Question 733236
The length of a rectangle is 3 times its width.
 The diagonal is 2 more than the length.
 Find the width of the rectangle.
:
Let L = the Length
Let W = the width
;
Write an equation for each statement:
:
"The length of a rectangle is 3 times its width."
L = 3W
:
"The diagonal is 2 more than the length."
let c = the diagonal
then
c = (L+2)
or, since L = 3W we can write it
c = (3W+2)
The diagonal is the hypotenuse of the two right triangles
formed by the rectangle:
:
Using Pythag a^2 + b^2 = c^2
then we have:
(3W)^2 + W^2 = (3W+2)^2
FOIL
9W^2 + W^2 = 9W^2 + 6W + 6W + 4
10W^2 = 9W^2 + 12W + 4
Combine like terms on the left
10W^2 - 9W^2 - 12W - 4 = 0
W^2 - 12W - 4 = 0
solve this using the quadratic formula
 {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
in this problem: x=w; a=1; b=-12; c=-4
 {{{w = (-(-12) +- sqrt(-12^2-4*1*-4 ))/(2*1) }}}
 {{{w = (12 +- sqrt(144+16 ))/2 }}}
the reasonable solution
 {{{w = (12 + 12.65)/2 }}}
w = {{{24.65/2}}}
w = 12.325 is the width
:
:
:
You can check this; find the hypotenuse with legs of 3(12.325)
and 12.325.
 It should be 2 units more than the Length which is 3 times the width