To solve the exercise, it is helpful first to draw the situation that the statement describes:
The total area of the figure will be
[tex]A_{\text{total}}=A_{\text{square}}+A_{\text{triangle}}[/tex]
Then, we can calculate the area of the square using the following formula:
[tex]\begin{gathered} A_{\text{square}}=s\cdot s \\ \text{ Where s is one side of the square} \end{gathered}[/tex]
So, we have:
[tex]\begin{gathered} s=9in \\ A_{\text{square}}=s\cdot s \\ A_{\text{square}}=9in\cdot9in \\ \boldsymbol{A}_{\boldsymbol{square}}\boldsymbol{=81in}^{\boldsymbol{2}} \end{gathered}[/tex]
Now, we can calculate the area of the triangle using the following formula:
[tex]\begin{gathered} A_{\text{triangle}}=\frac{b\cdot h}{2} \\ \text{ Where b is the base and} \\ h\text{ is the height of the triangle} \end{gathered}[/tex]
So, we have:
[tex]\begin{gathered} b=9in \\ h=7in \\ A_{\text{triangle}}=\frac{b\cdot h}{2} \\ A_{\text{triangle}}=\frac{9in\cdot7in}{2} \\ A_{\text{triangle}}=\frac{63in^2}{2} \\ \boldsymbol{A}_{\boldsymbol{triangle}}\boldsymbol{=31.5in}^{\boldsymbol{2}} \end{gathered}[/tex]
Finally, we calculate the total area of the figure
[tex]\begin{gathered} A_{\text{total}}=A_{\text{square}}+A_{\text{triangle}} \\ A_{\text{total}}=81in^2+31.5in^2 \\ \boldsymbol{A}_{\boldsymbol{total}}\boldsymbol{=112.5in}^{\boldsymbol{2}} \end{gathered}[/tex]
Therefore, the total area of the figure is 112.5 square inches, and the correct answer is option C.
