A flower pot

The concrete flower pot has the shape of a cube with an edge of 45 cm, and the thickness of the walls and bottom is 4 cm.

A) Calculate the clay volume in the flowerpot if it reaches 3 cm below the rim.
B) Calculate the weight of the flowerpot if the concrete density is 2.45 g/cm³

Final Answer:

V₁ =  52022 cm³
m =  85.7402 kg

Step-by-step explanation:

$$\begin{gathered} a=45\text{cm} \\ s=4\text{cm} \\ o=3\text{cm} \\ b=a-2\cdot s=45-2\cdot 4=37\text{cm} \\ c=a-s-o=45-4-3=38\text{cm} \\ {V}_1=b^2\cdot c=37^2\cdot 38=52022{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} d=a-s=45-4=41\text{cm} \\ {V}_2=b^2\cdot d=37^2\cdot 41=56129{\text{cm}}^3 \\ {V}_3=a^3=45^3=91125{\text{cm}}^3 \\ V=V_3-V_2=91125-56129=34996{\text{cm}}^3 \\ h=2.45{\text{g/cm}}^3 \\ {m}_1=h\cdot V=2.45\cdot 34996=\frac{428701}{5}=85740.2\text{g} \\ m=m_1\to \text{kg}=m_1:1000\text{kg}=85740.2:1000\text{kg}=85.74\text{kg}=85.7402\text{kg} \end{gathered}$$

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