check the picture below.
so the triangular prism is really just 3 rectangles and 2 right-triangles,
now, we know the base of one of the triangles is 2.6, what's its height?
since it's a right-triangle, we can simply use the pythagorean theorem to get "h".
so, we can now, simply get the area of both of the triangles and the three rectangles and sum them up, and that's the area of the triangular prism.
![\bf \stackrel{two~triangles}{2\left[ \cfrac{1}{2}(2.6)(4.5) \right]}~~+~~\stackrel{rectangle}{(2.6\cdot 4.3)}~~+~~\stackrel{rectangle}{(4.3\cdot 3.9)}~~+~~\stackrel{rectangle}{(4.3\cdot 5.2)} \\\\\\ 11.7+11.18+22.36\implies \blacktriangleright 45.24 \blacktriangleleft](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Btwo~triangles%7D%7B2%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%282.6%29%284.5%29%20%5Cright%5D%7D~~%2B~~%5Cstackrel%7Brectangle%7D%7B%282.6%5Ccdot%204.3%29%7D~~%2B~~%5Cstackrel%7Brectangle%7D%7B%284.3%5Ccdot%203.9%29%7D~~%2B~~%5Cstackrel%7Brectangle%7D%7B%284.3%5Ccdot%205.2%29%7D%0A%5C%5C%5C%5C%5C%5C%0A11.7%2B11.18%2B22.36%5Cimplies%20%5Cblacktriangleright%2045.24%20%5Cblacktriangleleft)
I think that (1/4)x+8 would be your algebraic expression.
Note that rotation will not affect the shape and size of an object.
Rotation with respect to a point preserves the corresponding sides and the corresponding angles of the original image.
Hence, the statements
The corresponding angle measurements in each triangle between the pre-image and the image are preserved and
The corresponding lengths, from the point of rotation, between the pre-image and the image are preserved
are true.
