Step-by-step explanation:
the concept what unit to use for what type of measurement is really simple :
if it is a line (does not have to be straight), then it is a normal measure of distance, like meters.
if it is an area, then we are using squared units, like square meters or meters².
if it is a volume, then we are using cubed units, like cubic meters or meters³.
so, what do you think the width of a rectangle is ?
a volume ? an area ? the whole rectangle is an area. but its sides are basic distances (from one corner of the rectangle to another).
so, we are using meters for that.
like when we are asking how far it is from a point A to a point B. we don't care about the areas of parks, places and so on that are on the way or we have to cross. we don't care about the volume of the buildings along the way or the amount of air above us. we only care about the distance. hence : meters.
we measure distances by finding how many pieces of a normed stick or string fit into the distance.
similarly, we measure areas by checking how many squares of a normed side length fit into the area.
and we measure volumes by checking how many cubes of a normed side length fit into a given space.
so, now back to the rectangle problem :
length = width + 5
and we know the area of a rectangle is length×width.
so, we have
length × width = 26
and we can use the first equation in here
(width + 5) × width = 26
width² + 5×width = 26
or
width² + 5×width - 26 = 0
the solution to such a quadratic equation
ax² + bx + c = 0
is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
width = (-5 ± sqrt(5² - 4×1×-26))/(2×1) =
= (-5 ± sqrt(25 + 104))/2 =
= (-5 ± sqrt(129))/2
width1 = (-5 + sqrt(129))/2 = 3.178908346... m
width2 = (-5 - sqrt(129))/2 = -8.178908346... m
negative solutions for an actual distance don't make any sense, so our solution is
width = 3.178908346... ≈ 3.2 m
length = width + 5 = 8.178908346... ≈ 8.2 m
