SOLUTION: The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width of the rectangle?
x^2 + (x +
Question 90301This question is from textbook Beginning Al
: The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width of the rectangle?
x^2 + (x + 1)^2 = 4^2
x^2+ x^2 + 2x + 1 = 16
2x^2 + 2x + 1 -16 = 0
2x^2 + 2x -15 = 0
x^2 + x -15/2 = 0
I think the above is correct, but now I don't know what to do... This question is from textbook Beginning Al
You can put this solution on YOUR website! YOU WERE ALMOST THERE!!!!!!!!!!!!!!!
Let x=width
Then x+1=length
x^2+(x+1)^2=4^2 get rid of parens
x^2+x^2+2x+1=16
2x^2+2x+1-16=0
2x^2+2x-15=0 quadratic in standard form. We can solve using the quadratic formula
We'll not worry about the negative value for x, since lengths and widths are positive cm----------------------width
cm-----------------length
CK
16.001~~16
