SOLUTION: Please help me solve this problem: The length of a rectangle is 5 in longer than its width. The diagonal is 5 in shorter than twice the width. Find the length, width, and diagonal

Algebra ->  Pythagorean-theorem -> SOLUTION: Please help me solve this problem: The length of a rectangle is 5 in longer than its width. The diagonal is 5 in shorter than twice the width. Find the length, width, and diagonal      Log On


   

Question 1004630: Please help me solve this problem:
The length of a rectangle is 5 in longer than its width. The diagonal is 5 in shorter than twice the width. Find the length, width, and diagonal measures of the triangle.
My attempt:
Length of rectangle= x + 5
The length of the diagonal = 2x-5
width = X


Answer by CubeyThePenguin(3113) About Me  (Show Source):
You can put this solution on YOUR website!
(width)^2 + (length)^2 = (diagonal)^2

x^2 + (x+5)^2 = (2x - 5)^2
x^2 + x^2 + 10x + 25 = 4x^2 - 20x + 25
2x^2 + 10x + 25 = 4x^2 - 20x + 25
0 = 2x^2 - 30x
0 = 2(x)(x - 15)

x = 15 (distance can't be nonpositive)

width = x = 15
length = x + 5 = 20
diagonal = 2x - 5 = 25