by Guest » Thu Jun 25, 2020 5:02 pm
That looks pretty straight forward. I presume you know that the area of a rectangle with side lengths a and b is ab while the perimeter is 2a+ 2b. So a rectangle with "perimeter of 24 cm and an area less than 30 cm^{2}" satisfies 2a+ 2b= 24 and ab< 30. From 2a+ 2b= 24, a+ b= 12 and b= 12- a. Then ab< 30 becomes a(12- a)= 12a- a^{2}< 30.
To solve that I would be inclined to "complete the square":
$12a- a^2= -(a^2- 12a+ 36- 36)= -(a- 6)^2+ 36< 30$
$6> (a- 6)^2$.
Since a has to be an whole number $0^2= 0$ $1^2= (-1)^2= 1$, $2^2= (-2)^2= 4$ are the only integers whose squares are less than 6 we must have x- 6= 0, x- 6= -1, x- 6= 1, x- 6= -2, or x- 6= 2.