If a haystack in the shape of a right circular cone has a base radius of 5 meters and contains 12 tons of hay, what will be the number of tons of hay in a similar conical haystack whose base diameter is 15 meters? Two right cones are similar if their heights and their base radii are in proportion. Express your answer as a decimal to the nearest tenth.
+9466 The radius of the first cone is 5 meters,
so the diameter of the first cone is 10 meters.
The diameter of the second cone is 15 meters.
\(\frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\Big(\frac{\text{diameter of second cone}}{\text{diameter of first cone}}\Big)^3\\~\\ \frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\Big(\frac{15}{10}\Big)^3\\~\\ \frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\Big(\frac{3}{2}\Big)^3\\~\\ \frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\Big(\frac{3}{2}\Big)\Big(\frac{3}{2}\Big)\Big(\frac{3}{2}\Big)\\~\\ \frac{\text{volume of second cone}}{\text{volume of first cone}}\,=\,\frac{27}{8}\\~\\ {\small\text{volume of second cone}}\,=\,\frac{27}{8}\cdot{\small\text{volume of first cone}}\)
Assuming that both haystacks have the same density, and each piece of hay weighs the same....
\({\small\text{weight of second cone}}\,=\,\frac{27}{8}\cdot{\small\text{weight of first cone}}\\~\\ {\small\text{weight of second cone}}\,=\,\frac{27}{8}\cdot12 \text{ tons}\\~\\ {\small\text{weight of second cone}}\,=\,40.5 \text{ tons}\)
