Answer:
l = 4, w = 14
Step-by-step explanation:
[tex]A_r = 56 = l * w[/tex]
l - length;
w - width;
l = w - 10;
We substitute the 'l' using the previous formula =>
[tex][tex]A_r = (w -10) \cdot w = w^2 - 10w = 56 =>\\w^2 - 10w - 56 = 0\\[/tex]
By the quadratic formula we solve for 'w':(we will use the positive value, because we're talking about lengths of planes in a Euclidean space)
[tex]w = \frac{10 + \sqrt{100+224} }{2} = \frac{10+18}{2} = \frac{28}{2} = 14[/tex]
l = w - 10 = 14 -10 = 4
