Question

The area of a rectangle is 33 m2 ,
And the length of the rectangle is 5 m less than the width .find the dimensions of the rectangle.

Answer 1

For this case we have that by definition, the area of ​​a rectangle is given by:

$A = w * l$

Where:

w: Is the width of the rectangle

l: is the length of the rectangle

According to the data of the statement we have:

$A = 33 \ m ^ 2\\l = w-5$

Substituting:

$w (w-5) = 33\\w ^ 2-5w-33 = 0\\\\So:\\w = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}\\a = 1\\b = -5\\c = -33$

Substituting the values:

$w = \frac {- (- 5) \pm \sqrt {(- 5) ^ 2-4 (1) (- 33)}} {2 (1)}\\\\w = \frac {5 \pm \sqrt {25 + 132}} {2}}\\\\w = \frac {5 \pm \sqrt {157}} {2}}$

Thus, we have two roots:

w_ {1} = \ frac {5+ \ sqrt {157} {2}} = 8.77

w_ {2} = \ frac {5+ \ sqrt {157} {2}} = - 3.77

We choose the positive value. So the width of the rectangle is approximately 8.77 meters and the length is 3.77 meters

Answer:

$l = 3.77 \ m\\w = 8.77 \ m$