second statement states "If the length is 6 meters less than 3 times the width..."
Therefore #m=3n-6#
Now you can create a system of equations:
#mn=27#
#m=3n-6#
Replace #m# in the first equation with #3n-6#:
#(3n-6)*n=27#
Expand the bracket:
#3n^2-6*n=27#
Make a quadratic equation:
#3n^2-6*n-27=0#
Simplify by dividing everything by 3:
#n^2-2*n-9=0#
Use #(-b+-sqrt(b^2-4ac))/(2a)#, where #a# is 1, #b# is -2, and #c# is -9:
=#(2+-sqrt(4+36))/(2)#
=#1+-sqrt10#
Since dimensions must be positive, #n# will be #1+sqrt10#, which to the nearest hundredths is 4.16.
Use #mn=27# to find #m#:
#m(1+sqrt10)=27#
#m=27/(1+sqrt10)#
#m=6.49#
