SOLUTION: The length of a rectangle is 1cm longer than its width. If the diaganol of rectangle is 4cm, what are the dimensions (length and width) of the rectangle? Thank you

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using the pathagorian theorum a^2+b^2=c^2 a=x, b=(x+1), c=4
x^2+(x+1)^2=4^2
x^2+x^2+2x+1=16
2x^2+2x-15=0 using the quadratic equation x=(-b+-sqrt[b^2-4ac])/2a
x=(-2+-sqrt[2^2-4*2*-15])/2*2
x=(-2+-sqrt[4+120])/4
x=(-2+-sqrt124)/4
x=(-2+-11.1355)/4
x=(-2+11.1355)/4
x=9.1355/4
x=2.2839 cm for the width.
length is 2.2839+1=3.2839
proof
2.2839^2+3.2839^2=4^2
5.216+10.784=16
16=16
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