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?
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Question 65775: 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 Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! 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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