Chapter 8.4, Problem 48PPS

To determine

To find: The area of the shaded region

Answer to Problem 48PPS

  $x^2+11x-6$

Explanation of Solution

Given:

  

Calculation:

The area of the square is given by

  $\begin{array}{l}{\left(x-3\right)}^2\\ =x^2-6x+9\end{array}$

The base of the triangle is given by

  $\begin{array}{l}x-3+x+5\\ =2x+2\end{array}$

And the height of the triangle is given by

  $\begin{array}{l}x+6+x-3\\ =2x+3\end{array}$

And the area of the triangle is given by

  $\begin{array}{l}\frac{1}{2}\left(2x+2\right)\left(2x+3\right)\\ =\frac{1}{2}\left(4x^2+6x+4x+6\right)\end{array}$

  $\begin{array}{l}=\frac{1}{2}\left(4x^2+10x+6\right)\\ =2x^2+5x+3\end{array}$

Hence, the area of the shaded region is given by

  $\begin{array}{l}2x^2+5x+3-\left(x^2-6x+9\right)\\ =2x^2+5x+3-x^2+6x-9\\ =x^2+11x-6\end{array}$