Question

A cylindrical block of radius 5 cm and height 9 cm is hollowed out from one end by removing a cone of radius 5 cm and slant height 10 cm. Find the total surface area and volume of the remaining solid.

Hint:

To calculate the total surface area and volume of a remaining solid, subtract the volume and area of the removed shape from the original solid.

Answer 1

Step 1: Calculate the volume of the cylindrical block.
The formula for the volume of a cylinder is: \[ V_{\text{cylinder}} = \pi r^2 h \] where \( r = 5 \, \text{cm} \) and \( h = 9 \, \text{cm} \). Thus, the volume of the cylinder is: \[ V_{\text{cylinder}} = \pi (5)^2 (9) = 225\pi \, \text{cm}^3. \]
Step 2: Calculate the volume of the cone.
The formula for the volume of a cone is: \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \] where \( r = 5 \, \text{cm} \) and \( h = \sqrt{10^2 - 5^2} = \sqrt{100 - 25} = \sqrt{75} \, \text{cm} \). Thus, the volume of the cone is: \[ V_{\text{cone}} = \frac{1}{3} \pi (5)^2 \times \sqrt{75} = \frac{1}{3} \pi (25) \times 5\sqrt{3} = \frac{125\pi \sqrt{3}}{3} \, \text{cm}^3. \]
Step 3: Calculate the remaining volume.
The remaining volume is the volume of the cylinder minus the volume of the cone: \[ V_{\text{remaining}} = V_{\text{cylinder}} - V_{\text{cone}} = 225\pi - \frac{125\pi \sqrt{3}}{3} \, \text{cm}^3. \]
Step 4: Calculate the total surface area of the remaining solid.
The total surface area consists of the curved surface area of the cylinder, the base area of the cylinder, and the area of the cone's slant height. The curved surface area of the cylinder is: \[ A_{\text{cylinder}} = 2\pi r h = 2\pi (5)(9) = 90\pi \, \text{cm}^2. \] The area of the cone's slant surface is: \[ A_{\text{cone}} = \pi r l = \pi (5)(10) = 50\pi \, \text{cm}^2. \] The remaining surface area is: \[ A_{\text{remaining}} = A_{\text{cylinder}} + A_{\text{cone}} = 90\pi + 50\pi = 140\pi \, \text{cm}^2. \]