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?

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Answer 1 · 2016-05-20T21:29:03

Length is 10
Width is 2

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.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(green)("Test "L=2W+x=2(2)+6=10)$

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