Question

From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$

Hint:

Use Pythagoras to find slant height in cones, and remember to adjust surface area when solids are carved or added.

Answer 1

Step 1: Volume of cube
Side of cube = 14 cm
Volume of cube = \( V_{\text{cube}} = a^3 = 14^3 = 2744 \, \text{cm}^3 \)

Step 2: Dimensions of cone
Largest cone that can be carved from one face will have:
Base radius \( r = \dfrac{14}{2} = 7 \) cm, height \( h = 14 \) cm

Step 3: Volume of cone
\[ V_{\text{cone}} = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \cdot \dfrac{22}{7} \cdot 7^2 \cdot 14 = \dfrac{1}{3} \cdot \dfrac{22 \cdot 49 \cdot 14}{7} = \dfrac{1}{3} \cdot 22 \cdot 7 \cdot 14 = \dfrac{2156}{3} \approx 718.67 \, \text{cm}^3 \]

Step 4: Volume of remaining solid
\[ V_{\text{remaining}} = V_{\text{cube}} - V_{\text{cone}} = 2744 - 718.67 \approx 2025.33 \, \text{cm}^3 \]

Step 5: Surface area of remaining solid
Surface area of cube = \(6a^2 = 6 \cdot 14^2 = 1176 \, \text{cm}^2\)
We remove one face of cube and replace it with base of cone and cone's curved surface.

Net surface area:
\[ \text{SA}_{\text{remaining}} = 5a^2 + \text{CSA}_{\text{cone}} + \text{Base area of cone} \]
CSA of cone = \( \pi r l \), where \( l = \sqrt{r^2 + h^2} = \sqrt{49 + 196} = \sqrt{245} \approx 15.4 \, \text{cm} \)
\[ \text{CSA} = \dfrac{22}{7} \cdot 7 \cdot 15.4 = 22 \cdot 15.4 = 338.8 \, \text{cm}^2 \] \[ \text{Base area} = \pi r^2 = \dfrac{22}{7} \cdot 49 = 154 \, \text{cm}^2 \]
Net surface area:
\[ \text{SA}_{\text{remaining}} = 5 \cdot 14^2 + 338.8 + 154 = 980 + 338.8 + 154 = 1472.8 \, \text{cm}^2 \]

Answer:
Volume of remaining solid ≈ 2025.33 cm³
Surface area of remaining solid ≈ 1472.8 cm²