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Posted by **Need Help** on Wednesday, January 11, 2012 at 10:06pm.

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: 2L-5

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 6:08amyes

yes

L^2 + (2L-5)^2 = 107^2

L^2 + 4L^2 - 20L + 25 = 11449

L = 49.84

- Math 11 -
**HelpPlease**, Thursday, January 12, 2012 at 9:38amthe next step is:

5L^2-20L=11424

right?

- Math 11 -
**Steve**, Thursday, January 12, 2012 at 10:44amAlmost. Set the equation to zero to solve:

5L^2-20L-11424 = 0

then use the quadratic formula to get L.

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