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: 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
yes
yes
L^2 + (2L-5)^2 = 107^2
L^2 + 4L^2 - 20L + 25 = 11449
L = 49.84
the next step is:
5L^2-20L=11424
right?
Almost. Set the equation to zero to solve:
5L^2-20L-11424 = 0
then use the quadratic formula to get L.
Yes, the substitutions you made are correct. Let's use the Pythagorean theorem to solve the problem.
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
In this case, the length and width of the painting form the two sides of a right triangle, with the diagonal distance being the hypotenuse.
Using the given information and the substitutions you made, we can write the equation as:
(L)^2 + (2L - 5)^2 = (107)^2
Now, we can start solving for L. First, expand and simplify the equation:
L^2 + 4L^2 - 20L + 25 = 11449
Combine like terms:
5L^2 - 20L + 25 = 11449
Move all terms to one side of the equation:
5L^2 - 20L + 25 - 11449 = 0
Now, we have a quadratic equation that we can solve. You can either use factoring, completing the square, or the quadratic formula to find the solutions for L.
Once you find the value for L, you can substitute it back into the equation 2L - 5 to find the corresponding width. Round both the length and the width to 2 decimal places as requested.
I hope this helps you get on the right track to solving the problem. Let me know if you have any further questions!