SOLUTION: Geometry. 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?

Algebra ->  Radicals -> SOLUTION: Geometry. 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?       Log On


   

Question 81697: Geometry. 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?
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Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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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 length
Then (x-1) = the width
:
Using pythag.
x^2 + (x-1)^2 = 4^2
:
x^2 + (x^2 - 2x + 1) = 16: FOILed (x-1)(x-1)
:
2x^2 - 2x + 1 - 16 = 0; subtracted 16 from both sides
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2x^2 - 2x - 15 = 0; our old friend, the quadratic equation
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Use the quadratic equation: a=2; b=-2; c=-15
x+=+%28-%28-2%29+%2B-+sqrt%28+-2%5E2+-+4+%2A+2+%2A+-15+%29%29%2F%282%2A2%29+
:
x+=+%28%2B2+%2B-+sqrt%28+4+%2B+120+%29%29%2F%284%29+
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Do the math here; you should get about: x = -2.28 and x = +3.28
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Use the positive solution for x: 3.28 is the length
:
Check using pythag and a good calc:
3.28^2 + 2.28^2 = 15.9 ~ 16 which is 4^2