document.write( "Question 61485: 6. The length of a rectangle is 1 cm longer than it's width. If the diagonal of the rectangle is 4cm, what are the dimensions (the length and the width) of the rectangle? \n" ); document.write( "

Algebra.Com's Answer #42338 by tutorcecilia(2152) $\"\"$ $\"About$

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a^2+b^2=c^2 [use the pythagorean theorem]
\n" ); document.write( "Let:
\n" ); document.write( "a=length
\n" ); document.write( "b=width
\n" ); document.write( "c=diagonal or hypotenuse
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\n" ); document.write( "(w+1)^2+w=4^2 [plug-in the values and solve]
\n" ); document.write( "w^2+2w+1+w=16
\n" ); document.write( "2w^2+2w+1-16=0 [set the equation equal to zero]
\n" ); document.write( "2w^2+2w-15=0 [factor using the quadratic formula]
\n" ); document.write( " $\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"$
\n" ); document.write( " $\"x+=+%28-2%29+%2B-+sqrt%28+2%5E2-%284%2A2%2A-15+%29%29%2F%284%29+\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"x+=+%28-2%29+%2B-+sqrt%28+4-%28-120+%29%29%2F%284%29+\"$
\n" ); document.write( " $\"x+=+%28-2%29+%2B-+sqrt%28+4%2B120+%29%29%2F%284%29+\"$
\n" ); document.write( " $\"x+=+%28-2%29+%2B-+sqrt%28+124+%29%29%2F%284%29+\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"x+=+%28-2%29+%2B+sqrt%28+124+%29%29%2F%284%29+\"$ =2.2838\r
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\n" ); document.write( " $\"x+=+%28-2%29+-+sqrt%28+124+%29%29%2F%284%29+\"$ =-3.284[eliminate a negative measurement]
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\n" ); document.write( "So, the width = 2.2838
\n" ); document.write( "Length = (w+1)=2.2838+1=3.2838\r
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\n" ); document.write( "check by plugging all of the values back into the pythagorean theorem and solve.
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