A cylindrical bucket (height 32 cm, base radius 18 cm) is filled with sand and emptied to form a conical heap of height 24 cm. Find the radius and slant height of the heap.
Radius = 36 cm, Slant height = \(\sqrt{36^2+24^2}=\sqrt{1872}\approx 43.3\,\text{cm}\)
Step 1: Write down the given dimensions.
Step 2: Volume of the cylinder.
The formula for volume of a cylinder is:
\(V_{cyl} = \pi r_c^2 h_c\)
Substitute values: \(V_{cyl} = \pi \times (18)^2 \times 32\)
\(V_{cyl} = \pi \times 324 \times 32 = \pi \times 10368\)
So, \(V_{cyl} = 10368\pi\,\text{cm}^3\).
Step 3: Volume of the cone.
The formula for volume of a cone is:
\(V_{cone} = \dfrac{1}{3}\pi R^2 h_{cone}\)
Substitute height: \(V_{cone} = \dfrac{1}{3}\pi R^2 (24)\)
\(V_{cone} = 8\pi R^2\).
Step 4: Since the sand volume is the same, equate volumes.
\(V_{cyl} = V_{cone}\)
\(10368\pi = 8\pi R^2\)
Step 5: Simplify the equation.
Cancel \(\pi\) from both sides:
\(10368 = 8R^2\)
Divide both sides by 8:
\(R^2 = 1296\)
So, \(R = \sqrt{1296} = 36\,\text{cm}\).
Step 6: Find the slant height of the cone.
Formula: \(l = \sqrt{R^2 + h_{cone}^2}\)
\(l = \sqrt{36^2 + 24^2} = \sqrt{1296 + 576} = \sqrt{1872}\)
\(l \approx 43.3\,\text{cm}\).
Final Answer: Radius = 36 cm, Slant height ≈ 43.3 cm.