SOLUTION: Can someone help me out with this problem? The length of a rectangle is 1cm longer than its width. If the diagonal of the rectangle is 4cm, what are the dimensions (the length a

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Question 50828This question is from textbook Beginning Algebra
: Can someone help me out with this problem?
The length of a rectangle is 1cm longer than its width. If the diagonal of the rectangle is 4cm, what are the dimensions (the length and the width)of the rectangle? Round the the nearest thousandth.
This question is from textbook Beginning Algebra

Answer by Cintchr(481) About Me  (Show Source):
You can put this solution on YOUR website!
A triange is 1/2 of a rectangle or square, and in this situation, is best defined by its altitude(width = x), base(length = x+1) and hypotenuse(diagonal = 4). Using the pythagorean theorm...
+a%5E2+%2B+b%5E2+=+c%5E2+
+x%5E2+%2B+%28x%2B1%29%5E2+=+4%5E2+
F: x * x = x^2
O: x * 1 = 1x
I: 1 * x = 1x
L: 1 * 1 = 1
ADD
+%28x%2B1%29%5E2+=+x%5E2+%2B+2x+%2B+1+
Substitute
+x%5E2+%2B+x%5E2+%2B+2x+%2B+1+=+4%5E2+
Combine like terms
+2x%5E2+%2B+2x+%2B+1+=+4%5E2+
Find 4^2
+2x%5E2+%2B+2x+%2B+1+=+16+
Subtract 16 from both sides
+2x%5E2+%2B+2x+-15+=+0+
Can not be factored .... Use the quadratic formula
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%2A2%2A%28-15%29+%29%29%2F%282%2A2%29+
x+=+%28-2+%2B-+sqrt%28+4-4%2A2%2A%28-15%29+%29%29%2F%282%2A2%29+
x+=+%28-2+%2B-+sqrt%28+4-8%2A%28-15%29+%29%29%2F%282%2A2%29+
x+=+%28-2+%2B-+sqrt%28+4%2B120+%29%29%2F%282%2A2%29+
x+=+%28-2+%2B-+sqrt%28+124+%29%29%2F%282%2A2%29+
x+=+%28-2+%2B-+sqrt%28+124+%29%29%2F%284%29+
x+=+%28-2+%2B-+sqrt%28+124+%29%29%2F%284%29+
x+=+%28-2+%2B-+11.14%29%2F%284%29+
From here you can only use the positive root because we are talking about a length.
x+=+%28-2+%2B+11.14%29%2F%284%29+
x+=+%289.14%29%2F%284%29+
x+=+%289.14%29%2F%284%29+
x=2.29
w = 2.29
l = 3.29
Remember, these are estimates. We rounded the square roots.