Question 89800
The length of a rectangle is 1 cm longer than its width. if the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width) of the rectangle?
:
Let x = the width
Then
(x+1) = the length
:
The diagonal given as 4:
:
From Pythag: a^2 + b^2 = c^2
:
x^2 + (x+1)^2 = 4^2
:
x^2 + x^2 + 2x + 1 = 16
:
2x^2 + 2x + 1 - 16 = 0
:
2x^2 + 2x - 15 = 0
:
Use the quadratic formula to find x: a=2; b=2; c=-15
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
:
{{{x = (-2 +- sqrt( 2^2 - 4 * 2 * -15 ))/(2*2) }}}
:
{{{x = (-2 +- sqrt( 4 - (-120 )))/(4) }}}
:
{{{x = (-2 +- sqrt( 4 + 120 ))/(4) }}}
:
{{{x = (-2 +- sqrt( 124))/(4) }}}
:
{{{x = (-2 + 11.1355)/(4) }}}; we only want the positive solution here
:
{{{x = 9.1355/4 }}}
:
x = 2.28388 is the width; the length = 3.28388
;
:
Check solution on calc: 2.28388^2 + 3.28388^2 = 15.99997571 ~ 16