A heap of rice is a cone of diameter 9 m and height 3.5 m. Find the volume of rice and the canvas required to just cover it.
Volume: \(\displaystyle \dfrac{23.625\pi}{1}\,\text{m}^3\approx 74.2\,\text{m}^3\); Canvas area: \(\pi r l=\pi\cdot4.5\cdot\sqrt{4.5^2+3.5^2}\approx 80.6\,\text{m}^2\).
Step 1: Identify the given values
Step 2: Formula for volume of a cone
\(V = \dfrac{1}{3}\pi r^2 h\)
Step 3: Substitute the values
\(V = \dfrac{1}{3}\pi (4.5)^2 (3.5)\)
\(= \dfrac{1}{3}\pi (20.25)(3.5)\)
\(= \dfrac{1}{3}\pi (70.875)\)
\(= 23.625\pi\,\text{m}^3\)
Approximate value: \(V \approx 74.2\,\text{m}^3\)
Step 4: Formula for curved surface area (canvas needed)
Curved surface area (CSA) = \(\pi r l\)
where \(l\) is the slant height, given by: \(l = \sqrt{r^2 + h^2}\)
Step 5: Find slant height
\(l = \sqrt{(4.5)^2 + (3.5)^2} = \sqrt{20.25 + 12.25} = \sqrt{32.5}\)
\(l \approx 5.7\,\text{m}\)
Step 6: Find CSA
CSA = \(\pi (4.5)(5.7)\)
≈ \(80.6\,\text{m}^2\)
Final Answer: