Question 686696
The length of a rectangle is 1cm more than twice it's width and 2cm less than the length of the diagonal of the rectangle.
 Find the dimensions of the rectangle.
:
Let a = the width
Let b = the length
Let c = the diagonal
;
Write an equation for each statement, get each equation in terms of a:
"The length of a rectangle is 1cm more than twice it's width"
b = 2a + 1
:
"the length is 2cm less than the length of the diagonal of the rectangle."
b = c - 2
replace b with (2a+1)
2a + 1 = c - 2
2a + 1 + 2 = c
c = 2a + 3
:
The diagonal is the hypotenuse (c) of the rectangle where 
a^2 + b^2 = c^2
Replace b with (2a+1) and replace c with (2a+3)
a^2 + (2a+1)^2  = (2a+3)^2
FOIL
a^2 + 4a^2 + 2a + 2a + 1 = 4a^2 + 6a + 6a + 9
Which is
5a^2 + 4a + 1 = 4a^2 + 12a + 9
Combine like terms on the left
5a^2 - 4a^2 + 4a - 12a + 1 - 9 = 0
a^2 - 8a - 8 = 0
Solve this quadratic equation using the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
where x = a; b = -8; c = -8
{{{a = (-(-8) +- sqrt(-8^2-4*1*-8 ))/(2*1) }}}
{{{a = (8 +- sqrt(64+32 ))/2 }}}
{{{a = (8 +- sqrt(96))/2 }}}
two solutions but we only need the positive solution which is approx
a = {{{(8 + 9.8)/2}}}
a = {{{17.8/2}}}
a = 8.9 cm is the width
then
b = 2(8.9) + 1
b = 18.8 cm is the length
:
:
Check this by finding the hypotenuse
c = 2(8.9) + 3
c = 20.8
:
Enter into calc {{{sqrt(8.9^2 + 18.8^2)}}} results: 20.800 confirms our solution
:
 Find the dimensions of the rectangle. 18.8 by 8.9