Question 718481
{{{x}}} = length of the sides of the original square (in centimeters).
The lengths of the sides of the rectangle will be
{{{(x-6)}}}{{{cm}}} and {{{(x-2)}}}{{{cm}}}
and the area of the rectangle (in square centimeters) would be
{{{(x-2)(x-6)=5}}}
We solve the quadratic equation:
{{{(x-2)(x-6)=5}}} --> {{{x^2-8x+12=5}}} --> {{{x^2-8x+7=0}}} --> {{{(x-7)(x-1)=5}}}
The solution to the equation above are {{{x=7}}} and {{{x=1}}}.
The solution to the problem is {{{highlight(7cm)}}}
because from a square with sides measuring 1cm
if you decrease by 2cm and 6cm the lengths of the side of such square,
you get negative measures of {{{-1cm}}} and {{{-5cm}}},
which cannot be the lengths of the sides of a rectangle,
even if they multiply to yield a product of {{{5cm^2}}}.
On the other hand, a square with sides measuring 7cm can have
two of its parallel sides shortened (by 6 centimeters) to 1 cm,
and the length of its other two parallel sides shortened by 2 centimeters to 5cm,
yielding a rectangle measuring 1cm by 5cm, with an area of 5 square centimeters