Answer:
The total surface area of the prism is [tex]SA=288\ in^{2}[/tex]
Step-by-step explanation:
we know that
The two diagonals of a rhombus are perpendicular and bisect each other
All sides are congruent
The surface area of a prism is equal to
[tex]SA=2B+Ph[/tex]
where
B is the area of the base of prism
P is the perimeter of the base of prism
h is the height of the prism
step 1
Find the length side of the rhombus
Applying Pythagoras Theorem
[tex]c^{2}=a^{2}+b^{2}[/tex]
we have
c is the length side of the rhombus
a and b are the semi diagonals of the rhombus
[tex]a=8/2=4\ in[/tex]
[tex]b=6/2=3\ in[/tex]
substitute
[tex]c^{2}=4^{2}+3^{2}[/tex]
[tex]c^{2}=25[/tex]
[tex]c=5\ in[/tex]
step 2
Find the perimeter of the base P
The perimeter of the rhombus is equal to
[tex]P=4c[/tex]
[tex]P=4(5)=20\ in[/tex]
step 3
Find the area of the base B
The area of the rhombus is
[tex]B=\frac{1}{2}[D1*D2][/tex]
D1 and D2 are the diagonals of the rhombus
substitute
[tex]B=\frac{1}{2}[8*6]=24\ in^{2}[/tex]
step 3
Find the surface area of the prism
[tex]SA=2B+Ph[/tex]
substitute the values
[tex]SA=2(24)+(20)(12)[/tex]
[tex]SA=288\ in^{2}[/tex]
