To solve the exercise, it is helpful first to draw the situation that the statement describes:
The total area of the figure will be
![A_{\text{total}}=A_{\text{square}}+A_{\text{triangle}}](https://img.qammunity.org/2023/formulas/mathematics/college/hhzxbmilnj15un97gf87fefgid4xo2tlbe.png)
Then, we can calculate the area of the square using the following formula:
![\begin{gathered} A_{\text{square}}=s\cdot s \\ \text{ Where s is one side of the square} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ze1x5ln7ydx7v0adq6idy6vqoci76naf4o.png)
So, we have:
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/be74b9uyniu1h5cycsw0ciokh8mf2zfovm.png)
Now, we can calculate the area of the triangle using the following formula:
![\begin{gathered} A_{\text{triangle}}=(b\cdot h)/(2) \\ \text{ Where b is the base and} \\ h\text{ is the height of the triangle} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/f85m8lslzrsx934xf0c8hmmjagznrwv24b.png)
So, we have:
![\begin{gathered} b=9in \\ h=7in \\ A_{\text{triangle}}=(b\cdot h)/(2) \\ A_{\text{triangle}}=(9in\cdot7in)/(2) \\ A_{\text{triangle}}=(63in^2)/(2) \\ \boldsymbol{A}_{\boldsymbol{triangle}}\boldsymbol{=31.5in}^{\boldsymbol{2}} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wxp54v2ou7bghz0xnxklzy31e7oc6zs4vb.png)
Finally, we calculate the total area of the figure
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/8bh3pn4h9lwyjbahphrqy5ffch3rogiow1.png)
Therefore, the total area of the figure is 112.5 square inches, and the correct answer is option C.