The length of a rectangle is 3ft more than twice its width, and the area of the rectangle is 77ft^2, how do you find the dimensions of the rectangle?

1 Answer
Mar 26, 2016

Answer:

Width = #11/2" ft = 5 foot 6 inches"#

Length = #14" feet"#

Explanation:

Breaking the question down into its component parts:

Let length be #L#
Let width be #w#
Let area be #A#

Length is 3 ft more than: #L=" "?+3#
twice#" "L=2?+3#
its width#" "L=2w+3#

Area #=A = 77="width "xx" Length"#

#A=77= wxx(2w+3)#

#2w^2+3w=77#

#2w^2+3w-77=0# This is a quadratic equation
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Standard form #y=ax^2+bx+c#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#a= 2"; "b=3"; "c=-77#

#x=(-(3)+-sqrt((-3)^2-4(2)(-77)))/(2(2))#

#x=(-(3)+-25)/4=-7 or 11/2#

As we can not have a negative area in this context the answer for #x# is #11/2#

But #color(blue)(x=w" so the width is "11/2)#

#color(blue)(L=2w+3 = 11+3=14)#