Posted by Need Help on Wednesday, January 11, 2012 at 10:06pm.
I need help on this math problem.
The width of a certain painting is 5cm less than twice the length. The length of the diagonal distance across the painting is 107cm. Find the length and width. Round your answer to 2 decimal places.
Substitute:
diagonal distance: 107cm
length: L
width: 2L5
Is the substitutes correct?
Should I use the pythagoras method?
Can someone put me in the right track please so I can work out the problem?
please help and thank you

Math 11  Steve, Thursday, January 12, 2012 at 6:08am
yes
yes
L^2 + (2L5)^2 = 107^2
L^2 + 4L^2  20L + 25 = 11449
L = 49.84

Math 11  HelpPlease, Thursday, January 12, 2012 at 9:38am
the next step is:
5L^220L=11424
right?

Math 11  Steve, Thursday, January 12, 2012 at 10:44am
Almost. Set the equation to zero to solve:
5L^220L11424 = 0
then use the quadratic formula to get L.
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