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3. A heap of wheat is in the form of a cone whose diametor is 10.5 m \& height in 3 m . Jindits volume. The heap is to be covered by canvas to proteet it from rain. Find the area of the canvas reanined.

Ask by Murphy Montgomery. in India
Feb 20,2025

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Answer

The volume of the wheat heap is approximately 86.6 cubic meters, and the area of the canvas needed to cover it is about 100 square meters.

Solution

To solve the problem, we need to find the volume of the cone-shaped heap of wheat and the surface area of the cone that will be covered by the canvas. ### Step 1: Extract Known Conditions - Diameter of the cone, \( d = 10.5 \, m \) - Radius of the cone, \( r = \frac{d}{2} = \frac{10.5}{2} = 5.25 \, m \) - Height of the cone, \( h = 3 \, m \) ### Step 2: Calculate the Volume of the Cone The volume \( V \) of a cone is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] ### Step 3: Calculate the Slant Height of the Cone The slant height \( l \) of the cone can be calculated using the Pythagorean theorem: \[ l = \sqrt{r^2 + h^2} \] ### Step 4: Calculate the Surface Area of the Cone The surface area \( A \) of the cone (excluding the base) is given by: \[ A = \pi r l \] Now, let's perform the calculations step by step. 1. Calculate the volume \( V \). 2. Calculate the slant height \( l \). 3. Calculate the surface area \( A \). Let's proceed with these calculations. Calculate the value by following steps: - step0: Calculate: \(\frac{1}{3}\pi \times 5.25^{2}\times 3\) - step1: Convert the expressions: \(\frac{1}{3}\pi \left(\frac{21}{4}\right)^{2}\times 3\) - step2: Multiply the terms: \(\pi \left(\frac{21}{4}\right)^{2}\) - step3: Evaluate the power: \(\pi \times \frac{441}{16}\) - step4: Multiply: \(\frac{441\pi }{16}\) Calculate or simplify the expression \( \sqrt(5.25^2 + 3^2) \). Calculate the value by following steps: - step0: Calculate: \(\sqrt{5.25^{2}+3^{2}}\) - step1: Convert the expressions: \(\sqrt{\left(\frac{21}{4}\right)^{2}+3^{2}}\) - step2: Add the numbers: \(\sqrt{\frac{585}{16}}\) - step3: Use the properties of radicals: \(\frac{\sqrt{585}}{\sqrt{16}}\) - step4: Simplify the expression: \(\frac{3\sqrt{65}}{4}\) Calculate or simplify the expression \( \pi * 5.25 * l \). Simplify the expression by following steps: - step0: Solution: \(\pi \times 5.25l\) - step1: Reorder the terms: \(5.25\pi l\) - step2: Simplify: \(16.493361l\) Calculate or simplify the expression \( \pi * 5.25 * 6.046693 \). Calculate the value by following steps: - step0: Calculate: \(\pi \times 5.25\times 6.046693\) - step1: Multiply the terms: \(\pi \times 31.74513825\) - step2: Reorder the terms: \(31.74513825\pi \) - step3: Simplify: \(99.730293\) ### Step 1: Calculate the Volume of the Cone The volume \( V \) of the cone is approximately: \[ V \approx 86.59 \, m^3 \] ### Step 2: Calculate the Slant Height of the Cone The slant height \( l \) of the cone is approximately: \[ l \approx 6.05 \, m \] ### Step 3: Calculate the Surface Area of the Cone The surface area \( A \) of the cone (excluding the base) is approximately: \[ A \approx 99.73 \, m^2 \] ### Conclusion - The volume of the heap of wheat is approximately \( 86.59 \, m^3 \). - The area of the canvas required to cover the heap is approximately \( 99.73 \, m^2 \). This area will protect the heap from rain.

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To find the volume of the cone-shaped heap of wheat, we can use the formula for the volume of a cone: V = (1/3)πr²h. The diameter is 10.5 m, so the radius r = 10.5/2 = 5.25 m. The height h is given as 3 m. Plugging these numbers into the formula gives us V = (1/3)π(5.25)²(3) ≈ 82.68 m³ (using π ≈ 3.14). Now, to find the area of the canvas needed to cover it, we need to calculate the slant height (l) using the Pythagorean theorem: l = √(r² + h²) = √(5.25² + 3²) ≈ 6.04 m. The surface area S of the cone, which is the area of the canvas, is S = πrl + πr² (for the base, but this can be omitted if the base doesn't require covering). Thus, S ≈ π(5.25)(6.04) + π(5.25)² ≈ 96.57 m² (approximately). So, the area of the canvas required is about 96.57 m²! Time to get that canvas ready for the rain!

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