Pit

Pit has shape of a truncated pyramid with rectangular bases and is 0.8 m deep. The length and width of the pit is the top 3 × 1.5 m bottom 1 m × 0.5 m. To paint one square meter of pit we use 0.6 l of green colour. How many liters of paint is needed when we paint only the sides and bottom of pit?

Result

V =  4.1 l

Solution:

x0=3 m y0=1.5 m  x=1 m y=0.5 m  h=0.8 m  h1=h2+((y0−y)/2)2=0.82+((1.5−0.5)/2)2≐0.9434 m h2=h2+((x0−x)/2)2=0.82+((3−1)/2)2≐1.2806 m  S1=x⋅ y=1⋅ 0.5=12=0.5 m2  S2=(x+x0)⋅ h12=(1+3)⋅ 0.94342≐1.8868 m2 S3=(y+y0)⋅ h22=(0.5+1.5)⋅ 1.28062≐1.2806 m2  S=S1+2⋅ (S2+S3)=0.5+2⋅ (1.8868+1.2806)≐6.8348 m2  V=0.6⋅ S=0.6⋅ 6.8348≐4.1009≐4.1 lx_{ 0 }=3 \ \text{m} \ \\ y_{ 0 }=1.5 \ \text{m} \ \\ \ \\ x=1 \ \text{m} \ \\ y=0.5 \ \text{m} \ \\ \ \\ h=0.8 \ \text{m} \ \\ \ \\ h_{ 1 }=\sqrt{ h^2 + ((y_{ 0 }-y)/2)^2 }=\sqrt{ 0.8^2 + ((1.5-0.5)/2)^2 } \doteq 0.9434 \ \text{m} \ \\ h_{ 2 }=\sqrt{ h^2 + ((x_{ 0 }-x)/2)^2 }=\sqrt{ 0.8^2 + ((3-1)/2)^2 } \doteq 1.2806 \ \text{m} \ \\ \ \\ S_{ 1 }=x \cdot \ y=1 \cdot \ 0.5=\dfrac{ 1 }{ 2 }=0.5 \ \text{m}^2 \ \\ \ \\ S_{ 2 }=(x+x_{ 0 }) \cdot \ \dfrac{ h_{ 1 } }{ 2 }=(1+3) \cdot \ \dfrac{ 0.9434 }{ 2 } \doteq 1.8868 \ \text{m}^2 \ \\ S_{ 3 }=(y+y_{ 0 }) \cdot \ \dfrac{ h_{ 2 } }{ 2 }=(0.5+1.5) \cdot \ \dfrac{ 1.2806 }{ 2 } \doteq 1.2806 \ \text{m}^2 \ \\ \ \\ S=S_{ 1 } + 2 \cdot \ (S_{ 2 }+S_{ 3 })=0.5 + 2 \cdot \ (1.8868+1.2806) \doteq 6.8348 \ \text{m}^2 \ \\ \ \\ V=0.6 \cdot \ S=0.6 \cdot \ 6.8348 \doteq 4.1009 \doteq 4.1 \ \text{l}

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Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

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Math student
the process of the answer is not enough clear try to associate it well with the graph

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Dr Math
yes, the image is only for illustration .... symbols have not the same meaning as in solutions... We re-write steps of solutions to be more clear.

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