The length of a rectangle is more than twice its width, and the area of the rectangle is 20. How do you find the dimension?

1 Answer
May 20, 2016

Answer:

Length is 10
Width is 2

Explanation:

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

Given that #L>2W#

Let #L=2W+x#

#A=LxxW# .......................(1)

But #L=2W+x# so by substituting for #L# in equation (1)

#A=(2W+x)xxW#

#A=2W^2+xW#...............(2)

But area is given as #A=20#

Substitute for #A# in equation (2)

#20=2W^2+xW#

#=>2W^2+xW-20=0#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
as #x# is a variable we can find it by default

Set #color(brown)(" "2W^2+xW-20" ")color(blue)( ->" "(2W -4)(W+5))#

Multiply out the brackets to determine the value of #x#

#color(brown)(2W^2+xW-20)color(blue)( -> 2W^2+10W-4W-20)#

#color(brown)(2W^2+xW-20)color(blue)( -> 2W^2+6W-20)#

#color(green)("Thus "x=6" and "W=2" and "cancel(-5)#

However, #W=-5# is not logical so discard it.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(green)("Test "L=2W+x=2(2)+6=10)#

#color(red)(=> area ->LxxW ->10xx2 =20" as given")#