Question

Find the length and width of a rectangle whose area is 56 units squared and whose length is 10 units less than it's width

Answer 1

Answer:

l = 4, w = 14

Step-by-step explanation:

$A_r = 56 = l * w$

l - length;

w - width;

l = w - 10;

We substitute the 'l' using the previous formula =>

$[tex]A_r = (w -10) \cdot w = w^2 - 10w = 56 =>\\w^2 - 10w - 56 = 0$

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)

$w = \frac{10 + \sqrt{100+224} }{2} = \frac{10+18}{2} = \frac{28}{2} = 14$

l = w - 10 = 14 -10 = 4