Question 341864
The width of a rectangle is 8 less than twice its length.
 If the area of the rectangle is 200 cm^2, what is the length of the diagonal?
:
Let x = the length
then
(2x-8) = the width
and
x(2x-8) = 200; the area
A quadratic equation
2x^2 - 8x - 200 = 0
Simplify, divide by 2
x^2 - 4x - 100 = 0
Find x using the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation: a=1; b=-4; c=-100
{{{x = (-(-4) +- sqrt(-4^2-4*1*-100 ))/(2*1) }}}
:
{{{x = (4 +- sqrt(16 + 400 ))/2 }}}
:
{{{x = (4 +- sqrt(416 ))/2 }}}
The positive solution is what we want here
{{{x = (4 + 20.4)/2 }}}
x = {{{24.4/2}}}
x = 12.2 is the length
then
2(12.2) - 8 = 16.4 is the width
:
Check solution by finding the area: 12.2 * 16.4 = 200.1 ~ 200, close enough
:
Find the diagonal (d)
d = {{{sqrt(12.2^2 + 16.4^2)}}}
c = 20.44 cm is the diagonal