SOLUTION: the length of a rectange is 1 cm longer than its width. if the diagonial of the rectangle is 4 cm, what are the dimensions (the length and the width) of the rectangle?

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Question 89800: the length of a rectange is 1 cm longer than its width. if the diagonial of the rectangle is 4 cm, what are the dimensions (the length and the width) of the rectangle?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
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+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
:
x+=+%28-2+%2B-+sqrt%28+2%5E2+-+4+%2A+2+%2A+-15+%29%29%2F%282%2A2%29+
:
x+=+%28-2+%2B-+sqrt%28+4+-+%28-120+%29%29%29%2F%284%29+
:
x+=+%28-2+%2B-+sqrt%28+4+%2B+120+%29%29%2F%284%29+
:
x+=+%28-2+%2B-+sqrt%28+124%29%29%2F%284%29+
:
x+=+%28-2+%2B+11.1355%29%2F%284%29+; we only want the positive solution here
:
x+=+9.1355%2F4+
:
x = 2.28388 is the width; the length = 3.28388
;
:
Check solution on calc: 2.28388^2 + 3.28388^2 = 15.99997571 ~ 16