Answer:
Part a) The exterior surface area is equal to [tex]160\ ft^{2}[/tex]
Part b) The volume is equal to [tex]240\ ft^{3}[/tex]
Part c) The volume water left in the trough will be [tex]84\ ft^{3}[/tex]
Step-by-step explanation:
Part a) we know that
The exterior surface area is equal to the area of both trapezoids plus the area of both rectangles
so
Find the area of two rectangles
[tex]A=2[12*5]=120\ ft^{2}[/tex]
Find the area of two trapezoids
[tex]A=2[\frac{1}{2}(8+2)h][/tex]
Applying Pythagoras theorem calculate the height h
[tex]h^{2}=5^{2}-3^{2}[/tex]
[tex]h^{2}=16[/tex]
[tex]h=4\ ft[/tex]
substitute the value of h to find the area
[tex]A=2[\frac{1}{2}(8+2)(4)]=40\ ft^{2}[/tex]
The exterior surface area is equal to
[tex]120\ ft^{2}+40\ ft^{2}=160\ ft^{2}[/tex]
Part b) Find the volume
We know that
The volume is equal to
[tex]V=BL[/tex]
where
B is the area of the trapezoidal face
L is the length of the trough
we have
[tex]B=20\ ft^{2}[/tex]
[tex]L=12\ ft[/tex]
substitute
[tex]V=20(12)=240\ ft^{3}[/tex]
Part c)
step 1
Calculate the area of the trapezoid for h=2 ft (the half)
the length of the midsegment of the trapezoid is (8+2)/2=5 ft
[tex]A=\frac{1}{2}(5+2)(2)=7\ ft^{2}[/tex]
step 2
Find the volume
The volume is equal to
[tex]V=BL[/tex]
where
B is the area of the trapezoidal face
L is the length of the trough
we have
[tex]B=7\ ft^{2}[/tex]
[tex]L=12\ ft[/tex]
substitute
[tex]V=7(12)=84\ ft^{3}[/tex]
