The triangular ends have a base of 11" and height 12". Area of the base=½×11×12=66 sq in.
There are two triangular ends so that's 132 sq in.
Three rectangular faces. The base is a rectangle 11" long and 11" wide, area=121 sg in.
We need to calculate the width of the other two rectangles, but we don't know what type of triangle makes up the ends, so let's assume the triangles are right-angled. That gives us one rectangle with height 12" and length 11", area=132 sq in. The other rectangle has height=√(122+112)=√(144+121)=√265=16.28" approx. Its length is 11" so area=179.07 sq in approx.
Total surface area=132+132+179.07=443.07 sq in.
If the end triangles are isosceles the two rectangles have the same width=√(122+(11/2)2)=13.20". Each rectangle has an area of 11×13.20=145.20 sq in, so the total surface area is:
132+145.20+145.20=422.41 sq in approx.