SOLUTION: 6. The length of a rectangle is 1 cm longer than it's width. If the diagonal of the rectangle is 4cm, what are the dimensions (the length and the width) of the rectangle?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 6. The length of a rectangle is 1 cm longer than it's width. If the diagonal of the rectangle is 4cm, what are the dimensions (the length and the width) of the rectangle?      Log On


   

Question 61485: 6. The length of a rectangle is 1 cm longer than it's width. If the diagonal of the rectangle is 4cm, what are the dimensions (the length and the width) of the rectangle?
Answer by tutorcecilia(2152) About Me  (Show Source):
You can put this solution on YOUR website!
a^2+b^2=c^2 [use the pythagorean theorem]
Let:
a=length
b=width
c=diagonal or hypotenuse
.
(w+1)^2+w=4^2 [plug-in the values and solve]
w^2+2w+1+w=16
2w^2+2w+1-16=0 [set the equation equal to zero]
2w^2+2w-15=0 [factor using the quadratic formula]
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-2%29+%2B-+sqrt%28+2%5E2-%284%2A2%2A-15+%29%29%2F%284%29+
x+=+%28-2%29+%2B-+sqrt%28+4-%28-120+%29%29%2F%284%29+
x+=+%28-2%29+%2B-+sqrt%28+4%2B120+%29%29%2F%284%29+
x+=+%28-2%29+%2B-+sqrt%28+124+%29%29%2F%284%29+
x+=+%28-2%29+%2B+sqrt%28+124+%29%29%2F%284%29+=2.2838
.
x+=+%28-2%29+-+sqrt%28+124+%29%29%2F%284%29+=-3.284[eliminate a negative measurement]
.
So, the width = 2.2838
Length = (w+1)=2.2838+1=3.2838
.
check by plugging all of the values back into the pythagorean theorem and solve.
.