SOLUTION: Problem #6 Solve each of the following applications give answers to the nearest thoudandth. The length of one leg of a rectangle is 1cm longer than its width. If the diagona

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Question 77328: Problem #6
Solve each of the following applications give answers to the nearest thoudandth.
The length of one leg 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?
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=width
Then x+1=length
When we draw the diagonal of a rectangle we, in effect, draw the hypotenuse of two identical right triangles whose sides are the length and width of the rectangle. We can therefore apply the Pythagorean Theorem:
a%5E2%2Bb%5E2=c%5E2 or
x%5E2%2B%28x%2B1%29%5E2=4%5E2 get rid of parens
x%5E2%2Bx%5E2%2B2x%2B1=16 subtract 16 from both sides
2x%5E2%2B2x%2B1-16=16-16
2x%5E2%2B2x-15=0 quadratic in standard form
We will solve using the quadratic formula:
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%2B120%29%29%2F%284%29+
x+=+%28-2+%2B-+%2811.136%29%29%2F%284%29+
x+=+%28-2+%2B%2811.136%29%29%2F%284%29+
x+=+%289.136%29%2F%284%29+

x=2.284 cm------------------------width
x%2B1=2.284%2B1=3.384cm-----------------------length
We will ignore the negative value for x ( lengths and widths are positive)

CK
%282.284%29%5E2%2B%283.284%29%5E2=4%5E2
5.217%2B10.784=16
16=16

Hope this helps------ptaylor