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 ->  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

Ad: You enter your algebra equation or inequality - Algebrator solves it step-by-step while providing clear explanations. Free on-line demo .
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

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(2144) 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.
.