Post a New Question

Math 11

posted by on .

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.

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 - ,


    L^2 + (2L-5)^2 = 107^2
    L^2 + 4L^2 - 20L + 25 = 11449
    L = 49.84

  • Math 11 - ,

    the next step is:



  • Math 11 - ,

    Almost. Set the equation to zero to solve:

    5L^2-20L-11424 = 0

    then use the quadratic formula to get L.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question