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?

Correct result:

V =  4.1 l

Solution:

$$\begin{gathered} x_0=3\text{ m } \\ {y}_0=1.5\text{ m } \\ x=1\text{ m } \\ y=0.5\text{ m } \\ h=0.8\text{ m } \\ {h}_1=\sqrt{h^2+((y_0-y)\mathrm{/}2{)}^2}=\sqrt{0.8^2+((1.5-0.5)\mathrm{/}2{)}^2}\doteq 0.9434\text{ m } \\ {h}_2=\sqrt{h^2+((x_0-x)\mathrm{/}2{)}^2}=\sqrt{0.8^2+((3-1)\mathrm{/}2{)}^2}\doteq 1.2806\text{ m } \\ {S}_1=x\cdot y=1\cdot 0.5={\displaystyle \frac{1}{2}}=0.5{\text{m}}^2 \\ {S}_2=(x+x_0)\cdot {\displaystyle \frac{h_1}{2}}=(1+3)\cdot {\displaystyle \frac{0.9434}{2}}\doteq 1.8868{\text{m}}^2 \\ {S}_3=(y+y_0)\cdot {\displaystyle \frac{h_2}{2}}=(0.5+1.5)\cdot {\displaystyle \frac{1.2806}{2}}\doteq 1.2806{\text{m}}^2 \\ S=S_1+2\cdot (S_2+S_3)=0.5+2\cdot (1.8868+1.2806)\doteq 6.8348{\text{m}}^2 \\ V=0.6\cdot S=0.6\cdot 6.8348=4.1\text{ l} \end{gathered}$$

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Showing 2 comments:

Math student

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

1 year ago  Like

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.

1 year ago  Like

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