![\text{area}=184\text{ square cm}](https://img.qammunity.org/2023/formulas/mathematics/high-school/yu3tx80duvdglreytsg3fp87nnr8bkqi19.png)
Step-by-step explanation
Step 1
to find the total area divide the figure
![\begin{gathered} \text{Area}=4\text{ squares+4 triangles} \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/hd8da9u4glffn850itaefpykrn47tauewv.png)
now, we need to find the area of the triangle
so, we need to find the hypotenuse of the triangle
![\begin{gathered} \text{hyp}^2=3^2+4^2 \\ \text{hyp}^2=9+16 \\ \text{hyp}^2=25 \\ \sqrt[]{\text{hyp}^2}=\sqrt[]{25} \\ \text{hyp}=5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/fbjqg520prf1l2fb3iregu45lmjjas87up.png)
so, the height of the triangle is 5 cm
Step 2
let
side of the square=6
heigth of the triangle=5
base of the triangle=6
![\begin{gathered} \text{Area}=4\text{ squares+4 triangles} \\ \text{Area}=4(side^2)\text{+4 }((b\cdot h)/(3)) \\ \text{replace} \\ \text{Area}=4(6^2)\text{+4 }((6\cdot5)/(3)) \\ \text{area}=4(36)+4(10) \\ \text{area}=144+40 \\ \text{area}=184\text{ square cm} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/inba9kn94sy31pc9w7e1io4jaihfh9song.png)
I hope this helps you