Solution
- The way to solve the question is to use a formula that relates the areas and volumes of similar solids.
- The formula is given below:
[tex]\begin{gathered} \text{ Let the dimension of the smaller solid be }l\text{ and the dimension of the bigger solid be }L \\ \text{ Let the area and volume of the smaller solid be }a\text{ and }v,\text{ and those of the bigger solide be }A\text{ and }V \\ \text{ Thus, the ratio of their areas and volumes are given as} \\ \frac{l^2}{L^2}=\frac{a}{A}\text{ \lparen Equation 1\rparen} \\ \\ \frac{l^3}{L^3}=(\frac{l}{L})^3=\frac{v}{V}\text{ \lparen Equation 2\rparen} \\ \\ \text{ From Equation 1, we have:} \\ \frac{l}{L}=\sqrt{\frac{a}{A}} \\ \\ \text{ We can substitute this expression into Equation 2 as follows:} \\ (\sqrt{\frac{a}{A}})^3=\frac{v}{V}\text{ \lparen Equation 3\rparen} \end{gathered}[/tex]
- Equation 3 gives us the formula that relates the areas and volumes of similar solids.
- Thus, we can apply it to solve the question as follows:
[tex]\begin{gathered} a=404yd^2,A=1232yd^2 \\ v=?,V=2568 \\ \\ \text{ Thus, we have:} \\ (\sqrt{\frac{404}{1232}})^3=\frac{v}{2568} \\ \\ \text{ Multiply both sides by 2568} \\ \\ \therefore v=2568\times(\sqrt{\frac{404}{1232}})^3 \\ \\ v=482.22642...\approx482yd^3 \end{gathered}[/tex]
Final Answer
The volume of the smaller solid is 482yd³
