let's recall that the area of a rectangle is simply width * length.
the area of the larger "pink" rectangle is therefore (3x+6)(2x+4), and that includes the area of the smaller rectangle inside it.
now, if we get the area of the smaller rectangle, namely (x-3)(x-1) and subtract it from the larger rectangle, we'll in effect be making a hole on the larger rectangle, and what's leftover is the shaded area.
[tex]\bf \begin{array}{llll} \stackrel{\textit{large rectangle}}{(3x+6)(2x+4)}&~~ - ~~&\stackrel{\textit{small rectangle}}{(x-3)(x-1)}\\\\ 6x^2+12x+12x+24&-&(x^2-x-3x+3)\\\\ 6x^2+24x+24&-&(x^2-4x+3)\\\\ 6x^2+24x+24&-&x^2+4x-3 \end{array} \\\\\\ ~\hfill \stackrel{\textit{shaded area}}{5x^2+28x+21}~\hfill[/tex]
Answer: the area of the shaded region is 5x² + 26x + 21
Step-by-step explanation:
The given figure contains a small rectangle and a large rectangle. The difference between the area of the large rectangle and the small rectangle is the shaded area.
The formula for determining the area of a rectangle is expressed as
Area = length × width.
Area of the large rectangle is
(3x + 6)(2x + 4). Expanding the brackets, it becomes
6x² + 12x + 12x + 24
= 6x² + 24x + 24
Area of the small rectangle is
(x - 3)(x - 1). Expanding the brackets, it becomes
x² + x - 3x + 3
= x² - 2x + 3
Therefore, the area of the shaded region is
6x² + 24x + 24 - (x² - 2x + 3)
= 6x² + 24x + 24 - x² + 2x - 3
= 6x² - x² + 24x + 2x + 24 - 3
= 5x² + 26x + 21
