Step 1: Find the radius of the cone
The radius \( r \) is half of the diameter: \[ r = \frac{10.5}{2} = 5.25 \, \text{m} \]
Step 2: Formula for the Volume of the Cone
The volume \( V \) of a cone is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] where \( r \) is the radius and \( h \) is the height of the cone. Substituting \( r = 5.25 \, \text{m} \) and \( h = 3 \, \text{m} \): \[ V = \frac{1}{3} \pi (5.25)^2 \times 3 \] First, calculate \( (5.25)^2 \): \[ (5.25)^2 = 27.5625 \] Now, calculate the volume: \[ V = \frac{1}{3} \times 3.1416 \times 27.5625 \times 3 \approx 195.44 \, \text{m}^3 \] Therefore, the volume of the heap is approximately \( 195.44 \, \text{m}^3 \).
Step 3: Find the Slant Height of the Cone
The slant height \( l \) of the cone can be found using the Pythagorean theorem: \[ l = \sqrt{r^2 + h^2} \] Substituting \( r = 5.25 \, \text{m} \) and \( h = 3 \, \text{m} \): \[ l = \sqrt{(5.25)^2 + (3)^2} = \sqrt{27.5625 + 9} = \sqrt{36.5625} \approx 6.05 \, \text{m} \]
Step 4: Formula for the Lateral Surface Area of the Cone
The lateral surface area \( A \) of a cone is given by the formula: \[ A = \pi r l \] Substituting \( r = 5.25 \, \text{m} \) and \( l = 6.05 \, \text{m} \): \[ A = \pi \times 5.25 \times 6.05 \approx 99.88 \, \text{m}^2 \] Therefore, the area of the canvas required to cover the heap is approximately \( 99.88 \, \text{m}^2 \).
The volume of the heap of wheat is approximately \( 195.44 \, \text{m}^3 \) and the area of the canvas required to cover it is approximately \( 99.88 \, \text{m}^2 \).