Answer:
The total area of the prism is
![SA=((9√(3))/(2)+54)\ in^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/k42nvueotqu3o02pr02ugwq1bmpiy2e86l.png)
Explanation:
we know that
The surface area of the triangular prism of the figure is equal to
![SA=2B+PL](https://img.qammunity.org/2020/formulas/mathematics/middle-school/6gudw6m8t4qhsr0dxamk3klgaijxkrb583.png)
where
B is the area of the triangular face
P is the perimeter of the triangular face
L is the length of the prism
Find the area of the base B
The base is an equilateral triangle
so
Applying the law of sines the area is equal to
![B=(1)/(2)(3)^(2)sin(60\°)](https://img.qammunity.org/2020/formulas/mathematics/high-school/tj85r1a1tl0v6c5a08h26qxtvw7g0ecgma.png)
![B=(9√(3))/(4)\ in^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/t2u9nj8zsfv8vvvc8tnwpz66r7npqupvht.png)
Find the perimeter P of the triangular face
![P=(3+3+3)=9\ in](https://img.qammunity.org/2020/formulas/mathematics/high-school/8wmo6vy6re0b5i4xdi0o3sfkae4mp865l0.png)
we have
![L=6\ in](https://img.qammunity.org/2020/formulas/mathematics/middle-school/b16jdhoexr2sl6o5o0zup4vlb9nyl96skf.png)
substitute
![SA=2((9√(3))/(4))+(9)(6)](https://img.qammunity.org/2020/formulas/mathematics/high-school/sl96oy2kv58xex9v8yciwu9ozrgczqh4ts.png)
![SA=((9√(3))/(2)+54)\ in^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/k42nvueotqu3o02pr02ugwq1bmpiy2e86l.png)