For this case, we have that the area of the surface of the figure shown is given by:
[tex]A_{1}[/tex]: The area of the upper rectangle (inclined)
[tex]A_{2}[/tex]: The area of the two triangles.
[tex]A_{3}[/tex]: The area of the base of the prism
[tex]A_ {4}[/tex]: The area of the rear rectangle
We apply Pythagoras to know the length of the inclined side of the rectangle:
[tex]h = \sqrt {15 ^ 2 + 8 ^ 2}\\h = \sqrt {289}\\h = 17 \ units[/tex]
Thus, the [tex]A_{1}[/tex] is given by:
[tex]A_ {1} = 10 * 17\\A_ {1} = 170 \ units ^ 2[/tex]
We look for the area given by the triangles, knowing that the area of a triangle is [tex]\frac {1} {2} b * h[/tex], then:
[tex]A_ {2} = \frac {1} {2} b * h + \frac {1} {2} b * h\\A_ {2} = b * h\\A _{2} = 15 * 8\\A_{2} = 120 \ units ^ 2[/tex]
We look for the area of the base of the prism:
[tex]A_ {3} = 10 * 15\\A_ {3} = 150 \ units ^ 2[/tex]
We look for the area of the posterior rectangle:
[tex]A_ {4} = 8 * 10\\\A_ {4} = 80 \ units ^ 2[/tex]
Adding we have:
[tex]A_ {t} = (170 + 120 + 150 + 80) \ units ^ 2\\A_ {t} = 520 \ units ^ 2[/tex]
Answer:
[tex]520 \ units ^ 2[/tex]
